4.7 integration by substitution homework answers

Active Calculus

Matthew Boelkins

Appendix C Answers to Selected Exercises

1 understanding the derivative 1.1 how do we measure velocity 1.1.4 exercises.

• \(s(15)-s(0) \approx -98.75\text{.}\)
• \begin{align*} AV_{[0,15]} &= \frac{s(15)-s(0)}{15-0} \approx -6.58\\ AV_{[0,2]} &= \frac{s(2)-s(0)}{2-0} \approx -47.63\\ AV_{[1,6]} &= \frac{s(6)-s(1)}{6-1} \approx -13.25\\ AV_{[8,10]} &= \frac{s(10)-s(8)}{10-8} \approx -7.35 \end{align*}
• Most negative average velocity on \([0,4]\text{;}\) most positive average velocity on \([4,8]\text{.}\)
• \(\frac{21.31+22.25}{2} = 21.78\) feet per second.
• The average velocities are negative; the instantaneous velocity was positive. Downward motion corresponds to negative average velocity; upward motion to positive average velocity.
• Sketch a plot where the diver’s height at time \(t\) is on the vertical axis. For instance, \(h(2.45) = 0\text{.}\)
• \(AV_{[2.45,7]} \approx \frac{-3.5-0}{7-2.45}=\frac{-3.5}{4.55}=-0.7692\) m/sec. The average velocity is not the same on every time interval within \([2.45,7]\text{.}\)
• When the diver is going upward, her velocity is positive. When she is going downward, her velocity is negative. At the peak of her dive and when her feet touch the bottom of the pool.
• It looks like when the position function is steep, the velocity function’s value is farther away from zero, and that whenever the height/position function is rising/increasing, the velocity function has a positive value. Similarly, whenever the position function is decreasing, the velocity is negative.
• \(15 957\) people.
• In an average year the population grew by about \(798\) people/year.
• The slope of a secant line through the points \((a,f(a))\) and \((b,f(b))\text{.}\)
• \(AV_{[0,20]} \approx 798\) people per year.
• \begin{align*} AV_{[5,10]} & \approx 734.50\\ AV_{[5,9]} & \approx 733.06\\ AV_{[5,8]} & \approx 731.62\\ AV_{[5,7]} & \approx 730.19\\ AV_{[5,6]} & \approx 728.7535 \end{align*}

1.2 The notion of limit 1.2.4 Exercises

• All real numbers except \(x = \pm 2\text{.}\)
• \(\lim_{x \to 2} \frac{16-x^4}{x^2-4} = -8\text{.}\)
• All real numbers except \(x = -3\text{.}\)
• If \(x \gt -3\text{,}\) \begin{equation*} -\frac{|x+3|}{x+3} = -\frac{x+3}{x+3} = -1; \end{equation*} if \(x \lt -3\text{,}\) it follows that \begin{equation*} -\frac{|x+3|}{x+3} = -\frac{-(x+3)}{x+3} = +1\text{.} \end{equation*} Hence the limit does not exist.
• \begin{equation*} AV_{[1,1+h]} = \frac{100\cos(0.75(1+h)) \cdot e^{-0.2(1+h)} - 100\cos(0.75) \cdot e^{-0.2}}{h} \end{equation*}
• \begin{equation*} \lim_{h \to 0} AV_{1, 1+h} \approx -53.837\text{.} \end{equation*}
• The instantaneous velocity of the bungee jumper at the moment \(t = 1\) is approximately \(-53.837\) ft/sec.

1.3 The derivative of a function at a point 1.3.3 Exercises

• \(AV_{[-3,-1]} \approx 1.15\text{;}\) \(AV_{[0,2]} \approx -0.4\text{.}\)
• \(f'(-3) \approx 3\text{;}\) \(f'(0) \approx -\frac{1}{2}\text{.}\)
• For instance, you could let \(f(-3) = 3\) and have \(f\) pass through the points \((-3,3)\text{,}\) \((-1,-2)\text{,}\) \((0,-3)\text{,}\) \((1,-2)\text{,}\) and \((3,-1)\) and draw the desired tangent lines accordingly.
• For instance, you could draw a function \(g\) that passes through the points \((-2,3)\text{,}\) \((-1,2)\text{,}\) \((1,0)\text{,}\) \((2,0)\text{,}\) and \((3,3)\) in such a way that the tangent line at \((-1,2)\) is horizontal and the tangent line at \((2,0)\) has slope \(1\text{.}\)
• \(AV_{[0,7]}=\frac{0.1175}{7} \approx 0.01679\) billion people per year; \(P'(7) \approx 0.1762\) billion people per year; \(P'(7) \gt AV_{[0,7]}\text{.}\)
• \(AV_{[19,29]} \approx 0.02234\) billion people/year.
• We will say that today’s date is July 1, 2015, which means that \(t = 22.5\text{;}\) \begin{equation*} P'(22.5) = \lim_{h \to 0} \frac{115(1.014)^{22.5+h}-115(1.014)^{22.5}}{h}; \end{equation*} \(P'(22.5) \approx 0.02186\) billions of people per year.
• \(y - 1.57236 = 0.02186(t-22.5)\text{.}\)
• All three approaches show that \(f'(2) = 1\text{.}\)
• All three approaches show that \(f'(1) = -1\text{.}\)
• All three approaches show that \(f'(1) = \frac{1}{2}\text{.}\)
• All three approaches show that \(f'(1)\) does not exist.
• The first two approaches show that \(f'(\frac{\pi}{2}) = 0\text{.}\)

1.4 The derivative function 1.4.3 Exercises

• See the figure below.
• One example of a formula for \(f\) is \(f(x) = \frac{1}{2}x^2 - 1\text{.}\)
• \(g'(x) = 2x - 1\text{.}\)
• \(p'(x) = 10x - 4\text{.}\)
• The constants \(3\) and \(12\) don’t seem to affect the results at all. The coefficient \(-4\) on the linear term in \(p(x)\) appears to make the `` \(-4\) ’’ appear in \(p'(x)= 10x - 4\text{.}\) The leading coefficient \(5\) in \((x) = 5x^2 - 4x + 12\) leads to the coefficient of `` \(10\) ’’ in \(p'(x) = 10x -4\text{.}\)
• \(g\) is linear.
• On \(-3.5 \lt x \lt -2\text{,}\) \(-2 \lt x \lt 0\) and \(2 \lt x \lt 3.5\text{.}\)
• At \(x = -2, 0, 2\text{;}\) \(g\) must have sharp corners at these points.

1.5 Interpreting, estimating, and using the derivative 1.5.4 Exercises

• \(F'(10) \approx -3.33592\text{.}\)
• The coffee’s temperature is decreasing at about \(3.33592\) degrees per minute.
• \(F'(20)\text{.}\)
• We expect \(F'\) to get closer and closer to \(0\) as time goes on.
• If a patient takes a dose of \(50\) ml of a drug, the patient will experience a body temperature change of \(0.75\) degrees F.
• ``degrees Fahrenheit per milliliter.’’
• For a patient taking a \(50\) ml dose, adding one more ml to the dose leads us to expect a temperature change that is about \(0.02\) degrees less than the temperature change induced by a \(50\) ml dose.
• \(t=0\text{.}\)
• \(v'(1) = -32\text{.}\)
• ``feet per second per second’’; \(v'(1) = -32\) tells us that the ball’s velocity is decreasing at a rate of 32 feet per second per second.
• The acceleration of the ball.
• \(AV_{[40000,55000]} \approx -0.153 \) dollars per mile.
• \(h'(55000) \approx -0.147\) dollars per mile. During \(55 0001\) st mile, we expect the car’s value to drop by \(0.147\) dollars.
• \(h'(30000) \lt h'(80000)\text{.}\)
• The graph of \(h\) might have the general shape of the graph of \(y = e^{-x}\) for positive values of \(x\text{:}\) always positive, always decreasing, and bending upwards while tending to \(0\) as \(x\) increases.

1.6 The second derivative 1.6.5 Exercises

• \(f\) is increasing and concave down near \(x=2\text{.}\)
• \(g'(2) \approx 1.4\text{.}\)
• At most one.
• \(9\text{.}\)
• \(g''(2) \approx 5.5 \text{.}\)
• \(h'(4.5) \approx 14.3\text{;}\) \(h'(5) \approx 21.2\text{;}\) \(h'(5.5) \approx = 23.9\text{;}\) rising most rapidly at \(t = 5.5\text{.}\)
• \(h'(5) \approx 9.6 \text{.}\)
• Acceleration of the bungee jumper in feet per second per second.
• \(0 \lt t \lt 2\text{,}\) \(6 \lt t \lt 10\text{.}\)

1.7 Limits, Continuity, and Differentiability 1.7.5 Exercises

• \(a = 0\text{.}\)
• \(a = 0, 3\text{.}\)
• \(a = -2, 0, 1, 2, 3\text{.}\)
• \(f(x) = |x-2|\text{.}\)
• Impossible.
• Let \(f\) be the function defined to be \(f(x) = 1\) for every value of \(x \ne -2\text{,}\) and such that \(f(-2) = 4\text{.}\)
• \(h\) must be piecewise linear with slope of \(1\) or \(-1\text{,}\) depending on the interval.
• \(h'(x)\) is not defined for \(x = -2, 0, 2\text{.}\)
• It is possible that \(h\) is not continuous at \(x = -2, 0, 2\text{.}\)
• At \(x = 0\text{.}\) \begin{align*} g'(0) & = \lim_{h \to 0} \frac{g(0+h) - g(0)}{h}\\ & = \lim_{h \to 0} \frac{\sqrt{|h|} - \sqrt{|0|}}{h}\\ & = \lim_{h \to 0} \frac{\sqrt{|h|}}{h} \end{align*}

1.8 The Tangent Line Approximation 1.8.4 Exercises

• \(p(3) = -1\) and \(p'(3) = -2\text{.}\)
• \(p(2.79) \approx -0.58\text{.}\)
• \(F'(60)\approx 1.56\) degrees per minute.
• \(L(t) \approx 1.56(t-60)+324.5\text{.}\)
• \(F(63)\approx L(63)\approx = 329.18\) degrees F.
• Overestimate.
• \(s(9.34) \approx L(9.34) = 3.592\text{.}\)
• underestimate.
• The object is slowing down as it moves toward toward its starting position at \(t=4\text{.}\)
• \(x=1\text{.}\)
• On \(-0.37 \lt x \lt 1.37\text{;}\) \(f\) is concave up.
• \(f(1.88) \approx -3.0022\text{,}\) and this estimate is larger than the true value of \(f(1.88)\text{.}\)

2 Computing Derivatives 2.1 Elementary derivative rules 2.1.5 Exercises

• \(h(2) = 27\text{;}\) \(h'(2) = -19/2\text{.}\)
• \(L(x) = 27 - \frac{19}{2}(x-2)\text{.}\)
• \(p\) is increasing at \(x=2\text{.}\)
• \(p(2.03) \approx -11.44\text{.}\)
• \(p\) is not differentiable at \(x=-1\) and \(x=1\text{;}\) \(q\) is not differentiable at \(x=-1\) and \(x=1\text{.}\)
• \(r\) is not differentiable at \(x=-1\) and \(x=1\text{.}\)
• \(r'(-2) = 4\text{;}\) \(r'(0) = \frac{1}{2}\text{.}\)
• \(y = 4\text{.}\)
• \(w'(t) = 3t^t(\ln(t) + 1) + 2\frac{1}{\sqrt{1-t^2}}\text{.}\)
• \(L(t) = (\frac{3}{\sqrt{2}} - \frac{2\pi}{3}) + (\frac{3}{\sqrt{2}}(\ln(\frac{1}{2}) + 1) + \frac{4}{\sqrt{3}})(t-\frac{1}{2})\text{.}\)
• \(v\) is decreasing at \(t = \frac{1}{2}\text{.}\)
• \begin{align*} f'(x) &= \lim_{h \to 0} \frac{a^{x+h}-a^x}{h}\\ &= \lim_{h \to 0} \frac{a^{x} \cdot a^{h} - a^x}{h}\\ &= \lim_{h \to 0} \frac{a^{x}(a^{h}-1)}{h}\text{,} \end{align*}
• Since \(a^x\) does not depend at all on \(h\text{,}\) we may treat \(a^x\) as constant in the noted limit and thus write the value \(a^x\) in front of the limit being taken.
• When \(a = 2\text{,}\) \(L \approx 0.6931\text{;}\) when \(a = 3\text{,}\) \(L \approx 1.0986\text{.}\)
• \(a \approx 2.71828\) (for which \(L \approx 1.000\) )
• \(\frac{d}{dx}[2^x] = 2^x \cdot \ln(2)\) and \(\frac{d}{dx}[3^x] = 3^x \cdot \ln(3)\)
• \begin{equation*} \frac{d}{dx}[e^x] = e^x\text{.} \end{equation*}

2.2 The sine and cosine functions 2.2.3 Exercises

• \(V'(2) = 24 \cdot 1.07^2 \cdot \ln(1.07) + 6 \cos(2) \approx -0.63778\) thousands of dollars per year.
• \(V''(2) = 24 \cdot 1.07^2 \cdot \ln(1.07)^2 - 6 \sin(2) \approx -5.33\) thousands of dollars per year per year. At this moment, \(V'\) is decreasing and we expect the derivative’s value to decrease by about \(5.33\) thousand dollars per year over the course of the next year.
• \(f'\left(\frac{\pi}{4}\right) = -5\left(\frac{\sqrt{2}}{2}\right)\text{.}\)
• \(L(x) = 3+2(x-\pi)\text{.}\)
• Decreasing.
• The tangent line to \(f\) lies above the curve at this point.
• Hint: in the numerator of the difference quotient, combine the first and last terms and remove a factor of \(\sin(x)\text{.}\)
• Hint: divide each part of the numerator by \(h\) and consider the sum of two separate limits.
• \(\lim_{h \to 0} \left( \frac{\cos(h)-1}{h} \right) = 0\) and \(\lim_{h \to 0} \left( \frac{\sin(h)}{h} \right) = 1 \text{.}\)
• \(f'(x) = \sin(x) \cdot 0 + \cos(x) \cdot 1\text{.}\)
• Hint: \(\cos(\alpha + \beta)\) is \(\cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)\text{.}\)

2.3 The product and quotient rules 2.3.5 Exercises

• \(h(2) = -15\text{;}\) \(h'(2) = 23/2\text{.}\)
• \(L(x) = -15 + 23/2(x-2)\text{.}\)
• Increasing.
• \(r(2.06) \approx -0.5796\text{.}\)
• \(w'(t) = t^t\left(\ln t+1\right)\cdot\left(\arccos t\right)+t^t\cdot\frac{-1}{\sqrt{1-t^2}} \text{.}\)
• \(L(t) \approx 0.740-0.589(t-0.5)\text{.}\)
• \(r'(-2) = 5\) and \(r'(0) = 1\text{.}\)
• At \(x = -1\) and \(x = 1\text{.}\)
• \(L(x) = 2\text{.}\)
• \(z'(0) = -\frac{1}{4}\) and \(z'(2) = -1\text{.}\)
• At \(x = -1\text{,}\) \(x = 1\text{,}\) \(x = -1.5\text{,}\) and \(x = 1\text{.}\)
• \(C(t) = A(t)Y(t)\) bushels in year \(t\text{.}\)
• \(1 190 000\) bushels of corn.
• \(C'(t) = A(t)Y'(t) + A'(t)Y(t)\text{.}\)
• \(C'(0) = 158 000\) bushels per year.
• \(C(1) \approx 1 348 000\) bushels.
• \(g(80) = 20\) kilometers per liter, and \(g'(80) = -0.16\text{.}\) kilometers per liter per kilometer per hour.
• \(h(80) = 4\) liters per hour and \(h'(80) = 0.082\) liters per hour per kilometer per hour.
• Think carefully about units and how each of the three pairs of values expresses fundamentally the same facts.

2.4 Derivatives of other trigonometric functions 2.4.3 Exercises

• \(h'(2) = \frac{-2\sin(2) - 2\cos(2) \ln(1.2)}{1.2^2} \approx -1.1575\) feet per second.
• \(h''(2) = \frac{\cos(2)(-2 + 2ln^2(1.2)) + 4\ln(1.2)\sin(2))}{1.2^2} \approx 1.0193\) feet per second per second.
• The object is falling and slowing down.
• \(f'(x) = \sin(x) \cdot (-\csc^2(x)) + \cot(x) \cdot \cos(x)\text{.}\)
• \(f'(x) = \frac{-\sin^2(x)}{\sin(x)} = -\sin(x)\) for \(x \ne \frac{\pi}{2} + k\pi\) for some integer value of \(k\text{.}\)
• \(\displaystyle p'(z) = \frac{\left(z^2\sec(z) +1 \right)\left(z\sec^2(z)+\tan(z)\right) - z\tan(z) \left(z^2\sec(z)\tan(z)+2z\sec(z)\right)}{\left(z^2\sec(z) + 1\right)^2} +3e^z\)
• \(y - 4 = 3(x-0)\text{.}\)

2.5 The chain rule 2.5.5 Exercises

• \(h'\left( \frac{\pi}{4} \right) = \frac{3}{2\sqrt{2}}\text{.}\)
• \(r'(0.25) = \cos(0.25^3) \cdot 3(0.25)^2 \approx 0.1875 \gt h'(0.25) = 3\sin^2(0.25) \cdot \cos(0.25) \approx 0.1779\text{;}\) \(r\) is changing more rapidly.
• \(h'(x)\) is periodic; \(r'(x)\) is not.
• \(p'(x) = e^{u(x)} \cdot u'(x)\text{.}\)
• \(q'(x) = u'(e^x) \cdot e^x\text{.}\)
• \(r'(x) = -\csc^2(u(x)) \cdot u'(x)\text{.}\)
• \(s'(x) = u'(\cot(x)) \cdot (-\csc^2(x))\text{.}\)
• \(a'(x) = u'(x^4) \cdot 4x^3\text{.}\)
• \(b'(x) = 4(u(x))^3 \cdot u'(x)\text{.}\)
• \(C'(0) = 0\) and \(C'(3) = -\frac{1}{2}\text{.}\)
• Consider \(C'(1)\text{.}\) By the chain rule, we’d expect that \(C'(1) = p'(q(1)) \cdot q'(1)\text{,}\) but we know that \(q'(1)\) does not exist since \(q\) has a corner point at \(x = 1\text{.}\) This means that \(C'(1)\) does not exist either.
• Since \(Y(x) = q(q(x))\text{,}\) the chain rule implies that \(Y'(x) = q'(q(x)) \cdot q'(x)\text{,}\) and thus \(Y'(-2) = q'(q(-2)) \cdot q'(-2) = q'(-1) \cdot q'(-2)\text{.}\) But \(q'(-1)\) does not exist, so \(Y'(-2)\) also fails to exist. Using \(Z(x) = q(p(x))\) and the chain rule, we have \(Z'(x) = q'(p(x)) \cdot p'(x)\text{.}\) Therefore \(Z'(0) = q'(p(0)) \cdot p'(0) = q'(-0.5) \cdot p'(0) = 0 \cdot 0.5 = 0\text{.}\)
• \(\left. \frac{dV}{dh} \right|_{h=1} = 7 \pi\) cubic feet per foot.
• \(h'(2) = \pi \cos(2\pi) = \pi\) feet per hour.
• \(\left. \frac{dV}{dt} \right|_{t=2} = 7 \pi^2\) cubic feet per hour.
• In (a) we are determining the instantaneous rate at which the volume changes as we increase the height of the water in the tank, while in (c) we are finding the instantaneous rate at which volume changes as we increase time.

2.6 Derivatives of Inverse Functions 2.6.6 Exercises

• \(f'(x) = \frac{1}{2\arctan(x) + 3\arcsin(x) + 5} \cdot \left(\frac{2}{1+x^2} + \frac{3}{\sqrt{1-x^2}}\right)\text{.}\)
• \(r'(z) = \frac{1}{1+\left(\ln(\arcsin(z))\right)^2} \cdot \left( \frac{1}{\arcsin(z)} \right) \cdot \frac{1}{\sqrt{1-z^2}}\text{.}\)
• \(q'(t) = \arctan^2(3t) \cdot \left[4\arcsin^3(7t) \left( \frac{7}{\sqrt{1-(7t)^2}} \right)\right] + \arcsin^4(7t) \cdot \left[2\arctan(3t) \left(\frac{3}{1+(3t)^2}\right) \right]\text{.}\)
• \(\displaystyle g'(v) = \frac{1}{\frac{\arctan(v)}{\arcsin(v) + v^2}} \cdot \frac{(\arcsin(v) + v^2) \cdot \frac{1}{1+v^2} - \arctan(v) \cdot \left(\frac{1}{\sqrt{1-v^2}} + 2v \right)}{(\arcsin(v) + v^2)^2} \)
• \(f'(1) \approx 2\text{.}\)
• \((f^{-1})'(-1) \approx 1/2\text{.}\)
• \(f\) passes the horizontal line test.
• \(f^{-1}(x) = g(x) = \sqrt[3]{4x-16}\text{.}\)
• \(f'(x) = \frac{3}{4}x^2\text{;}\) \(f'(2) = 3\text{.}\) \(g'(x) = \frac{1}{3}(4x-16)^{-2/3} \cdot 4\text{;}\) \(g'(6) = \frac{1}{3}\text{.}\) These two derivative values are reciprocals.
• \(h\) passes the horizontal line test.
• The equation \(y = x + \sin(x)\) can’t be solved for \(x\) in terms of \(y\text{.}\)
• \((h^{-1})'(\frac{\pi}{2} + 1) = 1\text{.}\)

2.7 Derivatives of Functions Given Implicitly 2.7.3 Exercises

• \(x = \frac{\ln(y)}{\ln(a)}\text{.}\)
• \(1 = \frac{1}{\ln(a)} \cdot \frac{1}{y}\frac{dy}{dx}\text{.}\)
• \(\frac{d}{dx}[a^x] = a^x \ln(a)\text{.}\)

2.8 Using Derivatives to Evaluate Limits 2.8.4 Exercises

• \(\ln(x^{2x}) = 2x \cdot \ln(x)\text{.}\)
• \(x = \frac{1}{\frac{1}{x}}\text{.}\)
• \(\lim_{x \to 0^+} h(x) = 0\text{.}\)
• \(\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} x^{2x} = 1\text{.}\)
• Show that \(\lim_{x \to \infty}\frac{\ln(x)}{\sqrt{x}} = 0\text{.}\)
• Show that \(\lim_{x \to \infty}\frac{\ln(x)}{\sqrt[n]{x}} = 0\text{.}\)
• Consider \(\lim_{x \to \infty} \frac{p(x)}{e^x} \) By repeated application of LHR, the numerator will eventually be simply a constant (after \(n\) applications of LHR), and thus with \(e^x\) still in the denominator, the overall limit will be \(0\text{.}\)
• Show that \(\lim_{x \to \infty} \frac{\ln(x)}{x^n} = 0\)
• For example, \(f(x) = 3x^2 + 1\) and \(g(x) = -0.5x^2 + 5x - 2\text{.}\)

3 Using Derivatives 3.1 Using derivatives to identify extreme values 3.1.4 Exercises

• \(f'\) is positive for \(-1 \lt x lt 1\) and for \(x \gt 1\text{;}\) \(f'\) is negative for all \(x \lt -1\text{.}\) \(f\) has a local minimum at \(x = -1\text{.}\)
• A possible graph of \(y = f''(x)\) is shown at right in the figure.
• \(f''(x)\) is negative for \(-0.35 \lt x \lt 1\text{;}\) \(f''(x)\) is positive everywhere else; \(f\) has points of inflection at \(x \approx -0.35\) and \(x = 1\text{.}\)
• \(g''(2) = 0\text{;}\) \(g''\) is negative for \(1 \lt x \lt 2\) and positive for \(2 \lt x \lt 3\text{.}\)
• \(g\) has a point of inflection at \(x = 2\text{.}\)
• One root is negative and the other positive.
• \(h\) will look like a line with slope \(3\text{.}\)
• \(h\) is concave up everywhere; \(h\) is almost linear for large values of \(|x|\text{.}\)
• \(p''(x)\) is negative for \(-1 \lt x \lt 2\) and positive for all other values of \(x\text{;}\) \(p\) has points of inflection at \(x = -1\) and \(x = 2\text{.}\)
• Local maximum.

3.2 Using derivatives to describe families of functions 3.2.3 Exercises

• \(x = 0\) and \(x = \frac{2a}{3}\text{.}\)
• \(x = \frac{a}{3}\text{;}\) \(p''(x)\) changes sign from negative to positive at \(x = \frac{a}{3}\text{.}\)
• As we increase the value of \(a\text{,}\) both the location of the critical number and the inflection point move to the right along with \(a\text{.}\)
• \(x=c\) is a vertical asymptote because \(\lim_{x \to c^+} \frac{e^{-x}}{x-c} = \infty\) and \(\lim_{x \to c^-} \frac{e^{-x}}{x-c} = -\infty\text{.}\)
• \(\lim_{x \to \infty} \frac{e^{-x}}{x-c} = 0\text{;}\) \(\lim_{x \to -\infty} \frac{e^{-x}}{x-c} = -\infty\text{.}\)
• The only critical number for \(q\) is \(x=c-1\text{.}\)
• When \(x \lt c-1\text{,}\) \(q'(x) \gt 0\text{;}\) when \(x \gt c-1\text{,}\) \(q'(x) \lt 0\text{;}\) \(q\) has a local maximum at \(x = c-1\text{.}\)
• \(x = m\text{.}\)
• \(E\) is increasing for \(x \lt m\) and decreasing for \(x \gt m\text{,}\) with a local maximum at \(x = m\text{.}\)
• \(x = m \pm s\text{.}\)
• \(\lim_{x \to \infty} E(x) = \lim_{x \to -\infty} E(x) = 0\text{.}\)

3.3 Global Optimization 3.3.4 Exercises

• Not enough information is given.
• Global minimum at \(x = b\text{.}\)
• Global minimum at \(x = a\text{;}\) global maximum at \(x = b\text{.}\)
• Not enough information is provided.
• Absolute maximum \(p(0) = p(a) = 0\text{;}\) absolute minimum \(p\left( \frac{a}{\sqrt{3}} \right) = -\frac{2a^3}{3\sqrt{3}}\text{.}\)
• Absolute max \(r\left( \frac{1}{b} \right) \approx 0.368 \frac{a}{b}\text{;}\) absolute min \(r\left( \frac{2}{b} \right) \approx 0.270 \frac{a}{b}\text{.}\)
• Absolute minimum \(g(b) = a(1-e^{-b^2})\text{;}\) absolute maximum \(g(3b) = a(1-e^{-3b^2})\text{.}\)
• Absolute max \(s\left( \frac{\pi}{2k} \right) = 1\text{;}\) absolute min \(s\left( \frac{5\pi}{6k} \right) = \frac{1}{2}\text{.}\)
• Global maximum at \(x=a\text{;}\) global minimum at \(x=b\text{.}\)
• Global maximum at \(x=c\text{;}\) global minimum at either \(x=a\) or \(x=b\text{.}\)
• Global minimum at \(x=a\) and \(x=b\text{;}\) global maximum somewhere in \((a,b)\text{.}\)
• Global minimum at \(x=c\text{;}\) global maximum value at \(x = a\text{.}\)
• Absolute max \(s(\frac{5\pi}{12}) = 8\text{;}\) absolute min \(s(\frac{11\pi}{12}) = 2\text{.}\)
• Absolute max \(s(\frac{5\pi}{12}) = 8\text{;}\) absolute min \(s(0) = 5 - \frac{3\sqrt{3}}{2} \approx 2.402\text{.}\)
• Absolute max \(s(\frac{5\pi}{12}) = 8\text{;}\) absolute min \(s(\frac{11\pi}{12}) = 2\text{.}\) (There are other points at which the function achieves these values on the given interval.)
• Absolute max \(s(\frac{5\pi}{12}) = 8\text{;}\) absolute min \(s(\frac{5\pi}{6}) \approx 2.402\text{.}\)

3.4 Applied Optimization 3.4.3 Exercises

3.5 related rates 3.5.3 exercises, 4 the definite integral 4.1 determining distance traveled from velocity 4.1.5 exercises.

• At time \(t = 1\text{,}\) \(12\) miles north of the lake.
• \(s(2) - s(0) = 1\) mile north of the lake.
• \(40\) miles.
• \(t = \frac{500}{32} = \frac{125}{8} = 15.625\) is when the rocket reaches its maximum height.
• \(A = 3906.25\text{,}\) the vertical distance traveled on \([0, 15.625]\text{.}\)
• \(s(t) = 500t - 16t^2\text{.}\)
• \(s(15.625) - s(0) = 3906.25\) is the change of the rocket’s position on \([0,15.625]\text{.}\)
• \(s(5) - s(1) = 1616\text{;}\) the rocket rose \(1616\) feet on \([1,5]\text{.}\)
• \(\frac{1}{2} + \frac{1}{4} \pi \approx 1.285\text{.}\)
• \(s(5)-s(2) = -2\) is the change in position of the object on \([2,5]\text{.}\)
• On the time interval \([5,7]\text{.}\)
• \(s\) is increasing on the intervals \((0,2)\) and \((5,7)\text{;}\) the position function has a relative maximum at \(t=2\text{.}\)
• Think about the product of the units involved: ``units of pollution per day’’ times ``days’’. Connect this to the area of a thin vertical rectangle whose height is given by the curve.
• An underestimate is \(336\) units of pollution.

4.2 Riemann Sums 4.2.5 Exercises

• \(A = \frac{87}{2}\text{.}\)
• For any linear function \(g\) of the form \(g(x) = mx + b\) such that \(g(x) \ge 0\) on the interval of interest.
• \(f(x) = x^2 + 1\) on the interval \([1,3]\text{.}\)
• If \(S\) is a left Riemann sum, \(f(x) = x^2 + 1\) on the interval \([1.4, 3.4]\text{.}\) If \(S\) is a middle Riemann sum, \(f(x) = x^2 + 1\) on the interval \([1.2, 3.2]\text{.}\)
• The area under \(f(x) = x^2 + 1\) on \([1,3]\text{.}\)
• \(R_{10} = \sum_{i=1}^{10} \left( (1+0.2i)^2 + 1 \right) \cdot 0.2\text{.}\)
• \(M_3 = 99.6\) feet.
• \(L_6 = 114 \text{,}\) \begin{equation*} R_6 = 84\text{,} \end{equation*} and \(\frac{1}{2}(L_6 + R_6) = 99\text{.}\)
• \(114\) feet.
• The total tonnage of pollution escaping the scrubbing process in the time interval \([0,4]\) weeks.
• \(6.4\) tons.

4.3 The Definite Integral 4.3.5 Exercises

• The total change in position is \(P = \int_0^{4} v(t) dt\text{.}\)
• \(P = -2.625\) feet.
• \(D = 3.375\) feet.
• \(AV = -0.65625\) feet per second.
• \(s(t) = -t^2+t\text{.}\)
• The total change in position, \(P\text{,}\) is \(P = \int_0^1 v(t) \, dt + \int_1^3 v(t) \, dt + \int_3^4 v(t) \, dt = \int_0^4 v(t) \, dt\text{.}\)
• \(P = \int_0^4 v(t) \, dt \approx 2.665\text{.}\)
• The total distance traveled, \(D\text{,}\) is \(D = \int_0^1 v(t) \, dt - \int_1^3 v(t) \, dt + \int_3^4 v(t) \, dt\text{.}\)
• \(D \approx 8.00016\text{.}\)
• \begin{equation*} v_{\operatorname{AVG} [0,4]} \approx 0.66625 \end{equation*} feet per second.
• \(\int_0^1 [f(x) + g(x)] \,dx = 1-\frac{\pi}{4}\text{.}\)
• \(\int_1^4 [2f(x) - 3g(x)] \, dx = -\frac{15}{2} - 3\pi\text{.}\)
• \(h_{\operatorname{AVG} [0,4]} = \frac{5}{8} + \frac{3\pi}{16}\text{.}\)
• \(c = -\frac{3}{8} + \frac{3\pi}{16}\text{.}\)
• The exact area between the two curves is \(\int_{-1}^{1} (3-x^2) \, dx - \int_{-1}^{1} 2x^2 \, dx\text{.}\)
• Use the sum rule for definite integrals over the same interval.
• Think about subtracting the area under \(q\) from the area under \(p\text{.}\)

4.4 The Fundamental Theorem of Calculus 4.4.5 Exercises

• \(20\) meters.
• \(\displaystyle v_{\operatorname{AVG} [12,24]} = 12.5\) meters per minute.
• The object’s maximum acceleration is \(3\) meters per minute per minute at the instant \(t = 2\text{.}\)
• \(c = 5\text{.}\)
• \(-\frac{5}{6}\text{.}\)
• \(\displaystyle f_{\operatorname{AVG} [0,5]} = \frac{1}{2}\text{.}\)
• The antiderivative function tells the total number of minutes it takes for the plane to climb to an altitude of \(h\) feet.
• \(M = \int_{0}^{10000} m(h) \, dh \text{.}\)
• It takes the plane aabout \(M_5 \approx 15.27\) minutes.
• \(G'(x) = \frac{1}{-x} \cdot (-1) = \frac{1}{x}\text{.}\)
• Since for those values of \(x\text{,}\) \(G'(x) = \frac{1}{x}\text{.}\)
• If \(x \lt 0\text{,}\) then \(H(x) = \ln(|x|) = \ln(-x) = G(x)\text{;}\) if \(x \gt 0\text{,}\) then \(H(x) = \ln(|x|) = \ln(x) = F(x)\text{.}\)
• For \(x \lt 0\text{,}\) \(H'(x) = G'(x) = \frac{1}{x}\text{;}\) for \(x \gt 0\text{,}\) \(H'(x) = F'(x) = \frac{1}{x}\) for all \(x \ne 0\text{.}\)

5 Evaluating Integrals 5.1 Constructing Accurate Graphs of Antiderivatives 5.1.5 Exercises

• \(s(1) = \frac{5}{3}\text{,}\) \(s(3) = -1\text{,}\) \(s(5) = -\frac{11}{3}\text{,}\) \(s(6) = -\frac{5}{2}\text{.}\)
• \(s\) is increasing on \(0 \lt t \lt 1\) and \(5 \lt t \lt 6\text{;}\) decreasing for \(1 \lt t \lt 5\text{.}\)
• \(s\) is concave down for \(t \lt 3\text{;}\) concave up for \(t \gt 3\text{.}\)
• \(s(t) = -2t + \frac{1}{6}(t-3)^3 + 5\text{.}\)
• \(C\) measures the total number of calories burned in the workout since \(t = 0\text{.}\)
• \(C(5) = 12.5\text{,}\) \(C(10) = 50\text{,}\) \(C(15) = 125\text{,}\) \(C(20) = 187.5\text{,}\) \(C(25) = 237.5\text{,}\) \(C(30) = 262.5\text{.}\)
• \(C(t) = 12.5 + 7.5(t-5)\) on this interval.
• \(A\text{,}\) \(B\text{,}\) and \(C\) are vertical translations of each other.
• \(A' = f\text{.}\)

5.2 The Second Fundamental Theorem of Calculus 5.2.5 Exercises

• The total sand removed on this time interval is \begin{equation*} \int_0^6 \left[2 + 5\sin \left( \frac{4\pi t}{25} \right) \right] \, dt\text{.} \end{equation*}
• The total amount of sand on the beach at time \(x\) is given by \begin{equation*} Y(x) = \int_0^x \left[ S(t) - R(t) \right] \, dt = \int_0^x \left[ \frac{15t}{1+3t} - \left( 2 + 5\sin \left( \frac{4\pi t}{25} \right) \right) \right] \, dt\text{.} \end{equation*}
• \(Y'(4) = S(4) - R(4) \approx -1.90875\) cubic yards per hour.
• \(Y\) has an absolute minimum on \([0,6]\) of \(Y(5.118) \approx 2492.368\text{.}\)
• The antiderivative function tells us the total number of minutes that it takes for the plane to climb to an altitude of \(h\) feet.
• The number of minutes required for the airplane to ascend to \(10{,}000\) feet of altitude is given by the definite integral \begin{equation*} M = \int_{0}^{10000} m(h) \, dh\text{.} \end{equation*}
• The number of minutes required for the airplane to ascend to \(h\) feet of altitude is given by the definite integral \begin{equation*} M(h) = \int_{0}^{h} m(t) \, dt\text{.} \end{equation*}
• Estimating the desired integral using \(3\) subintervals and midpoints, \begin{equation*} \int_{0}^{6000} m(h) \, dh \approx 7.77\text{.} \end{equation*} Using \(5\) subintervals and midpoints, \begin{equation*} \int_{0}^{10000} m(h) \, dh \approx 15.27\text{.} \end{equation*}

5.3 Integration by Substitution 5.3.5 Exercises

• \(\int \tan(x) \, dx = \ln\left(|\sec(x)|\right) + C\text{.}\)
• \(\int \cot(x) \, dx = -\ln\left(|\csc(x)|\right) + C\text{.}\)
• \(\int \frac{\sec^2(x) + \sec(x) \tan(x)}{\sec(x) + \tan(x)} \, dx = \ln\left(|\sec(x) + \tan(x)|\right) + C\text{.}\)
• \(\frac{\sec^2(x) + \sec(x) \tan(x)}{\sec(x) + \tan(x)} = \sec(x)\text{.}\)
• \(\int \sec(x) \, dx = \ln\left(|\sec(x) + \tan(x)|\right) + C\text{.}\)
• \(\int \csc(x) \, dx = -\ln\left(|\csc(x) + \cot(x)|\right) + C\text{.}\)
• \(\int x \sqrt{x-1} \, dx = \int (u+1) \sqrt{u} \, du\text{.}\)
• \(\int x \sqrt{x-1} \, dx = \frac{2}{5} (x-1)^{5/2} + \frac{2}{3} (x-1)^{3/2} + C\text{.}\)
• \(\int x^2 \sqrt{x-1} \, dx = \frac{2}{7} (x-1)^{7/2} + \frac{4}{5} (x-1)^{5/2} + \frac{2}{3} (x-1)^{3/2} + C\text{.}\) \(\int x \sqrt{x^2 - 1} \, dx = \frac{1}{3} (x^2-1)^{3/2} + C\text{.}\)
• We don’t have a function-derivative pair.
• \(\sin^3(x) = \sin(x) (1-\cos^2(x))\text{.}\)
• \(u = \cos(x)\) and \(du = -\sin(x) \, dx\text{.}\)
• \(\int \sin^3(x) \, dx = \frac{1}{3}\cos^3(x) - \cos(x) + C\text{.}\)
• \(\int \cos^3(x) \, dx = \sin(x) - \frac{1}{3}\sin^3(x) + C\text{.}\)
• The total energy consumed in \(24\) hours, measured in megawatt-hours.
• \(\int_0^{24} r(t) \, dt \approx 95.7809 \) megawatt-hours of power used in \(24\) hours.
• \(\displaystyle r_{\operatorname{AVG} [0,24]} \approx 3.99087\) megawatt-hours.

5.4 Integration by Parts 5.4.7 Exercises

• \(F'(x) = xe^{-2x}\text{.}\)
• \(\displaystyle F(x) = -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + \frac{1}{4}\)
• \(\int e^{2x} \cos(e^x) \, dx = \int z \cos(z) \, dz\text{.}\)
• \(\int e^{2x} \cos(e^x) \, dx = e^x \sin(e^x) + \cos(e^x) + C\text{.}\)
• \(\displaystyle \int e^{2x} \cos(e^{2x}) \, dx = \frac{1}{2} \sin(e^{2x}) + C\)
• \(\int e^{2x} \sin(e^x) \, dx = \sin(e^x) - z\cos(e^x) + C\text{.}\)
• \(\int e^{3x} \sin(e^{3x}) \, dx = -\frac{1}{3} \cos(e^{3x}) + C\text{.}\)
• \(\int xe^{x^2} \cos(e^{x^2}) \sin(e^{x^2}) \, dx = \frac{1}{4} \sin^2(e^{x^2}) + C\text{.}\)
• \(u\) -substitution; \(\int x^2 \cos(x^3) \ dx = \frac{1}{3} \sin(x^3) + C\text{.}\)
• Both are needed; \(\int x^5 \cos(x^3) \ dx = \frac{1}{3} \left(x^3\sin(x^3) + \cos(x^3) \right) + C\text{.}\)
• Integration by parts; \(\int x ln(x^2) \ dx = \frac{x^2}{2} \ln(x^2) - \frac{x^2}{2} + C\text{.}\)
• \(u\) -substitution; \(\int x^3 \sin(x^4) \ dx = -\frac{1}{4} \cos(x^4) + C\text{.}\)
• Both are needed; \(\int x^7 \sin(x^4) \ dx = -\frac{1}{4} x^4 \cos(x^4) + \frac{1}{4} \sin(x^4) + C\text{.}\)

5.5 Other Options for Finding Algebraic Antiderivatives 5.5.5 Exercises

• \begin{equation*} \int \frac{x^3 + x + 1}{x^4 - 1} \, dx = -\frac{1}{2} \arctan(x) + \frac{1}{4} \ln|x+1| + \frac{3}{4} \ln|x-1| + C \end{equation*}
• \begin{equation*} \int \frac{x^5 + x^2 + 3}{x^3 - 6x^2 + 11x - 6} \, dx = \frac{x^3}{3} + 3x^2 + 25x + \frac{255}{2} \ln|x-3| - 39 \ln|x-2| + \frac{5}{2} \ln|x-1| + C\text{.} \end{equation*}
• \begin{equation*} \int \frac{x^2 - x - 1}{(x-3)^3} \, dx = \ln|x-3| - \frac{5}{x-3} - \frac{5}{2(x-3)^2} + C\text{.} \end{equation*}
• \(\int \frac{1}{x \sqrt{9x^2 + 25}} \, dx = -\frac{1}{5} \ln \left| \frac{5+\sqrt{9x^2 + 5^2}}{3x} \right| + C\text{.}\)
• \(\displaystyle \int x \sqrt{1 + x^4} \, dx = \frac{1}{2} \left( \frac{x^2}{2}\sqrt{x^4 + 1} + \frac{1}{2}\ln|x^2 + \sqrt{x^4 + 1}| \right) + C\)
• \(\displaystyle \int e^x \sqrt{4 + e^{2x}} \, dx = \frac{e^x}{2}\sqrt{e^{2x} + 4} + 2\ln|e^x + \sqrt{e^{2x} + 4}| + C\)
• \(\int \frac{\tan(x)}{\sqrt{9 - \cos^2(x)}} \, dx = \frac{1}{3} \ln \left| \frac{3 + \sqrt{9 - \cos^2(x)}}{\cos(x)} \right| + C\text{.}\)
• Try \(u = 1+x^2\) or \(u = x + \sqrt{1+x^2}\text{.}\)
• Try \(u = \sqrt{x+\sqrt{1+x^2}}\) and \(dv = \frac{1}{x} \, dx\text{.}\)
• It appears that the function \(\frac{\sqrt{x+\sqrt{1+x^2}}}{x}\) does not have an elementary antiderivative.

5.6 Numerical Integration 5.6.6 Exercises

• \(u\) -substitution fails since there’s not a composite function present; try showing that each of the choices of \(u = x\) and \(dv = \tan(x) \, dx\text{,}\) or \(u = \tan(x)\) and \(dv = x \, dx\text{,}\) fail to produce an integral that can be evaluated by parts.
• \(\displaystyle L_4 = 0.25892\)
• \(\displaystyle R_4 = 0.64827\)
• \(\displaystyle M_4 = 0.41550\)
• \(\displaystyle T_4 = \frac{L_4 + R_4}{2} = 0.45360\)
• \(\displaystyle S_8 = \frac{2M_4 + T_4}{3} = 0.42820\)
• \(L_4\) and \(M_4\) are underestimates; \(R_4\) and \(M_4\) are overestimates.
• Concave down.
• \(\int_3^6 f(x) \approx 7.03\text{.}\)
• \(\int_0^{60} r(t) \, dt \text{.}\)
• \(\int_0^{60} r(t) \, dt \gt M_3 = 204000\text{.}\)
• \(\int_0^{60} r(t) \, dt \approx S_6 = \frac{619000}{3} \approx 206333.33\text{.}\)
• \(\frac{1}{60} S_6 \approx 3438.89 \text{;}\) \(\frac{2000+2100+2400+3000+3900+5100+6500}{7} = \frac{25000}{7} \approx 3571.43 \text{.}\) each estimates the average rate at which water flows through the dam on \([0,60]\text{,}\) and the first is more accurate.

6 Using Definite Integrals 6.1 Using Definite Integrals to Find Area and Length 6.1.5 Exercises

• \(A = \int_{\frac{3 - \sqrt{3}}{2}}^{\frac{3 + \sqrt{3}}{2}} -2y^2 + 6y - 3 \ dy = \sqrt{3} \text{.}\)
• \(A = \int_{\pi/4}^{3\pi/4} \sin(x) - \cos(x) \ dx = \sqrt{2} \text{.}\)
• \(A = \int_{-1}{5/2} \frac{y+1}{2} - (y^2-y-2) \ dy = \frac{343}{48} \text{.}\)
• \(A = \int_{\frac{m - \sqrt{m^2+4}}{2}}^{\frac{m + \sqrt{m^2+4}}{2}} mx - \left(x^2-1\right) \ dx = - \frac{1}{3}\left(\frac{m + \sqrt{m^2+4}}{2}\right)^3 - \frac{1}{3}\left(\frac{m - \sqrt{m^2+4}}{2}\right)^3 + \frac{m}{2}\left(\frac{m + \sqrt{m^2+4}}{2}\right)^2 - \frac{m}{2}\left(\frac{m - \sqrt{m^2+4}}{2}\right)^2 + \left(\frac{m + \sqrt{m^2+4}}{2} - \frac{m - \sqrt{m^2+4}}{2} \right) \text{.}\)
• \(r = \frac{4}{3} \text{.}\)
• \(A_1 = A_2 = \frac{4 \sqrt{6}}{27}\text{.}\)

6.2 Using Definite Integrals to Find Volume 6.2.5 Exercises

• \(L = \int_0^{1.84257} \sqrt{1+\left( -3 \sin \left(\frac{x^3}{4}\right) \cdot \frac{3}{4}x^2 \right)^2} \, dx \approx 4.10521\text{.}\)
• \(A = \int_0^{1.84527} 3 \cos\left( \frac{x^3}{4} \right) \, dx \approx 4.6623 \text{.}\)
• \(V = \int_0^{1.84527} \pi \cdot 9 \cos^2 \left( \frac{x^3}{4} \right) \, dx \approx 40.31965 \text{.}\)
• \(V = \int_0^3 \pi \left( 4\arccos \left(\frac{y}{3} \right) \right)^{2/3} \, dy \approx 23.29194 \text{.}\)
• \(A = \int_0^{\frac{\pi}{4}} ( \cos(x) - \sin(x) ) \, dx \text{.}\)
• \(V = \int_0^{\frac{\pi}{4}} \pi (\cos^2(x) - \sin^2(x)) \, dx \text{.}\)
• \(\displaystyle V = \int_0^{\frac{\sqrt{2}}{2}} \pi \arcsin^2(y) \, dy + \int_{\frac{\sqrt{2}}{2}}^1 \pi \arccos^2(y) \, dy\)
• \(V = \int_0^{\frac{\pi}{4}} \pi [(2 - \sin(x))^2 - (2 - \cos(x))^2] \, dx \text{.}\)
• \(\displaystyle V = \int_0^{\frac{\sqrt{2}}{2}} \pi [ (1+\arcsin(y))^2 - 1^2 ] \, dy + \int_{\frac{\sqrt{2}}{2}}^1 \pi [ (1+\arccos(y))^2 - 1^2 ] \, dy\)
• \(A = \int_0^{1.5} 1+\frac{1}{2}(x-2)^2 - \frac{1}{2}x^2 \ dx = 2.25\text{.}\)
• \(\displaystyle V = \int_0^{1.5} \pi\left[\left(2+\frac{1}{2}(x-2)^2\right)^2 - \left(1+\frac{1}{2}x^2\right)^2 \right] \ dx = \frac{315}{32} \pi\)
• \(V = \int_{0}^{1.125} \pi \left(\sqrt{2y}\right)^2 \ dy + \int_{1.125}^3 \pi \left(2 - \sqrt{2(y-1)}\right)^2 \ dy \approx 7.06858347 \text{.}\)
• \(P = 3 + \int_0^{1.5} \sqrt{1+(x-2)^2} + \sqrt{1+x^2} \ dx \approx 7.387234642 \text{.}\)

6.3 Density, Mass, and Center of Mass 6.3.5 Exercises

• \(a = -10 \ln(0.7) \approx 3.567 \text{ cm}\text{.}\)
• Left of the midpoint.
• \(\overline{x} \approx \frac{50.3338}{30} \approx 1.687\text{.}\)
• \(q = -10\ln(0.85) \approx 1.625\) cm.
• \(M_1 = \arctan(10) \approx 1.47113\text{;}\) \(M_2 = 10 - 10e^{-1} \approx 6.32121\text{.}\)
• \(\overline{x_1} \approx 1.56857 \text{;}\) \(\overline{x_2} \approx 4.18023 \text{.}\)
• \begin{equation*} M = \int_0^{10} \rho(x) \, dx + \int_0^{10} p(x) \, dx \approx 1.47113 + 6.32121 = 7.79234\text{.} \end{equation*}
• \begin{equation*} \int_0^{10} x(\rho(x) + p(x))) \, dx = 28.73167\text{.} \end{equation*}
• \(V = \int_0^{30} \pi (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2 \, dx \approx 52.0666\) cubic inches.
• \(W \approx 0.6 \cdot 52.0666 = 31.23996\) ounces.
• At a given \(x\) -location, the amount of weight concentrated there is approximately the weight density ( \(0.6\) ounces per cubic inch) times the volume of the slice, which is \(V_{text{slice}} \approx \pi f(x)^2\text{.}\)
• \(\displaystyle \overline{x} \approx 23.21415\)

6.4 Physics Applications: Work, Force, and Pressure 6.4.5 Exercises

• \(W = \int_0^h 3744 x \cos \left( \frac{x^3}{4} \right) \, dx \text{.}\)
• \(F \approx 462.637\) pounds.
• \(W = 1404(19\pi - 8) \approx 305179.3 \) foot-pounds.
• \begin{equation*} F \approx 1123.2 \end{equation*} pounds.

6.5 Improper Integrals 6.5.5 Exercises

• Converges to \(1\text{.}\)
• \(\int_e^{\infty} \frac{1}{x(\ln(x))^p} \, dx\) diverges if \(p \leq 1\) and converges to \(\frac{1}{p-1}\) if \(p \gt 1\text{.}\)
• Converges to \(-1\text{.}\)

7 Differential Equations 7.1 An Introduction to Differential Equations 7.1.5 Exercises

• \(\frac{dT}{dt}\vert_{T=105} = -2\text{;}\) when \(T = 105\text{,}\) the coffee’s temperature is decreasing at an instantaneous rate of \(-2\) degrees F per minute.
• \(T\) decreasing at \(t=0\text{.}\)
• \(T(1) \approx 103\) degrees F.
• For \(T \lt 75\text{,}\) \(T\) increases. For \(T \gt 75\text{,}\) \(T\) decreases.
• Room temperature is \(75\) degrees F.
• Substitute \(T(t) = 75 + 30e^{-t/15}\) in for \(T\) in the differential equation \(\frac{dT}{dt}= -\frac1{15}T+5\) and verify the equality holds; \(T(0) = 75 + 30e^0 = 75 + 30 = 105\text{;}\) \(T(t) = 75 + 30e^{-t/15} \to 75\) as \(t \to \infty\text{.}\)
• \(1 \lt P \lt 3\text{.}\)
• \(P \lt 1\) and \(3 \lt P \lt 4\text{.}\)
• \(P\) will not change at all.
• The population will decrease toward \(P = 0\) with \(P\) always being positive.
• The population will increase toward \(P = 3\) with \(P\) always being between \(1\) and \(3\text{.}\)
• The population will decrease toward \(P = 3\) with \(P\) always being above \(3\text{.}\)
• There’s a maximum threshold of \(P = 3\text{.}\)
• \(y(t) = t + 1 + 2e^t\) is a solution to the DE.
• \(y(t) = t + 1\) is a solution to the DE.
• \(y(t) = t + 2\) is a not solution to the DE.
• \(k = 9\text{.}\)

7.2 Qualitative behavior of solutions to DEs 7.2.4 Exercises

• Sketch curves through appropriate points in the slope field above.
• \(y(t) = t-1\text{.}\)
• \(t\) and \(y\) are equal.
• Any solution curve that starts with \(P(0) \gt 3\) will decrease to \(P(t) = 3\) as \(t \to \infty\text{;}\) any curve that starts with \(1 \lt P(0) \lt 3\) will increase to \(P(t) = 3\text{;}\) any curve that starts with \(0 \lt P(0) \lt 1\) will decrease to \(P(t) = 0\text{.}\)
• \(P = 0\text{,}\) \(P = 1\text{,}\) and \(P = 3\text{.}\) \(P = 1\) is unstable; \(P = 0\) and \(P = 3\) are stable.
• The population will stabilize either at the value \(P = 3\) or at \(P = 0\text{.}\)
• \(P(t) = 1\) is the threshold.
• \(\frac{dP}{dt} = g(P) = P(6-P)-1\) ; the equilibrium at \(P\approx 0.172\) is unstable; the equilibrium at \(P \approx 5.83\) is stable.
• If \(P \lt \frac{6-\sqrt{32}}{2}\text{,}\) then the fish population will die out. If \(\frac{6-\sqrt{32}}{2} \lt P\text{,}\) then the fish population will approach \(\frac{6+\sqrt{32}}{2}\) thousand fish.
• \(\frac{dP}{dt} = g(P) = P(6-P)-h \text{;}\) equilibrium solutions \(P = \frac{6+\sqrt{36-4h}}{2}, \ \frac{6-\sqrt{36-4h}}{2} \text{.}\)
• \(9000\) fish; harvesting at that rate will maintain the number of fish we start with, provided it’s at least \(3000\text{.}\)
• \(\frac{dy}{dt} = 20y\text{.}\)
• \(\displaystyle \frac{dy}{dt} = 20y - C\frac{y}{2+y}\)
• For positive \(y\) near \(0\text{,}\) \(M(y) = \frac{y}{2+y} \approx 0\text{;}\) for large values of \(y\text{,}\) \(M(y) = \frac{y}{2+y} \approx 1\text{.}\)
• The only equilibrium solution is \(y = 0\text{,}\) which is unstable.
• The equilibrium solutions are \(y = 0\) (stable) and \(y = 1\) (unstable).
• At least \(41\) cats.

7.3 Euler’s method 7.3.4 Exercises

• Alice’s coffee: \(\frac{dT_A}{dt} \vert_{T = 100} = -0.5(30) = -15\) degrees per minute; Bob’s coffee: \(\frac{dT_B}{dt} \vert_{T = 100} = -0.1(30) = -3\) degrees per minute.
• Consider the insulation of the containers.
• Compare the rate of initial decrease and amplitude of oscillation.
• \(K = 1.054\text{;}\) \(y(1) = 2.6991\text{.}\)
• \(K = 1.272\text{;}\) \(y(1) = 2.7169\text{.}\)
• \(K = 0.122\) and \(y(0.3) = 0.0412\text{.}\)
• \(y(1) \approx y_5 = 2.7027\text{.}\)
• \(y(1) \approx y_{10} = 2.7141\text{.}\)
• The square of \(\Delta t\text{.}\)

7.4 Separable differential equations 7.4.3 Exercises

• \(\frac{dM}{dt} = kM \text{.}\)
• \(M(t) = M_0e^{kt} \text{.}\)
• \(\displaystyle M(t) = M_0e^{-\frac{\ln(2)}{5730}t} \approx M_0e^{-0.000121t}\)
• \(t = \frac{5730\ln(4)}{\ln(2)} \approx 11460\) years.
• \(t = -\frac{5730\ln(0.3)}{\ln(2)} \approx 9952.8\) years.
• \(y = \sqrt{64 - t^2} \text{.}\)
• \(-8 \le t \le 8\text{.}\)
• \(y(8) = 0\text{.}\)
• \(\frac{dy}{dt} = -\frac ty\) is not defined when \(y = 0\text{.}\)
• \(\frac{dh}{dt} = k \sqrt{h} \text{.}\)
• The tank with \(k = -10\) has water leaving the tank much more rapidly.
• \(k = -2\text{.}\)
• \(h(t) = \left( 10 - t \right)^2 \text{.}\)
• \(10\) minutes.
• \(P = 3\) is stable.
• \(P(t) = 3e^{\ln \left(\frac{1}{3} \right) e^{-t}} \text{.}\)
• \(P(t) = 3e^{\ln \left( 2 \right) e^{-t}} \text{.}\)

7.5 Modeling with differential equations 7.5.3 Exercises

• \(\displaystyle \frac{dA}{dt} = 1 + 0.05A\)
• \(A(25) = 49.80686\) million dollars.
• \(A(25) = 34.90343\) million dollars.
• \(t = 20 \ln(2) \approx 13.86 \ \text{years} \text{.}\)
• \(\displaystyle \frac{dv}{dt} = 9.8 - kv\)
• \(v = \frac{9.8}{k}\) is a stable equilibrium.
• \(\displaystyle v(t) = \frac{9.8 - 9.8e^{-kt}}{k}\)
• \(k = 9.8/54 \approx 0.181481\text{.}\)
• \(t = \frac{\ln(0.5)}{-0.181481} \approx 3.1894\) seconds.
• \(\frac{dw}{dt} = \frac{k}{w} \text{.}\)
• \(w(t) = \sqrt{17t+64} \text{;}\) \(w(12) = \sqrt{268} \approx 16.37\) pounds.
• The model is unrealistic.
• The inflow and outflow are at the same rate.
• \(60\) grams per minute.
• \(\displaystyle \frac{S(t)}{100} \frac{\text{grams}}{\text{gallon}}\)
• \(\frac{3S(t)}{100} \frac{\text{grams}}{\text{minute}} \text{.}\)
• \(\frac{dS}{dt} = 60 - \frac{3}{100} S \text{.}\)
• \(S = 2000\) is a stable equilibrium solution.
• \(S(t) = 2000 - 2000e^{-\frac{3}{100}t} \text{.}\)
• \(S(t) \to 2000\text{.}\)

7.6 Population Growth and the Logistic Equation 7.6.4 Exercises

• \(p(t) \to 1\) as \(t \to \infty\) provided \(p(0) \gt 0\text{.}\)
• \(p(t) = \frac{1}{9e^{-0.2t} + 1} \text{.}\)
• \(t = -5 \ln(1/9) \approx 10.986\) days.
• \(t = -5 \ln(0.25/9) \approx 17.19\) days.
• \(\displaystyle \frac{db}{dt} = \frac{1}{3000} b(15000 - b)\)
• \(b = 15000\text{.}\)
• When \(b = 7500\text{.}\)
• \(t = -\frac{1}{5} \ln(1/70) \approx 0.8497\) days.
• 10000 fish.
• \(\frac{dP}{dt} = 0.1P(10 - P) - 0.2P \text{.}\)
• \(8000\) fish.
• \(P(1) \approx 8.7899\) thousand fish.
• \(t = -1.25 \ln(5/11) \approx 0.986\) years.

8 Sequences and Series 8.1 Sequences 8.1.3 Exercises

• If \(\lim_{x \to \infty} f(x) = L\text{,}\) then \(\lim_{n \to \infty} \frac{\ln(n)}{n} = L\) as well.
• \begin{equation*} \lim_{n \to \infty} \frac{\ln(n)}{n} = \lim_{x \to \infty} \frac{\ln(x)}{x} = 0\text{.} \end{equation*}
• \(P_1\left(\frac{r}{12}\right)\) in interest in the second month; at the end of the second month, \(P_2 = P\left( 1 + \frac{r}{12}\right)^2 \text{.}\)
• \(P_3 = P\left( 1 + \frac{r}{12}\right)^3 \text{.}\)
• \(P_n = P\left( 1 + \frac{r}{12}\right)^n\) is a pattern to these calculations.
• \(A_1 = \frac{1}{2}(100)\text{.}\)
• \(A_2 = \left( \frac{1}{2} \right)^2 \cdot 100\text{.}\)
• \(A_3 = \left( \frac{1}{2} \right)^3 \cdot 100\text{.}\)
• \(A_4 = \left( \frac{1}{2} \right)^4 \cdot 100\text{.}\)
• \(A_n = \left( \frac{1}{2} \right)^n \cdot 100 \text{.}\)
• \(A_n \to 0\) as time goes on.
• It takes about \(6.6439\) half-lives to elapse to get down to \(1\) gram remaining, or \(5 \cdot 6.6439 = 33.2193\) minutes.
• At least \(13\) samples, so at least every \(10/13 \approx 0.76923\) seconds.
• \(44100\) Hz is slightly more than double \(20\) KHz.

8.2 Geometric Series 8.2.3 Exercises

• \(30 \cdot 500 = 1500\) dollars.
• \(\dollar0.01\left(2^{30}-1\right) = \dollar10,737,418.23 \text{.}\)
• \(h_1 = \left(\frac{3}{4}\right)h \text{.}\)
• \(h_2 = \left(\frac{3}{4}\right)h_1 = \left(\frac{3}{4}\right)^2h \text{.}\)
• \(h_3 = \left(\frac{3}{4}\right)h_2 = \left(\frac{3}{4}\right)^3h \text{.}\)
• \(h_n = \left(\frac{3}{4}\right)h_{n-1} = \left(\frac{3}{4}\right)^nh \text{.}\)
• The distance traveled by the ball is \(7h\text{,}\) which is finite.
• There are \(6\) equally possible outcomes when we roll one die.
• The three rolls are independent so the probability of the overall outcome is the product of the three probabilities.
• \(P = \frac{6}{11} \text{.}\)
• \(0.75P\) dollars spent.
• \(0.75P + 0.75(0.75P) = 0.75P(1+0.75)\) dollars.
• \(0.75P + 0.75^2P + 0.75^2P + \cdots = 0.75P(1+0.75+0.75^2+ \cdots )\) dollars.
• A stimulus of \(200\) billion dollars adds \(600\) billion dollars to the economy.
• \(\left(\frac{r}{12}\right)P_1\) dollars.
• \(P_2 = \left(1 + \frac{r}{12}\right)P_1 - M \text{.}\)
• \(P_2 = \left(1 + \frac{r}{12}\right)^2P - \left[1 + \left(1+\frac{r}{12}\right)\right] M \text{.}\)
• \(P_3 = \left(1 + \frac{r}{12}\right)P_2 - M \text{.}\) \(P_3 = \left(1 + \frac{r}{12}\right)^3P - \left[1 + \left(1+\frac{r}{12}\right) + \left(1+\frac{r}{12}\right)^2 \right] M \text{.}\)
• \(P_n = P \left(1+\frac{r}{12}\right)^n - \left(\frac{12M}{r}\right) \left( \left(1+\frac{r}{12}\right)^n - 1\right) \text{.}\)
• \(\displaystyle P(t) = P \left(1+\frac{r}{12}\right)^{12t} - \left(\frac{12M}{r}\right) \left( \left(1+\frac{r}{12}\right)^{12t} - 1\right)\)
• \(A(t) = \left(1000 - \frac{12(25)}{0.2}\right)\left(1+\frac{0.2}{12}\right)^{12t} + \frac{12(25)}{0.2} \text{.}\) \(t \approx 5.5 \text{.}\) We pay $659 dollars in interest on our $1000 loan.
• $291.74 each month to complete the loan in 5 years; we pay $2,504.40 in interest.

8.3 Series of Real Numbers 8.3.7 Exercises

• \(\sum \frac{10^k}{k!}\) converges.
• The sequence \(\left\{\frac{b^n}{n!}\right\}\) has to converge to 0.
• \(\sqrt[n]{a_n} \approx r\) for large \(n\text{.}\)
• \(\frac{a_{n+1}}{a_n} \approx r\text{.}\)
• \(0 \lt r \lt 1\text{.}\)
• \(a_n = 1 + \frac{1}{2^n} \to 1 \ne 0\) and \(b_n = -1 \to -1 \ne 0\text{.}\)
• The series is geometric with \(r = \frac{1}{2}\text{.}\)
• Since the two individual series diverge, neither sum is a finite number, so it doesn’t make any sense to add them.
• Note that \(A_n + B_n = (a_1 + b_1) + (a_2 + b_2) + \cdots + (a_n + b_n) \text{.}\)
• Note that \(\lim_{n \to \infty} \sum_{k=1}^n (a_k+b_k) = \lim_{n \to \infty} \left( \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} b_k \right) \text{.}\)
• \(\sum_{k=0}^{\infty} \frac{2^k+3^k}{5^k} = \frac{25}{6} \text{.}\)
• \(S_1 = 1\) and \(T_1 = \frac{1}{2}\text{.}\)
• \(S_2 \gt T_2\text{.}\)
• \(S_3 \gt T_3\text{.}\)
• \(S_n \gt T_n\text{.}\)
• \(\sum \frac{1}{k^2} \gt \sum \frac{1}{k^2+k} \text{;}\) \(\sum \frac{1}{k^2+k}\) converges.
• Note that \(0 \lt \frac{1}{k} \lt \frac{1}{k-1}\text{.}\)
• Note that \(\frac{1}{k^3} \gt \frac{1}{k^3+1}\text{.}\)

8.4 Alternating Series 8.4.6 Exercises

• \(\sum_{k=0}^{\infty} \frac{1}{2k+1}\) diverges by comparison to the Harmonic series.
• \(|S_{100} - \sum_{k=0}^{\infty} (-1)^{k} \frac{1}{2k+1}| \lt \approx 0.0049 \text{.}\)
• \(\displaystyle n \gt 4{,}999{,}999{,}999.5\)
• \(\frac{S_n+S_{n+1}}{2} = \frac{S_n+S_{n} + (-1)^{n+2}a_{n+1}}{2} \text{.}\)
• \(S_{20} = 0.668771403 \ldots \text{;}\) \(\frac{S_{20} + S_{21}}{2} = \frac{161227687}{232792560} = 0.692580926 \ldots \text{,}\) accurate to within about \(0.0006\text{.}\)
• \(\left\{\frac{1}{n}\right\}\) and \(\left\{-\frac{1}{n^2}\right\}\) converge to 0.
• Notice that \(\frac{1}{k} - \frac{1}{k^2} = \frac{k-1}{k^2}\) and compare to the Harmonic series.
• It is possible for a series to alternate, have the terms go to zero, have the terms not decrease to zero, and the series diverge.

8.5 Taylor Polynomials and Taylor Series 8.5.6 Exercises

• \(P_3(x) = -1 + 3x - \frac{4}{2!}x^2 + \frac{6}{3!}x^3 \text{,}\) which is the same polynomial as \(f(x)\text{.}\)
• For \(n \ge 3\text{,}\) \(P_n(x) = f(x)\text{.}\)
• For \(n \ge m\text{,}\) \(P_n(x) = f(x)\text{.}\)
• \begin{align*} P_1(x) &= 1 + 0\left(x - \frac{\pi}{2} \right) = 1\\ P_2(x) &= 1 + 0\left(x - \frac{\pi}{2} \right) - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 = 1-\frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2\\ P_3(x) &= 1 + 0\left(x - \frac{\pi}{2} \right) - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 + \frac{0}{3!}\left(x - \frac{\pi}{2} \right)^3\\ &= 1-\frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 = P_2(x)\\ P_4(x) &= 1 + 0\left(x - \frac{\pi}{2} \right) - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 + \frac{0}{3!}\left(x - \frac{\pi}{2} \right)^3 + \frac{1}{4!}\left(x - \frac{\pi}{2} \right)^4\\ &= 1 - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 + \frac{1}{4!}\left(x - \frac{\pi}{2} \right)^4\\ P(x) &= 1 - \frac{1}{2!}\left(x - \frac{\pi}{2} \right)^2 + \frac{1}{4!}\left(x - \frac{\pi}{2} \right)^4 - \frac{1}{6!}\left(x - \frac{\pi}{2} \right)^6 + \cdots \end{align*}
• \begin{equation*} P_4(x) = 0 + 1(x-1) - \frac{1}{2!}(x-1)^2 + \frac{2}{3!}(x-1)^3 - \frac{6}{4!}(x-1)^4\text{.} \end{equation*} \begin{equation*} P(x) = 1(x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3 - \frac{1}{4}(x-1)^4 + \frac{1}{5}(x-1)^5 - \cdots \end{equation*}
• \(\displaystyle P_{101}(1) \approx 0.698073\)
• \(P_4(x) = x^2 \text{.}\)
• \(g(x) = P(x^2) = x^2 - \frac{1}{3!}x^6 + \frac{1}{5!}x^10 - \cdots \text{.}\)
• All real numbers.

8.6 Power Series 8.6.4 Exercises

• \(\sin(x^2) = \sum_{k=0}^{\infty} (-1)^k\frac{x^{2(2k+1)}}{(2k+1)!} \text{,}\) with interval of convergence \((-\infty, \infty)\text{.}\)
• \(\int \sin(x^2) \, dx = \sum_{k=0}^{\infty} (-1)^k\frac{x^{4k+3}}{(2k+1)!(4k+3)} + C \text{.}\)
• \(\int_0^1 \sin(x^2) \, dx = \sum_{k=0}^{\infty} (-1)^k\frac{1}{(2k+1)!(4k+3)} \text{.}\) Use \(n = 1\) to generate the desired estimate.
• Then \begin{align*} f'(x) \amp = \sum_{k=1}^{\infty} ka_kx^{k-1}\\ f''(x) \amp = \sum_{k=2}^{\infty} k(k-1)a_kx^{k-2}\\ f^{(3)}(x) \amp = \sum_{k=3}^{\infty} k(k-1)(k-2)a_kx^{k-3}\\ \vdots \amp \ \qquad \vdots\\ f^{(n)}(x) \amp = \sum_{k=n}^{\infty} k(k-1)(k-2) \cdots (k-n+1) a_kx^{k-n}\\ \vdots \amp \ \qquad \vdots \end{align*} So \begin{align*} f(0) \amp = a_0\\ f'(0) \amp = a_1\\ f''(0) \amp = 2!a_2\\ f^{(3)}(0) \amp = 3!a_3\\ \vdots \amp \ \qquad \vdots\\ f^{(k)}(0) \amp = k!a_k\\ \vdots \amp \ \qquad \vdots \end{align*} and \begin{equation*} a_k = \frac{f^{(k)}(0)}{k!} \end{equation*} for each \(k \geq 0\text{.}\) But these are just the coefficients of the Taylor series expansion of \(f\text{,}\) which leads us to the following observation.

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Active Calculus

Matthew Boelkins

Solutions C Answers to Selected Exercises

This appendix contains answers to all non- WeBWorK exercises in the text. For WeBWorK exercises, please use the HTML version of the text for access to answers and solutions.

Chapter 1 Limits, Continuity and Derivatives

Section 1.1 the notion of limit, exercises 1.1.4 exercises, 1.1.4.1. limits on a piecewise graph., 1.1.4.2. estimating a limit numerically., 1.1.4.3. limits for a piecewise formula., 1.1.4.4. evaluating a limit algebraically., section 1.2 the derivative, exercises 1.2.4 exercises, 1.2.4.1. the derivative function graphically., 1.2.4.2. applying the limit definition of the derivative., 1.2.4.3. sketching the derivative., 1.2.4.4. comparing function and derivative values., 1.2.4.5. limit definition of the derivative for a rational function., section 1.3 the second derivative, exercises 1.3.5 exercises, 1.3.5.1. comparing \(f, f', f''\) values., 1.3.5.2. signs of \(f, f', f''\) values., 1.3.5.3. acceleration from velocity., 1.3.5.4. rates of change of stock values., 1.3.5.5. interpreting a graph of \(f'\)., section 1.4 limits, continuity, and differentiability, exercises 1.4.5 exercises, 1.4.5.1. limit values of a piecewise graph., 1.4.5.2. limit values of a piecewise formula., 1.4.5.3. continuity and differentiability of a graph., 1.4.5.4. continuity of a piecewise formula., section 1.5 the tangent line approximation, exercises 1.5.4 exercises, 1.5.4.1. approximating \(\sqrt{x}\)., 1.5.4.2. local linearization of a graph., 1.5.4.3. estimating with the local linearization., 1.5.4.4. predicting behavior from the local linearization., chapter 2 computing derivatives, section 2.1 elementary derivative rules, exercises 2.1.5 exercises, section 2.2 the sine and cosine functions, exercises 2.2.3 exercises, section 2.3 the product and quotient rules, exercises 2.3.5 exercises, 2.3.5.1. derivative of a basic product., 2.3.5.2. derivative of a product., 2.3.5.3. derivative of a quotient of linear functions., 2.3.5.4. derivative of a rational function., 2.3.5.5. derivative of a product of trigonometric functions., 2.3.5.6. derivative of a product of power and trigonmetric functions., 2.3.5.7. derivative of a sum that involves a product., 2.3.5.8. product and quotient rules with graphs., 2.3.5.9. product and quotient rules with given function values., section 2.4 derivatives of other trigonometric functions, exercises 2.4.3 exercises, 2.4.3.1. a sum and product involving \(\tan(x)\)., 2.4.3.2. a quotient involving \(\tan(t)\)., 2.4.3.3. a quotient of trigonometric functions., 2.4.3.4. a quotient that involves a product., 2.4.3.5. finding a tangent line equation., section 2.5 the chain rule, exercises 2.5.5 exercises, 2.5.5.1. mixing rules: chain, product, sum., 2.5.5.2. mixing rules: chain and product., 2.5.5.3. using the chain rule repeatedly., 2.5.5.4. derivative involving arbitrary constants \(a\) and \(b\)., 2.5.5.5. chain rule with graphs., 2.5.5.6. chain rule with function values., 2.5.5.7. a product involving a composite function., section 2.6 derivatives of inverse functions, exercises 2.6.6 exercises, 2.6.6.1. composite function involving logarithms and polynomials., 2.6.6.2. composite function involving trigonometric functions and logarithms., 2.6.6.3. product involving \(\arcsin(w)\)., 2.6.6.4. derivative involving \(\arctan(x)\)., 2.6.6.5. composite function from a graph., 2.6.6.6. composite function involving an inverse trigonometric function., 2.6.6.7. mixing rules: product, chain, and inverse trig., 2.6.6.8. mixing rules: product and inverse trig., section 2.7 derivatives of functions given implicitly, exercises 2.7.3 exercises, 2.7.3.1. implicit differentiaion in a polynomial equation., 2.7.3.2. implicit differentiation in an equation with logarithms., 2.7.3.3. implicit differentiation in an equation with inverse trigonometric functions., 2.7.3.4. slope of the tangent line to an implicit curve., 2.7.3.5. equation of the tangent line to an implicit curve., chapter 3 using derivatives, section 3.1 using derivatives to evaluate limits, exercises 3.1.4 exercises, 3.1.4.1. l'hôpital's rule with graphs., 3.1.4.2. l'hôpital's rule to evaluate a limit., 3.1.4.3. determining if l'hôpital's rule applies., 3.1.4.4. using l'hôpital's rule multiple times., section 3.2 using derivatives to identify extreme values, exercises 3.2.4 exercises, 3.2.4.1. finding critical points and inflection points., 3.2.4.2. finding inflection points., 3.2.4.3. matching graphs of \(f,f',f''\)., section 3.3 using derivatives to describe families of functions, exercises 3.3.3 exercises, 3.3.3.1. drug dosage with a parameter., 3.3.3.2. using the graph of \(g'\)., section 3.4 applied optimization, exercises 3.4.3 exercises, 3.4.3.1. maximizing the volume of a box., 3.4.3.2. minimizing the cost of a container., 3.4.3.3. maximizing area contained by a fence., 3.4.3.4. minimizing the area of a poster., 3.4.3.5. maximizing the area of a rectangle., section 3.5 related rates, exercises 3.5.3 exercises, 3.5.3.1. height of a conical pile of gravel., 3.5.3.2. movement of a shadow., 3.5.3.3. a leaking conical tank., chapter 4 the definite integral, section 4.1 determining distance traveled from velocity, exercises 4.1.5 exercises, 4.1.5.1. estimating distance traveled from velocity data., 4.1.5.2. distance from a linear veloity function., 4.1.5.3. change in position from a linear velocity function., 4.1.5.5. finding average acceleration from velocity data., 4.1.5.6. change in position from a quadratic velocity function., section 4.2 riemann sums, exercises 4.2.5 exercises, 4.2.5.1. evaluating riemann sums for a quadratic function., 4.2.5.2. estimating distance traveled with a riemann sum from data., 4.2.5.3. writing basic riemann sums., section 4.3 the definite integral, exercises 4.3.5 exercises, 4.3.5.1. evaluating definite integrals from graphical information., 4.3.5.2. estimating definite integrals from a graph., 4.3.5.3. finding the average value of a linear function., 4.3.5.4. finding the average value of a function given graphically., 4.3.5.5. estimating a definite integral and average value from a graph., 4.3.5.6. using rules to combine known integral values., section 4.4 the fundamental theorem of calculus, exercises 4.4.5 exercises, 4.4.5.1. finding exact displacement., 4.4.5.2. evaluating the definite integral of a rational function., chapter 5 evaluating integrals, section 5.1 constructing accurate graphs of antiderivatives, exercises 5.1.5 exercises, 5.1.5.1. definite integral of a piecewise linear function., 5.1.5.2. a smooth function that starts out at 0., 5.1.5.3. a piecewise constant function., 5.1.5.4. another piecewise linear function., section 5.2 the second fundamental theorem of calculus, exercises 5.2.5 exercises, 5.2.5.1. a definite integral starting at 3., 5.2.5.2. variable in the lower limit., 5.2.5.3. approximating a function with derivative \(e^{-x^2/5}\)., section 5.3 integration by substitution, exercises 5.3.5 exercises, 5.3.5.1. product involving 4th power of a polynomial., 5.3.5.2. product involving \(\sin(x^6)\)., 5.3.5.3. fraction involving \(\ln^9\)., 5.3.5.4. fraction involving \(e^{5 x}\)., 5.3.5.5. fraction involving \(e^{5 \sqrt{y}}\)., 5.3.5.6. definite integral involving \(e^{-cos(q)}\)., section 5.4 integration by parts, exercises 5.4.7 exercises, 5.4.7.2. product involving \(\cos(5 x)\)., 5.4.7.3. product involving \(e^{8 z}\)., 5.4.7.4. definite integral of \(t e^{-t}\)., section 5.5 other options for finding algebraic antiderivatives, exercises 5.5.5 exercises, 5.5.5.1. partial fractions: linear over difference of squares., 5.5.5.2. partial fractions: constant over product., 5.5.5.3. partial fractions: linear over quadratic., 5.5.5.4. partial fractions: cubic over 4th degree., 5.5.5.5. partial fractions: quadratic over factored cubic., section 5.6 numerical integration, exercises 5.6.6 exercises, 5.6.6.1. various methods for \(e^x\) numerically., 5.6.6.2. comparison of methods for increasing concave down function., chapter 6 using definite integrals, section 6.1 using definite integrals to find area and length, exercises 6.1.5 exercises, 6.1.5.1. area between two power functions., 6.1.5.2. area between two trigonometric functions., 6.1.5.3. area between two curves., 6.1.5.4. arc length of a curve., section 6.2 using definite integrals to find volume, exercises 6.2.3 exercises, 6.2.3.1. solid of revolution from one function about the \(x\)-axis., 6.2.3.2. solid of revolution from one function about the \(y\)-axis., 6.2.3.3. solid of revolution from two functions about the \(x\)-axis., 6.2.3.4. solid of revolution from two functions about a horizontal line., 6.2.3.5. solid of revolution from two functions about a different horizontal line., 6.2.3.6. solid of revolution from two functions about a vertical line., section 6.3 density, mass, and center of mass, exercises 6.3.5 exercises, 6.3.5.1. center of mass for a linear density function., 6.3.5.2. center of mass for a nonlinear density function., 6.3.5.3. interpreting the density of cars on a road., 6.3.5.4. center of mass in a point-mass system., section 6.4 physics applications: work, force, and pressure, exercises 6.4.5 exercises, 6.4.5.1. work to empty a conical tank., 6.4.5.2. work to empty a cylindrical tank., 6.4.5.3. work to empty a rectangular pool., 6.4.5.4. work to empty a cylindrical tank to differing heights., 6.4.5.5. force due to hydrostatic pressure., section 6.5 improper integrals, exercises 6.5.5 exercises, 6.5.5.1. an improper integral on a finite interval., 6.5.5.2. an improper integral on an infinite interval., 6.5.5.3. an improper integral involving a ratio of exponential functions., 6.5.5.4. a subtle improper integral., 6.5.5.5. an improper integral involving a ratio of trigonometric functions., chapter 7 differential equations, section 7.1 an introduction to differential equations, exercises 7.1.5 exercises, 7.1.5.2. finding constant to complete solution., section 7.2 qualitative behavior of solutions to des, exercises 7.2.4 exercises, 7.2.4.1. graphing equilibrium solutions., 7.2.4.2. sketching solution curves., 7.2.4.4. describing equilibrium solutions., section 7.3 euler's method, exercises 7.3.4 exercises, 7.3.4.1. a few steps of euler's method., 7.3.4.2. using euler's method for a solution of \(y'=4y\)., 7.3.4.3. using euler's method with different time steps., section 7.4 separable differential equations, exercises 7.4.3 exercises, 7.4.3.1. initial value problem for \(dy/dx=x^8 y\)., 7.4.3.2. initial value problem for \(dy/dt=0.9(y-300)\)., 7.4.3.3. initial value problem for \(dy/dt=y^2(8+t)\)., 7.4.3.4. initial value problem for \(du/dt=e^{6u+10t}\)., 7.4.3.5. initial value problem for \(dy/dx=170yx^{16}\)., section 7.5 modeling with differential equations, exercises 7.5.3 exercises, 7.5.3.1. mixing problem., 7.5.3.2. mixing problem., 7.5.3.3. population growth problem., 7.5.3.4. radioactive decay problem., 7.5.3.5. investment problem., section 7.6 population growth and the logistic equation, exercises 7.6.4 exercises, 7.6.4.1. analyzing a logistic equation., 7.6.4.2. analyzing a logistic model., 7.6.4.3. finding a logistic function for an infection model., 7.6.4.4. analyzing a population growth model., chapter 8 sequences and series, section 8.1 sequences, exercises 8.1.3 exercises, 8.1.3.2. formula for a sequence, given first terms., 8.1.3.3. divergent or convergent sequences., 8.1.3.4. terms of a sequence from sampling a signal., section 8.2 geometric series, exercises 8.2.3 exercises, 8.2.3.1. seventh term of a geometric sequence., 8.2.3.2. a geometric series., 8.2.3.3. a series that is not geometric., 8.2.3.4. two sums of geometric sequences., section 8.3 series of real numbers, exercises 8.3.7 exercises, 8.3.7.1. convergence of a sequence and its series., 8.3.7.2. two partial sums., 8.3.7.3. convergence of a series and its sequence., 8.3.7.4. convergence of an integral and a related series., section 8.4 alternating series, exercises 8.4.6 exercises, 8.4.6.2. estimating the sum of an alternating series., 8.4.6.3. estimating the sum of a different alternating series., 8.4.6.4. estimating the sum of one more alternating series., section 8.5 taylor polynomials and taylor series, exercises 8.5.6 exercises, 8.5.6.1. determining taylor polynomials from a function formula., 8.5.6.2. determining taylor polynomials from given derivative values., 8.5.6.3. finding the taylor series for a given rational function., 8.5.6.4. finding the taylor series for a given trigonometric function., 8.5.6.5. finding the taylor series for a given logarithmic function., section 8.6 power series, exercises 8.6.4 exercises, 8.6.4.1. finding coefficients in a power series expansion of a rational function., 8.6.4.2. finding coefficients in a power series expansion of a function involving \(\arctan(x)\)..

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4.1E: Exercises

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This page is a draft and is under active development. 

Exercise \(\PageIndex{1}\)

1. Why is u -substitution referred to as change of variable ?

2. If \(\displaystyle f=g∘h\), when reversing the chain rule, \(\displaystyle \frac{d}{d}x(g∘h)(x)=g′(h(x))h′(x)\), should you take \(\displaystyle u=g(x)\) or u= \(\displaystyle h(x)?\)

\(\displaystyle u=h(x)\).

Exercise \(\PageIndex{2}\)

In the following exercises, verify each identity using differentiation. Then, using the indicated u-substitution, identify f such that the integral takes the form \(\displaystyle∫f(u)du.\)

1. \(\displaystyle∫x\sqrt{x+1}x=\frac{2}{15}(x+1)^{3/2}(3x−2)+C;u=x+1\)

2. \(\displaystyle∫\frac{x^2}{\sqrt{x−1}}dx(x>1)=\frac{2}{15}\sqrt{x−1}(3x^2+4x+8)+C;u=x−1\)

\(\displaystyle f(u)=\frac{(u+1)^2}{\sqrt{u}}\)

3. \(\displaystyle∫x\sqrt{4x^2+9}dx=\frac{1}{12}(4x^2+9)^{3/2}+C;u=4x^2+9\)

4. \(\displaystyle∫\frac{x}{\sqrt{4x^2+9}}dx=\frac{1}{4}\sqrt{4x^2+9}+C;u=4x^2+9\)

\(\displaystyle du=8xdx;f(u)=\frac{1}{8\sqrt{u}}\)

5. \(\displaystyle∫\frac{x}{(4x^2+9)^2}dx=−\frac{1}{8(4x^2+9)};u=4x^2+9\)

\(\displaystyle du=8xdx;f(u)=\frac{1}{8u^2}\)

Exercise \(\PageIndex{3}\)

In the following exercises, find the antiderivative using the indicated substitution.

1. \(\displaystyle ∫(x+1)^4dx;u=x+1\)

\(\displaystyle \frac{1}{5}(x+1)^5+C\)

2. \(\displaystyle∫(x−1)^5dx;u=x−1\)

3. \(\displaystyle∫(2x−3)^{−7}dx;u=2x−3\)

\(\displaystyle−\frac{1}{12(3−2x)^6}+C\)

4. \(\displaystyle∫(3x−2)^{−11}dx;u=3x−2\)

5. \(\displaystyle∫\frac{x}{\sqrt{x^2+1}}dx;u=x^2+1\)

\(\displaystyle\sqrt{x^2+1}+C\)

6. \(\displaystyle∫\frac{x}{\sqrt{1−x^2}}dx;u=1−x^2\)

7. \(\displaystyle∫(x−1)(x^2−2x)^3dx;u=x^2−2x\)

\(\displaystyle\frac{1}{8}(x^2−2x)^4+C\)

8. \(\displaystyle∫(x^2−2x)(x^3−3x^2)^2dx;u=x^3=3x^2\)

9. \(\displaystyle∫cos^3θdθ;u=sinθ (Hint:cos^2θ=1−sin^2θ)\)

\(\displaystylesinθ−\frac{sin^3θ}{3}+C\)

10. \(\displaystyle∫sin^3θdθ;u=cosθ (Hint:sin^2θ=1−cos^2θ)\)

Exercise \(\PageIndex{4}\)

In the following exercises, use a suitable change of variables to determine the indefinite integral.

1. \(\displaystyle∫x(1−x)^{99}dx\)

\(\displaystyle\frac{(1−x)^{101}}{101}−\frac{(1−x)^{100}}{100}+C\)

2. \(\displaystyle∫t(1−t^2)^{10}dt\)

3. \(\displaystyle∫(11x−7)^{−3}dx\)

\(\displaystyle−\frac{1}{22(7−11x^2)}+C\)

4. \displaystyle∫(7x−11)^4dx\)

5. \(\displaystyle∫cos^3θsinθdθ\)

\(\displaystyle−\frac{cos^4θ}{4}+C\)

6. \(\displaystyle∫sin^7θcosθdθ\)

7. \(\displaystyle∫cos^2(πt)sin(πt)dt\)

\(\displaystyle−\frac{cos^3(πt)}{3π}+C\)

8. \(\displaystyle∫sin^2xcos^3xdx (Hint:sin^2x+cos^2x=1)\)

9. \(\displaystyle∫tsin(t^2)cos(t^2)dt\)

\(\displaystyle−\frac{1}{4}cos^2(t^2)+C\)

10. \(\displaystyle∫t^2cos^2(t^3)sin(t^3)dt\)

11. \(\displaystyle∫\frac{x^2}{(x^3−3)^2}dx\)

\(\displaystyle−\frac{1}{3(x^3−3)}+C\)

12. \(\displaystyle∫\frac{x^3}{\sqrt{1−x^2}}dx\)

13. \(\displaystyle∫\frac{y^5}{(1−y^3)^{3/2}}dy\)

\(\displaystyle−\frac{2(y^3−2)}{3\sqrt{1−y^3}}\)

14. \(\displaystyle∫cosθ(1−cosθ)^{99}sinθdθ\)

15. \(\displaystyle∫(1−cos^3θ)^{10}cos^2θsinθdθ\)

\(\displaystyle\frac{1}{33}(1−cos^3θ)^{11}+C\)

16. \(\displaystyle∫(cosθ−1)(cos^2θ−2cosθ)^3sinθdθ\)

17. \(\displaystyle∫(sin^2θ−2sinθ)(sin^3θ−3sin^2θ^)3cosθdθ\)

\(\displaystyle\frac{1}{12}(sin^3θ−3sin^2θ)^4+C\)

Exercise \(\PageIndex{5}\)

1) \(\displaystyle \int \frac{ x^3\ dx}{\sqrt{1-x^2}} \)

2) \(\displaystyle\int (\csc x)(\cot x)(e^{\csc x})\,\,dx\)

3)\(\displaystyle \int \frac{1}{e^x+e^{-x}}dx\)

4) \(\displaystyle \int \frac {1}{\sqrt{4-x^2}} \ dx \)

5) \(\displaystyle \int \frac{\sqrt{x^2-4}}{x} \ dx \)

6) \(\displaystyle \int \frac{1}{x^2 + 6x + 13} dx\)

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Contributors

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org .

Pamini Thangarajah  (Mount Royal University, Calgary, Alberta, Canada)

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