integration by substitution homework

Integration by Substitution

"Integration by Substitution" (also called "u-Substitution" or "The Reverse Chain Rule") is a method to find an integral , but only when it can be set up in a special way.

The first and most vital step is to be able to write our integral in this form:

Like in this example:

When our integral is set up like that, we can do this substitution :

Then we can integrate f(u) , and finish by putting g(x) back as u .

Example: ∫ cos(x 2 ) 2x dx

We know (from above) that it is in the right form to do the substitution:

Now integrate:

∫ cos(u) du = sin(u) + C

And finally put u=x 2 back again:

sin(x 2 ) + C

So ∫ cos(x 2 ) 2x dx = sin(x 2 ) + C

That worked out really nicely! (Well, I knew it would.)

Let's just run through that again in a step-by-step manner:

But this method only works on some integrals of course, and it may need rearranging:

Example: ∫ cos(x 2 ) 6x dx

Oh no! It is 6x , not 2x like before. Our perfect setup is gone.

Never fear! Just rearrange the integral like this:

∫ cos(x 2 ) 6x dx = 3 ∫ cos(x 2 ) 2x dx

(We can pull constant multipliers outside the integration, see Rules of Integration .)

Then go ahead as before:

3 ∫ cos(u) du = 3 sin(u) + C

Now put u=x 2 back again:

3 sin(x 2 ) + C

Now let's try a slightly harder example:

Example: ∫ x/(x 2 +1) dx

Let me see ... the derivative of x 2 +1 is 2x ... so how about we rearrange it like this:

∫ x/(x 2 +1) dx = ½ ∫ 2x/(x 2 +1) dx

Then we have:

Then integrate:

½ ∫ 1/u du = ½ ln | u | + C

Now put u=x 2 +1 back again:

½ ln(x 2 +1) + C

And how about this one:

Example: ∫ (x+1) 3 dx

Let me see ... the derivative of x+1 is ... well it is simply 1.

So we can have this:

∫ (x+1) 3 dx = ∫ (x+1) 3 · 1 dx

∫ u 3 du = u 4 4 + C

Now put u=x+1 back again:

(x+1) 4 4 + C

We can take that idea further like this:

Example: ∫ (5x+2) 7 dx

If it was in THIS form we could do it:

∫ (5x+2) 7 5 dx

So let's make it so by doing this:

1 5 ∫ (5x+2) 7 5 dx

The 1 5 and 5 cancel out so all is fine.

And now we can have u=5x+2

And then integrate:

1 5 ∫ u 7 du = 1 5 u 8 8 + C

Now put u=5x+2 back again, and simplify:

(5x+2) 8 40 + C

Calculus Tutorials

Computing integrals by substitution.

Many integrals are most easily computed by means of a change of variables, commonly called a $u$-substitution .

Let’s compute $\displaystyle\int\! 2x(x^2-1)^4\, dx$ by making the substitution \begin{eqnarray*} u&=&x^2-1\\ du&=&2x\, dx. \end{eqnarray*} Then \[ \int 2x(x^2-1)^4\, dx=\int (x^2-1)^4(2x\, dx)=\int u^4\, du=\frac{u^5}{5}+C=\frac{(x^2-1)^5}{5}+C.\] We may check this result by differentiating using the Chain Rule: \[\frac{d}{dx}\left(\frac{(x^2-1)^5}{5}+C\right)=\frac{5(x^2-1)^4}{5}(2x) =2x(x^2-1)^4.\qquad\qquad \surd\]

The substitution method amounts to applying the Chain Rule in reverse:

To compute $\displaystyle\int\! f(g(x))g'(x)\, dx$, we let \begin{eqnarray*} u&=&g(x)\\ du&=&g'(x)\, dx. \end{eqnarray*} Then \[\int f(g(x))g'(x)\, dx=\int f(u)\, du=F(u)=F(g(x))\] where $F$ is an antiderivative of $f$.

To compute $\displaystyle\int\! \sin (2x)\cos (2x)\, dx$, let \begin{eqnarray*} u&=&\sin (2x)\\ du&=&2\cos (2x)\, dx. \end{eqnarray*} Then \[\int \sin (2x)\cos (2x)\, dx=\int\frac{1}{2}\sin (2x)[2\cos (2x)\, dx]=\int \frac{1}{2}u\, du=\frac{1}{4}u^2+C=\frac{1}{4}\sin^2 (2x)+C.\]

With practice, you will often be able to write down the result immediately.

We can evaluate $\displaystyle\int\! \frac{dx}{(4x-3)^2}$ by letting \begin{eqnarray*} u&=&4x-3\\ du&=&4\, dx\quad\longrightarrow\quad dx=\frac{1}{4}\, du. \end{eqnarray*} Then \[\int \frac{dx}{(4x-3)^2}=\int \frac{\frac{1}{4}\, du}{u^2}=-\frac{1}{4u}+C=\frac{-1}{4(4x-3)}+C.\]

It is not always apparent until you try it whether or not a substitution will work.

To compute $\displaystyle\int\! x\sqrt{x-3}\, dx$, we will try \begin{eqnarray*} u&=&x-3\quad\longrightarrow\quad x=u+3\\ du&=&dx. \end{eqnarray*} So \begin{eqnarray*} \int x\sqrt{x-3}\, dx&=&\int (u+3)\sqrt{u}\, du=\int \left(u^{3/2}+3u^{1/2}\right)\, du\\ &=&\frac{2}{5}u^{5/2}+2u^{3/2}+C=\frac{2}{5}(x-3)^{5/2}+2(x-3)^{3/2}+C. \end{eqnarray*}

We can also compute a definite integral using a substitution.

Let’s evaluate $\displaystyle\int^2_0\! xe^{x^2}\, dx$. Let \begin{eqnarray*} u&=&x^2\\ du&=&2x\, dx. \end{eqnarray*} First, we will compute the indefinite integral: \[\int xe^{x^2}\, dx=\int \left(\frac{1}{2}e^{x^2}\right)(2x\, dx)=\int\frac{1}{2}e^u\, du=\frac{1}{2}e^u+C=\frac{1}{2}e^{x^2}+C.\] Now we have two approaches for the definite integral:

Thus, we find that \[\int^2_0 xe^{x^2}\, dx=\frac{1}{2}(e^4-1).\]

Approach 2 works provided certain conditions on $f$ and $g$ are met: \[\int^b_a f(g(x))\, dx=\int^{g(b)}_{g(a)} f(u)\, du\] if

• $g’$ is continuous on $[a,b]$.
• $f$ is continuous on the set of values taken by $g$ on $[a,b]$.

Substitutions are useful or necessary for a huge range of integrals. You will find yourself either implicitly or explicitly using a substitution in virtually every integral you compute!

Key Concepts

To compute $\displaystyle\int\! f(g(x))g'(x)\, dx$, we let $u = g(x)$ $du = g'(x) dx$.

Then $\displaystyle{\int f(g(x))g'(x)\, dx = \int f(u)\, du = F(u) = F(g(x))}$ where $F$ is an antiderivative of $f$.

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Integration by Substitution

Integration by substitution is an important method of integration, which is used when a function to be integrated, is either a complex function or if the direct integration of the function is not feasible. The integral of a function is simplified by this method of integration by substitution , by reducing the given function into a simplified function.

Let us learn the process of integration by substitutions, check some of the important substitutions, and also check the solved examples.

What Is Integration by Substitution?

Integration by substitution is used when the integration of the given function cannot be obtained directly, as the given algebraic function is not in the standard form. Further, the given function can be reduced to the standard form by appropriate substitution. Let us consider the indefinite integral  of a function f(x), \(\int f(x).dx\). Here this integral can be transformed to another form by replacing x with g(t) and by substituting x = g(t).

I = \(\int f(x).dx\)

x = g(t) such that dx/dt = g'(t)

dx = g'(t).dt

I =\(\int f(x).dx = \int f(g(t)).g'(t).dt\)

Steps to Integration by Substitution

The following are the steps that are helpful in performing this method of integration by substitution.

Step - 1: Choose a new variable t for the given function to be reduced.

Step - 2: Determine the value of dx, of the given integral, where f(x) is integrated with respect to x.

Step - 3: Make the required substitution in the function f(x), and the new value dx.

Step - 4: Integrate the function obtained after substitution

Step - 5: Substitute back the initial variable x to obtain the final answer.

Important Substituions in Integration by Substitution

The following are some of the important substitutions which are helpful in simplifying the given expression and easily performing the integration process. Let us check the following specific substitutions for integration by substitution.

• For the integral function \(f(\sqrt{a^2 - x^2})\) we use x = aSinθ or x = aCosθ.
• For the integral of the function \(f(\sqrt {x^2 - a^2})\) we use x = a Secθ or x = aCosecθ.
• For the integral of the function \(f(x^2 + a^2)\). \(f(\sqrt{x^2 + a^2})\) we use x = aTanθ, or x = aCotθ.
• For the integral of the functions \(f(\sqrt {\dfrac{a - x}{a + x}})\) , \(f(\sqrt {\dfrac{a + x}{a - x}})\), we use x = a Cos2θ.

Related Topics

The following topics help in a better understanding of this concept of integration by substitution.

• Integration by Partial Fractions
• Integration by Parts
• Differentiation
• Differential Equations Formula
• Differentiation and Integration Formula

Examples on Integration by Substitution

Example 1: Find the integral of \(\dfrac {e^{Tan^{-1}x}}{1 + x^2}\).

The given expression for integration is \(\int \dfrac {e^{Tan^{-1}x}}{1 + x^2}.dx\)

Here let us take \(Tan^{-1}x = t\), and we can differentiate it further.

\(\dfrac{d}{dx}.Tan^{-1}x = \dfrac{d}{dx}.t\)

\(\dfrac{1}{1 + x^2} = \dfrac{dt}{dx}\)

\(dx =(1 + x^2).dt\)

Let us now substitute the 't' value and dx value in the given expression for integration.

\(\int \dfrac {e^{Tan^{-1}x}}{1 + x^2}.dx = \int \dfrac{e^t}{1+ x^2}.(1 + x^2).dt\)

= \(\int e^t.dt\)

= \(e^t + C\)

= \(e^{Tan^{-1}x} + C\)

Therefore, the integration of \(\dfrac {e^{Tan^{-1}x}}{1 + x^2}\) is \(e^{Tan^{-1}x} + C\).

Example 2: Find the integral of \(2xSec^2(x^2 + 1)\).

The given expression for integration is \(\int 2xSec^2(x^2 + 1).dx\).

Here we substitute \(x^2 + 1 = t\) and differentiate it further.

\(\dfrac{d}{dx}.(x^2 + 1) = \dfrac{d}{dx}.t\)

Let us now substitute t and dx value in the above expression of integration.

\(\int.2xSec^2(x^2 + 1).dx = \int 2xSec^2t.\dfrac{dt}{2x}\)

=\(\int Sec^2t.dt\)

= Tan(x 2 + 1) + C

Therefore the integration of \(\int 2xSec^2(x^2 + 1).dx\) is Tan(x 2 + 1) + C.

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Practice Questions on Integration by Substitution

Faqs on integration by substitution.

Integration by substitution is an important method of integration, which is used when a function to be integrated, is either a complex function or if the direct integration of the function is not feasible.

How Do You Integrate by Substitution?

Integration by substitution can be performed through a set of sequential steps. First, choose a new variable for the part of the function to be substituted. Secondly, determine the value of differentiation of x, ie dx from this new variable substitution. The third steps involve the process of integration involving this new variable. Finally, substitute back the initial variable to obtain the final answer.

How Do You Know When To Use Integration by Substitution?

The process of integration by substitution is used if the given function to be integrated has one of the following three characteristics.

• The given function has a sub-function.
• The function to be integrated is a complex number -based function.
• The direct integration of the function is not possible.

What Is the Formula For Integration by Substitution?

There is no defined formula for integration by substitution. Based on the given function, the part of the function which is to be substituted is substituted with a new variable.

How Do You Use Integration By Substitution for Trigonometric Formulas?

The integration by substitution is used for trigonometric functions , similar to any other function. Here the trigonometric function is replaced with a new variable, to transform it into an algebraic expression, which is easy to integrate.

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5.3: Integration by Substitution

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• Page ID 4322

• Matthew Boelkins, David Austin & Steven Schlicker
• Grand Valley State University via ScholarWorks @Grand Valley State University

Learning Objectives

In this section, we strive to understand the ideas generated by the following important questions:

• How can we begin to find algebraic formulas for antiderivatives of more complicated algebraic functions?
• What is an indefinite integral and how is its notation used in discussing antiderivatives?
• How does the technique of u-substitution work to help us evaluate certain indefinite integrals, and how does this process rely on identifying function-derivative pairs?

In Section 4.4, we learned the key role that antiderivatives play in the process of evaluating definite integrals exactly. In particular, the Fundamental Theorem of Calculus tells us that if \(F\) is any antiderivative of \(f\), then

\[\int^b_a f (x) dx = F(b) − F(a).\]

Furthermore, we realized that each elementary derivative rule developed in Chapter 2 leads to a corresponding elementary antiderivative, as summarized in Table 4.1. Thus, if we wish to evaluate an integral such as

\[\int_0^1 x 3 − √ x + 5 x dx \label{eq5.3}\]

it is straightforward to do so, since we can easily antidifferentiate f (x) = x 3 − √ x + 5 x . In particular, since a function \(F\) whose derivative is \(f\) is given by

\[F(x) = 1 4 x 4− 2 3 x 3/2+ 1 \ln (5) 5 x \]

the Fundamental Theorem of Calculus tells us that

\[\int_0^1 x 3 − √ x + 5 x dx = 1 4 x 4 − 2 3 x 3/2 + 1 \ln (5) 5 x 1 0 = 1 4 (1) 4 − 2 3 (1) 3/2 + 1 \ln (5) 5 1 ! − 0 − 0 + 1 \ln (5) 5 0 ! = − 5 12 + 4 \ln (5) .\]

Because an algebraic formula for an antiderivative of f enables us to evaluate the definite integral

\[\int ^b_a f (x) dx\]

exactly, we see that we have a natural interest in being able to find such algebraic antiderivatives. Note that we emphasize algebraic antiderivatives, as opposed to any antiderivative, since we know by the Second Fundamental Theorem of Calculus that

\[G(x) = \int^x_a f (t) dt\]

is indeed an antiderivative of the given function \(f\), but one that still involves a definite integral. One of our main goals in this section and the one following is to develop understanding, in select circumstances, of how to “undo” the process of differentiation in order to find an algebraic antiderivative for a given function.

Preview Activity \(\PageIndex{1}\)

In Section 2.5, we learned the Chain Rule and how it can be applied to find the derivative of a composite function. In particular, if \(u\) is a differentiable function of \(x\), and \(f\) is a differentiable function of \(u(x)\), then

\[\dfrac{d}{dx} [ f (u(x))] = f' (u(x)) · u 0 (x).\]

In words, we say that the derivative of a composite function c(x) = f (u(x)), where f is considered the “outer” function and u the “inner” function, is “the derivative of the outer function, evaluated at the inner function, times the derivative of the inner function.”

(a) For each of the following functions, use the Chain Rule to find the function’s derivative. Be sure to label each derivative by name (e.g., the derivative of g(x) should be labeled g' (x)).

• g(x) = e 3x
• h(x) = sin(5x + 1)
• p(x) = arctan(2x)
• q(x) = (2 − 7x) 4
• r(x) = 3 4−11x

(b) For each of the following functions, use your work in (a) to help you determine the general antiderivative3 of the function. Label each antiderivative by name (e.g., the antiderivative of m should be called M). In addition, check your work by computing the derivative of each proposed antiderivative.

• m(x) = e 3x
• n(x) = cos(5x + 1)
• s(x) = 1 1+4x 2 3Recall that the general antiderivative of a function includes “+C” to reflect the entire family of functions that share the same derivative.
• v(x) = (2 − 7x) 3 v. w(x) = 3 4−11x

(c) Based on your experience in parts (a) and (b), conjecture an antiderivative for each of the following functions. Test your conjectures by computing the derivative of each proposed antiderivative.

• a(x) = cos(πx)
• b(x) = (4x + 7) 11
• c(x) = xex 2 ./

Reversing the Chain Rule: First Steps

In Preview Activity \(\PageIndex{1}\), we saw that it is usually straightforward to antidifferentiate a function of the form h(x) = f (u(x)), whenever f is a familiar function whose antiderivative is known and u(x) is a linear function. For example, if we consider h(x) = (5x − 3) 6 , in this context the outer function f is f (u) = u 6 , while the inner function is u(x) = 5x − 3. Since the antiderivative of f is

\[F(u) = 1 7 u 7 + C,\]

we see that the antiderivative of h is

\[H(x) = 1 7 (5x − 3) 7 · 1 5 + C = 1 35 (5x − 3) 7 + C.\]

The inclusion of the constant 1 5 is essential precisely because the derivative of the inner function is u 0 (x) = 5. Indeed, if we now compute \(H' (x)\), we find by the Chain Rule (and Constant Multiple Rule) that

\[H ' (x) = 1 35 · 7(5x − 3) 6 · 5 = (5x − 3) 6 = h(x), \]

and thus H is indeed the general antiderivative of \(h\). Hence, in the special case where the outer function is familiar and the inner function is linear, we can antidifferentiate composite functions according to the following rule. If \(h(x) = f (ax + b)\) and \(F\) is a known algebraic antiderivative of f , then the general antiderivative of h is given by

\[H(x) = 1 a F(ax + b) + C. \]

When discussing antiderivatives, it is often useful to have shorthand notation that indicates the instruction to find an antiderivative. Thus, in a similar way to how the notation d dx [ f (x)] represents the derivative of f (x) with respect to x, we use the notation of the indefinite integral, Z f (x) dx to represent the general antiderivative of \(f\) with respect to x. For instance, returning to the earlier example with h(x) = (5x − 3) 6 above, we can rephrase the relationship between h and its antiderivative H through the notation

\[\int (5x − 3) 6 dx = 1 35 (5x − 6) 7 + C.\]

When we find an antiderivative, we will often say that we evaluate an indefinite integral; said differently, the instruction to evaluate an indefinite integral means to find the general antiderivative. Just as the notation d dx [] means “find the derivative with respect to x of ,” the notation R dx means “find a function of x whose derivative is .”

Activity \(\PageIndex{2}\)

Evaluate each of the following indefinite integrals. Check each antiderivative that you find by differentiating.

• \(\displaystyle \int sin(8 − 3x) dx\)
• \(\displaystyle \int sec2 (4x) dx\)
• \(\displaystyle \int 1 11x−9 dx\)
• \(\displaystyle \int csc(2x + 1) cot(2x + 1) dx\)
• \(\displaystyle \int 1 √ 1−16x 2 dx\)
• \(\displaystyle \int 5 −x dx\)

Reversing the Chain Rule

u-substitution Of course, a natural question arises from our recent work: what happens when the inner function is not a linear function? For example, can we find antiderivatives of such functions as \(g(x) = xex^2\) and \(h(x) = e x^2\)? It is important to explicitly remember that differentiation and antidifferentiation are essentially inverse processes; that they are not quite inverse processes is due to the +C that arises when antidifferentiating. This close relationship enables us to take any known derivative rule and translate it to a corresponding rule for an indefinite integral. For example, since d dx x 5 = 5x 4 , we can equivalently write Z 5x 4 dx = x 5 + C. Recall that the Chain Rule states that

\[\dfrac{d}{dx} [ f (g(x))] = f' (g(x)) · g' (x).\]

Restating this relationship in terms of an indefinite integral,

\[\int f' (g(x))g' (x) dx = f (g(x)) + C. \label{5.5} \]

Hence, Equation \ref{5.5} tells us that if we can take a given function and view its algebraic structure as f' (g(x))g' (x) for some appropriate choices of f and g, then we can antidifferentiate the function by reversing the Chain Rule. It is especially notable that both g(x) and g' (x) appear in the form of f' (g(x))g' (x); we will sometimes say that we seek to identify a function-derivative pair when trying to apply the rule in Equation \ref{5.5}. In the situation where we can identify a function-derivative pair, we will introduce a new variable u to represent the function g(x). Observing that with u = g(x), it follows in Leibniz notation that du dx = g' (x), so that in terms of differentials4 , du = g' (x) dx. Now converting the indefinite integral of interest to a new one in terms of u, we have

\[\int f' (g(x))g' (x) dx = \int f' (u) du.\]

Provided that f' is an elementary function whose antiderivative is known, we can now 4 If we recall from the definition of the derivative that du dx ≈ 4u 4x and use the fact that du dx = g' (x), then we see that g' (x) ≈ 4u 4x . Solving for 4u, 4u ≈ g' (x)4x. It is this last relationship that, when expressed in “differential” notation enables us to write

\[du = g' (x) dx\]

in the change of variable formula.

easily evaluate the indefinite integral in u, and then go on to determine the desired overall antiderivative of f' (g(x))g' (x). We call this process u-substitution. To see u-substitution at work, we consider the following example.

Example \(\PageIndex{1}\):

Evaluate the indefinite integral

\[\int x^3 · sin(7x 4 + 3) dx\]

and check the result by differentiating.

We can make two key algebraic observations regarding the integrand, x 3 · sin(7x 4 + 3). First, sin(7x 4 + 3) is a composite function; as such, we know we’ll need a more sophisticated approach to antidifferentiating.

Second, x 3 is almost the derivative of (7x 4 + 3); the only issue is a missing constant. Thus, x 3 and (7x 4 + 3) are nearly a function-derivative pair. Furthermore, we know the antiderivative of f (u) = sin(u). The combination of these observations suggests that we can evaluate the given indefinite integral by reversing the chain rule through u-substitution. Letting u represent the inner function of the composite function sin(7x 4 + 3), we have u = 7x 4 + 3, and thus du dx = 28x 3 . In differential notation, it follows that

\[du = 28x 3 dx,\]

\[x 3 dx = 1 28 du.\]

We make this last observation because the original indefinite integral may now be written Z sin(7x 4 + 3) · x 3 dx, and so by substituting the expressions in u for x (specifically u for 7x 4 + 3 and 1 28 du for x 3 dx), it follows that

\[\int sin(7x 4 + 3) · x 3 dx = Z sin(u) · 1 28 du.\]

Now we may evaluate the original integral by first evaluating the easier integral in u, followed by replacing u by the expression 7x 4 + 3. Doing so, we find

\[\int sin(7x 4 + 3) · x 3 dx = Z sin(u) · 1 28 du = 1 28 Z sin(u) du = 1 28 (− cos(u)) + C = − 1 28 cos(7x 4 + 3) + C.\]

To check our work, we observe by the Chain Rule that

\[\dfrac{d}{dx} − 1 28 cos(7x 4 + 3) + C = − 1 28 · (−1)sin(7x 4 + 3) · 28x 3 = sin(7x 4 + 3) · x 3\]

which is indeed the original integrand. An essential observation about our work in Example 5.2 is that the u-substitution only worked because the function multiplying sin(7x 4 + 3) was x 3 . If instead that function was x 2 or x 4 , the substitution process may not (and likely would not) have worked. This is one of the primary challenges of antidifferentiation: slight changes in the integrand make tremendous differences. For instance, we can use u-substitution with u = x 2 and du = 2xdx to find that

\[\int xex 2 dx = Z e u · 1 2 du = 1 2 Z e u du = 1 2 e u + C = 1 2 e x 2 + C.\]

If, however, we consider the similar indefinite integral

\[\int e x 2 dx,\]

the missing x to multiply e x 2 makes the u-substitution u = x 2 no longer possible. Hence, part of the lesson of u-substitution is just how specialized the process is: it only applies to situations where, up to a missing constant, the integrand that is present is the result of applying the Chain Rule to a different, related function.

Activity \(\PageIndex{3}\)

Evaluate each of the following indefinite integrals by using these steps:

• Find two functions within the integrand that form (up to a possible missing constant) a function-derivative pair;
• Make a substitution and convert the integral to one involving u and du;
• Evaluate the new integral in u;
• Convert the resulting function of u back to a function of x by using your earlier substitution;
• Check your work by differentiating the function of x. You should come up with the integrand originally given.
• \(displaystyle \int x 2 5x 3 + 1 dx\)
• \(displaystyle \int e x sin(e x ) dx\)
• \(displaystyle \int cos( √ x) √ x dx C\)

Evaluating Definite Integrals via u-substitution

We have just introduced u-substitution as a means to evaluate indefinite integrals of functions that can be written, up to a constant multiple, in the form f (g(x))g' (x). This same technique can be used to evaluate definite integrals involving such functions, though we need to be careful with the corresponding limits of integration. Consider, for instance, the definite integral

\[\int^5_2 xex 2 dx.\]

Whenever we write a definite integral, it is implicit that the limits of integration correspond to the variable of integration. To be more explicit, observe that

\[\int^5_2 xex 2 dx = Z x=5 x=2 xex 2 dx.\]

When we execute a u-substitution, we change the variable of integration; it is essential to note that this also changes the limits of integration. For instance, with the substitution u = x 2 and du = 2x dx, it also follows that when x = 2, u = 2 2 = 4, and when x = 5, u = 5 2 = 25. Thus, under the change of variables of u-substitution, we now have

\[\int^{x=5}_{x=2} xex 2 dx = Z u=25 u=4 e u · 1 2 du = 1 2 e u u=25 u=4 = 1 2 e 25 − 1 2 e 4 \]

Alternatively, we could consider the related indefinite integral R xex 2 dx, find the antiderivative 1 2 e x 2 through u-substitution, and then evaluate the original definite integral.

From that perspective, we’d have

\[\int^5_2 xex 2 dx = 1 2 e x 2 5 2 = 1 2 e 25 − 1 2 e 4 \]

which is, of course, the same result.

Activity \(\PageIndex{1}\)

Evaluate each of the following definite integrals exactly through an appropriate usubstitution.

• Z 2 1 x 1 + 4x 2 dx
• \int_0^1 e −x (2e −x + 3) 9 dx
• Z 4/π 2/π cos 1 x x 2 dx C

In this section, we encountered the following important ideas:

• To begin to find algebraic formulas for antiderivatives of more complicated algebraic functions, we need to think carefully about how we can reverse known differentiation rules. To that end, it is essential that we understand and recall known derivatives of basic functions, as well as the standard derivative rules.
• The indefinite integral provides notation for antiderivatives. When we write “R f (x) dx,” we mean “the general antiderivative of f .” In particular, if we have functions f and F such that f' = f , the following two statements say the exact thing: d dx [F(x)] = f (x) and Z f (x) dx = F(x) + C. That is, f is the derivative of F, and F is an antiderivative of f.
• The technique of R u-substitution helps us evaluate indefinite integrals of the form f (g(x))g' (x) dx through the substitutions u = g(x) and du = g' (x) dx, so that Z f (g(x))g' (x) dx = Z f (u) du. A key part of choosing the expression in x to be represented by u is the identification of a function-derivative pair. To do so, we often look for an “inner” function g(x) that is part of a composite function, while investigating whether g' (x) (or a constant multiple of g' (x)) is present as a multiplying factor of the integrand.