Integration by Change of Variables or Substitution

Section 7.4 Integration by Change of Variables or Substitution

At the end of the last section, we warned that the symbolic integration techniques we have developed only work for problems that exactly fit our formulas. When we tried integrating an exponential function where the exponent was a constant times

t,

we had to change the base to get a function with only

t

in the exponent. We want to develop one more technique of integration, called change of variables or substitution, to handle integrals that are pretty close to our stated rules. This technique is often called

u

-substitution and is related to the chain rule for differentiation.

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Subsection 7.4.1 Change of variables for indefinite integrals

We start by exploring some examples where we can get the desired result by the guess and check technique.

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Example 7.4.1. Power of a linear by guess and check.

Find

\int (3 x + 5)^{7} d x .

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Solution.

We could do this problem by rewriting the integrand as an explicit seventh degree polynomial and then using the power and sum rules, but that is too much work. Instead, I will notice the integrand looks almost like a power, and thus guess an answer of

\frac{1}{8} (3 x + 5)^{8} + C .

I then check by differentiating. Using the chain rule,

\frac{d}{d x} (\frac{1}{8} (3 x + 5)^{8} + C) = \frac{1}{8} * 8 (3 x + 5)^{8 - 1} * 3 = 3 (3 x + 5)^{7} .

Thus our guess was off by a factor of 3 and the correct antiderivative is

\frac{1}{3} * \frac{1}{8} (3 x + 5)^{8} + C = \frac{1}{24} (3 x + 5)^{8} + C .

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We can easily use the same trick to produce a rule for powers of a linear polynomial.

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Example 7.4.2. Power of a generic linear by guess and check.

Find

\int (a x + b)^{n} d x .

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Solution.

As we did in the previous example, we first guess the antiderivative to be

\frac{1}{n + 1} (a x + b)^{n + 1} + C .

We then take the derivative of that expression and obtain

a (a x + b)^{n} .

This misses our integrand by a factor of

a .

We adjust by that factor and find the antiderivative is

\frac{1}{a} \frac{1}{(n + 1)} (a x + b)^{n + 1} + C .

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We can use the same trick to produce a rule for functions that are the exponential of a linear function.

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Example 7.4.3. Antidifferentiation of an exponential function by guess and check.

Find

\int e^{a x + b} d x .

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Solution.

As we did in the last example, our first guess uses the basic rule without worrying about the linear term, so we guess

e^{a x + b} + C .

We then take the derivative of that expression and obtain

a e^{a x + b} .

This misses our integrand by a factor of

a .

We adjust by that factor and find the antiderivative is

\frac{1}{a} e^{a x + b} + C .

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We run into a problem if we try to extend this method with quadratic terms. If we start with

(x^{2} + 5)^{3}

and guess an antiderivative of

\frac{1}{4} (x^{2} + 5)^{4},

when we differentiate we get

(x^{2} + 5)^{3} 2 x

and are off by a factor of

8 x .

However, when we divide by that factor to get

\frac{(x^{2} + 5)^{4}}{8 x}

as a proposed antiderivative, and then differentiate again, we get

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\frac{4 * 2 x (x^{2} + 5)^{3} * 8 x - (x^{2} + 5)^{4} * 8}{8 x},

which is not what we want. The key is to start by recalling the chain rule:

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\frac{d}{d x} (f (g (x))) = f^{'} (g (x)) g^{'} (x) .

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We want to use the same rule with a different notation, using implicit differentiation and a new variable

u :

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\frac{d}{d x} (f (u)) = f^{'} (u) \frac{d u}{d x} .

By the fundamental theorem of calculus, we can convert this to an integration formula:

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\int f^{'} (u) \frac{d u}{d x} d x = f (u) + C .

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We will generally simplify

\frac{d u}{d x} d x

d u,

so our substitution rule is

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\int f^{'} (u) d u = f (u) + C .

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Let us rework some earlier examples with this method and then illustrate the method with a more difficult problem.

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Example 7.4.4. Power of linear example redone with change of variables.

Find

\int (3 x + 5)^{7} d x .

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Solution.

The obvious candidate for

u

3 x + 5 .

Then

d u = 3 d x .

Thus

\begin{aligned} \int (3 x + 5)^{7} d x & = \frac{1}{3} \int (3 x + 5)^{7} (3 d x) (Make u and d u explicit.) \\ = \frac{1}{3} \int (u)^{7} d u (Do the substitution.) \\ = \frac{1}{3 * 8} (u)^{8} + C (Find the integral in terms of u .) \\ = \frac{1}{24} (3 x + 5)^{8} + C (Substitute back.) . \end{aligned}

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This is easy to generalize for a power of a linear term.

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Example 7.4.5. Power of generic linear example redone with change of variables.

Find

\int (a x + b)^{n} d x .

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Solution.

The obvious candidate for

u

a x + b .

Then

d u = a d x .

Hence

\begin{aligned} \int (a x + b)^{n} d x & = \frac{1}{a} \int (a x + b)^{n} (a d x) (Make u and d u explicit.) \\ = \frac{1}{a} \int (u)^{n} d u (Do the substitution.) \\ = \frac{1}{a * (n + 1)} (u)^{n + 1} + C (Find the integral in terms of u .) \\ = \frac{1}{a * (n + 1)} (a x + b)^{n + 1} + C (Substitute back.) . \end{aligned}

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To use this method with

u

replacing something more complicated than a linear term, we need to have

d u

available, with the possible addition of multiplying by a scalar constant.

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Example 7.4.6. Power of cubic function with change of variables.

Find

\int (2 x^{3} + 11)^{7} x^{2} d x .

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Solution.

The obvious candidate for

u

2 x^{3} + 11,

since it is an expression taken to a large power. Then

d u = 6 x^{2} d x .

Thus

\begin{aligned} \int (2 x^{3} + 11)^{7} x^{2} d x & = \frac{1}{6} \int (2 x^{3} + 11)^{7} (6 x^{2} d x) (Make u and d u explicit.) \\ = \frac{1}{6} \int (u)^{7} d u (Do the substitution.) \\ = \frac{1}{6 * 8} (u)^{8} + C (Find the integral in terms of u .) \\ = \frac{1}{48} (2 x^{3} + 11)^{8} + C (Substitute back.) . \end{aligned}

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By convention,

u

is often used the new variable used with this change of variables technique, so the technique is often called

u

-substitution.

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Subsection 7.4.2 Change of variables for definite integrals

In the definite integral, we understand that

a

and

b

are the

x

-values of the ends of the integral. We could be more explicit and write

x = a

and

x = b .

The last step in solving a definite integral is to substitute the endpoints back into the antiderivative we have found. We can either change the variables for the endpoints as well, or we can convert the antiderivative back to the original variables before substituting. Consider the following example.

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Example 7.4.7. A definite integral with change of variables.

Evaluate

\int_{1}^{3} e^{2 x + 5} d x .

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Solution 1.

Solution 1: Convert everything to

u .

The obvious candidate for

u

2 x + 5 .

Then

d u = 2 d x .

For the lower endpoint,

x = 1

becomes

u = 2 (1) + 5 = 7 .

For the upper endpoint

x = 3

becomes

u = 2 (3) + 5 = 11 .

Substituting,

\begin{aligned} \int_{1}^{3} e^{2 x + 5} d x & = (\frac{1}{2}) (2) \int_{1}^{3} e^{2 x + 5} d x) (Add needed factors.) \\ = \frac{1}{2} \int_{1}^{3} e^{2 x + 5} (2 d x) (Make u and d u explicit.) \\ = \frac{1}{2} \int_{u = 7}^{u = 11} e^{u} d u (Do the substitution.) \\ = {\frac{1}{2} e^{u} |}_{7}^{11} (Find the antiderivative.) \\ = \frac{1}{2} e^{11} - \frac{1}{2} e^{7} . (Evaluate.) . \end{aligned}

Solution 2.

Solution 2: Keeping, but labeling, the endpoints. We have the same

u

and

d u,

but do not convert the endpoints. To reduce confusion we make sure to label the variable when we are using both

x

and

u .

Thus,

\begin{aligned} \int_{1}^{3} e^{2 x + 5} d x & = \frac{1}{2} \int_{1}^{3} e^{2 x + 5} (2 d x) (Make u and d u explicit.) \\ = \frac{1}{2} \int_{x = 1}^{x = 3} e^{u} d u (Do the substitution.) \\ = {\frac{1}{2} e^{u} |}_{x = 1}^{x = 3} (Find the antiderivative.) \\ = {\frac{1}{2} e^{2 x + 3} |}_{x = 1}^{x = 3} (Convert back.) \\ = \frac{1}{2} e^{11} - \frac{1}{2} e^{7} . (Evaluate.) . \end{aligned}

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It should be noted that when we change variables we may find ourselves looking at an integral from

a

b

where the

b < a .

We do not change the order of the endpoints.

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Example 7.4.8. A second definite integral with change of variables.

Evaluate

\int_{- 2}^{1} x e^{(x^{2})} d x

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Solution.

(Convert everything to

u .

) The obvious candidate for

u

x^{2} .

Then

d u = 2 x d x .

For the lower endpoint,

x = - 2

becomes

u = (- 2)^{2} = 4 .

For the upper endpoint

x = 1

becomes

u = 1^{2} = 1 .

Substituting,

\begin{aligned} \int_{- 2}^{1} x e^{(x^{2})} d x & = \frac{1}{2} \int_{- 2}^{1} e^{(x^{2})} (2 x d x) (Make u and d u explicit.) \\ = \frac{1}{2} \int_{4}^{1} e^{u} d u (Do the substitution.) \\ = {\frac{1}{2} e^{u} |}_{4}^{1} (Find the antiderivative.) \\ = \frac{1}{2} (1 - e^{4}) . & (Evaluate.) . \end{aligned}

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Reading Questions 7.4.3 Reading Check

1. Reading check, Integration by Change of Variable or Substitution.

This question checks your reading comprehension of the material is section 7.4, Integration by Change of Variable or Substitution, of Business Calculus with Excel. Based on your reading, select all statements that are correct. There may be more than one correct answer. The statements may appear in what seems to be a random order.

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For the method of substitution, we want to make both $u$ and $d u$ explicit.
Substitution for integration is related to the chain rule for integration.
We can pick any expression for $u$ when doing substitution.
We can also do change of variables with definite integrals.
One of the methods in this section is called Guess and Check.
None of the above

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Exercises 7.4.4 Exercises: Integration by Change of Variables or Substitution Problems

Exercise Group.

Evaluate the following integrals. In each case identify the term that will be treated as

\int_{0}^{2} x^{2} e^{(1 - 0.2 x^{3})} d x .

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19.

Find an antiderivative

F (x)

for

f (x) = x^{2} (x^{3} + 7)^{3}

such that

F (0) = 5 .

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Solution.

Let

u = x^{3} + 7 .

Then

d u = 6 x^{2} d x .

\begin{aligned} \int (x^{2}) (x^{3} + 7)^{3} d x & = \frac{1}{6} \int u^{3} d u \\ = \frac{u^{4}}{24} + c \\ = \frac{(x^{3} + 7)^{4}}{24} + c \\ F (0) & = 5 = \frac{(7)^{4}}{24} + c \\ c & = 5 - \frac{(7)^{4}}{24} 4 \\ F (x) & = \frac{(x^{3} + 7)^{4}}{24} + 5 - \frac{(7)^{4}}{24} . \end{aligned}

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20.

Find an antiderivative

F (x)

for

f (x) = (4 x^{3} + 5) \exp (x^{4} + 5 x - 9)

such that

F (0) = 2 .

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21.

An investment stream pays out at a rate of $10,000 per year. In computing present value, I assume an investment return rate of 5% compounded continuously. What is the present value of the first 10 years of the payout?

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Solution.

\begin{aligned} Value & = \int_{0}^{10} 10000 e^{- 0.05 t} d t \\ = {\frac{- 10000 e^{- 0.05 t}}{0.05} |}_{0}^{10} \\ = \frac{- 10000}{0.05} (e^{- 0.5} - 1) \approx 78693.87 . \end{aligned}

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22.

My gas well is returning a payout of $10,000. The well output is expected to decay exponentially with half as much output in 7 years. How much do I make over the next 10 years?

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23.

The sales rate on a book is

s (t) = 1000 t \exp (- t^{2} / 4),

with time in years.

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What are the total sales over 10 years?
When does the sales rate drop to 10?
What is the maximum sales rate?

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Solution.

Let $u = - \frac{t^{2}}{4} .$ Then $d u = - \frac{t}{2} d t .$

$\begin{aligned} S a l e s & = \int_{0}^{10} 1000 t e^{- t^{2} / 4} d t = \int_{0}^{- 25} - 2000 e^{u} d u \\ = {- 2000 e^{u}) |}_{0}^{- 25} \\ = 2000 ((1 - e^{- 25}) \approx 2000 . \end{aligned}$
$\begin{aligned} 10 & = 1000 t e^{- t^{2} / 4} \\ t & \approx 5.0028 . \end{aligned}$
At $t = 1.454,$ sales are approximately 857.4

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24.

The marginal profit on an item is

M p (q) = \frac{(100 q)}{(q^{2} + 1)} - 2,

measured in thousands of dollars per thousands of units.

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How much should I produce to maximize profits?
What is my profit function if my start up cost is $60,000?
What is the maximum profit?

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