Calculus Of One Real Variable By Pheng Kim Ving
Chapter 10: Techniques Of Integration Section 10.2: The Method Of Substitution

 

10.2
The Method Of Substitution

 

 

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In the previous section, Section 10.1, we discussed integration by inspection. There are lots of integrals that are hard to evaluate
or can't be evaluated by inspection but that can each be found by an integration technique or method. This section and several
subsequent ones present some integration techniques.

 

1. The Method Of Substitution

 

Example 1.1

 

Calculate this indefinite integral:

 

 

Solution
Let u = x2 1. Then du = 2x dx. So:

 


EOS

 

We have 2x(x2 1)10 dx = (x2 1)10 (2x dx) = u10 du.

 

Recall that du is the differential of u (see Section 4.3 Definitions 2.1): du = u'(x) dx = 2x dx. In the integral, we spot
the factors 2x and x2 1, and we know that 2x is the derivative of x2 1, so the differential d(x2 1) of x2 1 has the
factor 2x in it: d(x2 1) = 2x dx. Thus we substitute u = x2 1, calculate du, and transform the integral in x into one in
u. The integral in u is ready for an integration formula to be applied to it. After finding the integral in u, we have to return
to the original variable, x in this case, in our answer.

 

We substitute u = x2 1. The technique used is therefore called the method of substitution. If we want to check to
see that our answer is correct, then we just differentiate it:

 

 

Example 1.2

 

Compute:

 

 

Solution

Let u = x 1, so that du = dx. Then:

 

EOS

 

Here d(x 1) = 1 . dx = dx. So we substitute u = x 1.

 

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2. The Method Of Substitution And The Chain Rule

 

 

{2.1} Example 1.1.

 

The method of substitution is valid because it's derived from the chain rule.

 

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3. Difference By A Constant Factor

 

Example 3.1

 

Evaluate:

 

 

Solution 1
Let u = x2 1. Then du = 2x dx, so that x dx = (1/2) du. Thus:

 


EOS

 

Solution 2
Let u = x2 1. Then du = 2x dx. Thus:

 

EOS

 

Solution 3

EOS

 

 

In this example we have x instead of 2x, so 1 term, x, isn't exactly the derivative, 2x, of the other, x2 1, but the
difference is only by a constant factor, namely 2. The difference by a constant factor can be removed easily without
making the integral more complicated, as shown in each of the solutions.

 

The bypassing of explicit substitution as done in Solution 3 can be applied to any integral formula.

 

Example 3.2

 

Find:

 

 

Solution

EOS

 

We think of x3 as u and write the integrand in the form eu du so that we can apply the integral formula for exponential
functions. In general this can easily be done when the substitution and the formation of du in terms of x and dx can
easily be performed mentally.

 

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4. Integrals Involving Radicals

 

Example 4.1

 

Calculate:

 

 

Solution 1
Let u = 5 x2. Then du = 2x dx, so that x dx = (1/2) du. Thus:

 


EOS

 

We can eliminate the radical, as shown in Solution 2 below.

 

Solution 2

EOS

 

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5. Handling The Definite Integrals

 

Example 5.1

 

Compute this definite integral:

 

 

Solution 1
Let u = x3 + 1. Then du = 3x2 dx, so that x2 dx = (1/3) du. Thus:

 


EOS

 

We must write the limits of integration as x = 0 and x = 2 whenever the variable of integration is u, not just as 0 and 2,
because it's x, not u, that goes from 0 to 2. Here we keep the x-limits of integration, so we must return to x before
substituting in the limits. Another approach is shown in Solution 2 below.

 

Solution 2
Let u = x3 + 1. Then du = 3x2 dx, so that x2 dx = (1/3) du. When x = 0 we have u = 03 + 1 = 1, and when x = 2 we
have u = 23 + 1 = 9. Thus:

 


EOS

 

Here we transform the x-limits into the u-limits, and we evaluate the integral as soon as we've found it in terms of u
using these u-limits.

 

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6. When The Method Of Substitution Is Likely To Work

 

 

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Problems & Solutions

 

1. Calculate the following indefinite integrals.

 

 

Solution

 

a. Let u = 3x4 + 5. Then du = 12x3 dx, so x3 dx = (1/12) du. Thus:

 

 

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2. Compute the following indefinite integrals.

 

 

Solution

 

a. Let u = s2. Then du = 2s ds, so s ds = (1/2) du. Thus:

 

 

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3. Show that:

 

 

Solution

 

We have 10 + 6x + x2 = 1 + (x + 3)2. Let u = x + 3. So du = dx. Thus:

 

 

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4. Evaluate the definite integral:

 

 

handling the limits of integration in 2 ways:
a. Keep the x-limits.
b. Change to the limits for the substitution variable.

 

Solution

 

 

 

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5. Find the area of the plane region bounded by the graph of y = x/(x4 + 16), the x-axis, the y-axis, and the vertical line
x = 2.

 

Solution

 

 

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