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Calculus 1 Problems & Solutions – Chapter 6 – Section 6.5.1.1

6.5.1.1
The Method Of Substitution

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Review

In the previous section, Section 6.4, 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 2.6 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 e^u 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: