INTEGRALS

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5. INTEGRALS. INTEGRALS. 5.5 The Substitution Rule. In this section, we will learn: To substitute a new variable in place of an existing expression in a function, making integration easier. INTRODUCTION. Equation 1.

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5

INTEGRALS

INTEGRALS

5.5The Substitution Rule

• In this section, we will learn:
• To substitute a new variable in place of an existing
• expression in a function, making integration easier.
INTRODUCTION

Equation 1

• Antidifferentiation formulas don’t tell us how to evaluate integrals such as
INTRODUCTION
• To find this integral, we use the problem-solving strategy of introducing something extra.
• The ‘something extra’ is a new variable.
• We change from the variable x to a newvariableu.
INTRODUCTION
• Suppose we let u be the quantity under the root sign in Equation 1, u = 1 + x2.
• Then, the differential of u is du = 2xdx.
INTRODUCTION

Equation 2

• So, formally, without justifying our calculation, we could write:
INTRODUCTION
• However, now we can check that we have the correct answer by using the Chain Ruleto differentiate the final function of Equation 2:
INTRODUCTION

Equation 3

• Observe that, if F’ = f, then
• because, by the Chain Rule,
INTRODUCTION
• If we make the ‘change of variable’ or ‘substitution’ u = g(x), from Equation 3, we have:
SUBSTITUTION RULE

Equation 4

• If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then ∫f(g(x))g’(x) dx = ∫f(u) du
SUBSTITUTION RULE
• Notice that the Substitution Rule for integration was proved using the Chain Rule for differentiation.
• Notice also that, if u = g(x), then du = g’(x) dx.
• So,a way to remember the Substitution Rule is to think of dx and du in Equation 4 as differentials.
SUBSTITUTION RULE

Example 1

• Find ∫x3 cos(x4 + 2) dx
• We make the substitutionu = x4 + 2.
• This is because its differential is du = 4x3 dx, which, apart from the constant factor 4, occurs in the integral.
Solution:

Example 1

• Thus, using x3dx = du/4 and the Substitution Rule, we have:
SUBSTITUTION RULE
• Finding the right substitution is a bit of an art.
• It’s not unusual to guess wrong.
• If your first guess doesn’t work, try another substitution.
SUBSTITUTION RULE

E. g. 2—Solution 1

• Evaluate
• Let u = 2x + 1.
• Then, du = 2 dx.
• So, dx = du/2.
Solution1:

E. g. 2—Solution 1

• Thus, the rule gives:
Solution2:

E. g. 2—Solution 2

• Another possible substitution is
• Then,
• So,
• Alternatively, observe that u2 = 2x + 1.
• So, 2udu = 2 dx.
Solution2:

E. g. 2—Solution 2

• Thus,
SUBSTITUTION RULE

Example 3

• Find
• Let u = 1 – 4x2.
• Then, du = -8x dx.
• So, x dx = -1/8 du and
Solution:
• Here, we have used a computer to graph both the integrand and its indefinite integral
• We take the case C = 0.

Figure 5.5.1, p. 335

Solution:
• Notice that g(x):
• Decreases when f(x) is negative
• Increases when f(x) is positive
• Has its minimum value when f(x) = 0

Figure 5.5.1, p. 335

SUBSTITUTION RULE

Example 4

• Calculate ∫ cos 5xdx
• If we let u = 5x, then du = 5 dx.
• So, dx = 1/5 du.
• Therefore,
SUBSTITUTION RULE

Example 5

• Find
• An appropriate substitution becomes more obvious if we factor x5 as x4 . x.
• Let u = 1 + x2.
• Then, du = 2x dx.
• So, x dx = du/2.
Solution:

Example 5

• Also, x2 = u – 1; so, x4 = (u – 1)2:
DEFINITE INTEGRALS
• Evaluate the indefinite integral first and then use the FTC for definite integral.
• For instance, using the result of Example 2, we have:
SUB. RULE FOR DEF. INTEGRALS

Equation 5

• If g’ is continuous on [a, b] and fis continuous on the range of u = g(x), then
SUB. RULE FOR DEF. INTEGRALS

Proof

• Let F be an antiderivative of f.
• Then, by Equation 3, F(g(x)) is an antiderivative of f(g(x))g’(x).
• So, by Part 2 of the FTC (FTC2), we have:
SUB. RULE FOR DEF. INTEGRALS

Proof

• However, applying the FTC2 a second time, we also have:
SUB. RULE FOR DEF. INTEGRALS

Example 6

• Evaluate using Equation 5.
• Using the substitution from Solution 1 of Example 2, we have: u = 2x + 1 and dx = du/2

Figure 5.5.2a, p. 337

Solution:

Example 6

• To find the new limits of integration, we note that:
• When x = 0, u = 2(0) + 1 = 1
• When x = 4, u = 2(4) + 1 = 9

Figure 5.5.2b, p. 337

Solution:

Example 6

• Thus,
SUB. RULE FOR DEF. INTEGRALS

Example 7

• Evaluate
• Let u = 3 – 5x.
• Then, du = –5 dx, so dx =–du/5.
• When x = 1, u = –2, and when x = 2, u = –7.
Solution:

Example 7

• Thus,
SYMMETRY
• The next theorem uses the Substitution Rule for Definite Integrals to simplify the calculation of integrals of functions that possess symmetry properties.
INTEGS. OF SYMM. FUNCTIONS

Theorem 6

• Suppose f is continuous on [–a , a].
• If f is even, [f(–x) = f(x)], then
• If f is odd, [f(-x) = -f(x)], then
INTEGS. OF SYMM. FUNCTIONS

Proof—Equation 7

• We split the integral in two:
INTEGS. OF SYMM. FUNCTIONS

Proof

• In the first integral in the second part, we make the substitution u = –x .
• Then, du = –dx, and when x = –a, u = a.
INTEGS. OF SYMM. FUNCTIONS

Proof—Equation 8

• So, Equation 7 becomes:
INTEGS. OF SYMM. FUNCTIONS

Proof a

• If f is even, then f(–u) = f(u).
• So, Equation 8 gives:
INTEGS. OF SYMM. FUNCTIONS

Proof b

• If f is odd, then f(–u) = –f(u).
• So, Equation 8 gives:
INTEGS. OF SYMM. FUNCTIONS
• Theorem 6 is illustrated here.

Figure 5.5.3, p. 338

INTEGS. OF SYMM. FUNCTIONS
• For the case where f is positive and even, part (a) says that the area under y = f(x) from –a to a is twice the area from 0 to a because of symmetry.

Figure 5.5.3a, p. 338

INTEGS. OF SYMM. FUNCTIONS
• Therefore, part (b) says the integral is 0 because the areas cancel.

Figure 5.5.3b, p. 338

INTEGS. OF SYMM. FUNCTIONS

Example 8

• As f(x) = x6 + 1 satisfies f(–x) = f(x), it is even.
• So,
INTEGS. OF SYMM. FUNCTIONS

Example 9

• As f(x) = (tan x)/ (1 + x2 + x4) satisfies f(–x) = –f(x), it is odd.
• So,