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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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slide1

5

INTEGRALS

slide2

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
INTRODUCTION

Equation 1

  • Antidifferentiation formulas don’t tell us how to evaluate integrals such as
introduction1
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.
introduction2
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.
introduction3
INTRODUCTION

Equation 2

  • So, formally, without justifying our calculation, we could write:
introduction4
INTRODUCTION
  • However, now we can check that we have the correct answer by using the Chain Ruleto differentiate the final function of Equation 2:
introduction5
INTRODUCTION

Equation 3

  • Observe that, if F’ = f, then
  • because, by the Chain Rule,
introduction6
INTRODUCTION
  • If we make the ‘change of variable’ or ‘substitution’ u = g(x), from Equation 3, we have:
substitution rule
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 rule1
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 rule2
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
Solution:

Example 1

  • Thus, using x3dx = du/4 and the Substitution Rule, we have:
    • Notice that, at the final stage, we had to return to the original variable x.
substitution rule3
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 rule4
SUBSTITUTION RULE

E. g. 2—Solution 1

  • Evaluate
    • Let u = 2x + 1.
    • Then, du = 2 dx.
    • So, dx = du/2.
solution1
Solution1:

E. g. 2—Solution 1

  • Thus, the rule gives:
solution2
Solution2:

E. g. 2—Solution 2

  • Another possible substitution is
  • Then,
  • So,
    • Alternatively, observe that u2 = 2x + 1.
    • So, 2udu = 2 dx.
solution21
Solution2:

E. g. 2—Solution 2

  • Thus,
substitution rule5
SUBSTITUTION RULE

Example 3

  • Find
    • Let u = 1 – 4x2.
    • Then, du = -8x dx.
    • So, x dx = -1/8 du and
solution3
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

solution4
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 rule6
SUBSTITUTION RULE

Example 4

  • Calculate ∫ cos 5xdx
    • If we let u = 5x, then du = 5 dx.
    • So, dx = 1/5 du.
    • Therefore,
substitution rule7
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.
solution5
Solution:

Example 5

  • Also, x2 = u – 1; so, x4 = (u – 1)2:
definite integrals
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
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 integrals1
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 integrals2
SUB. RULE FOR DEF. INTEGRALS

Proof

  • However, applying the FTC2 a second time, we also have:
sub rule for def integrals3
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

solution6
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

solution7
Solution:

Example 6

  • Thus,
sub rule for def integrals4
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.
solution8
Solution:

Example 7

  • Thus,
symmetry
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
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 functions1
INTEGS. OF SYMM. FUNCTIONS

Proof—Equation 7

  • We split the integral in two:
integs of symm functions2
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 functions4
INTEGS. OF SYMM. FUNCTIONS

Proof—Equation 8

  • So, Equation 7 becomes:
integs of symm functions5
INTEGS. OF SYMM. FUNCTIONS

Proof a

  • If f is even, then f(–u) = f(u).
  • So, Equation 8 gives:
integs of symm functions6
INTEGS. OF SYMM. FUNCTIONS

Proof b

  • If f is odd, then f(–u) = –f(u).
  • So, Equation 8 gives:
integs of symm functions7
INTEGS. OF SYMM. FUNCTIONS
  • Theorem 6 is illustrated here.

Figure 5.5.3, p. 338

integs of symm functions8
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 functions9
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 functions10
INTEGS. OF SYMM. FUNCTIONS

Example 8

  • As f(x) = x6 + 1 satisfies f(–x) = f(x), it is even.
  • So,
integs of symm functions11
INTEGS. OF SYMM. FUNCTIONS

Example 9

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