Chapter 5 | Integration
589
Theorem 5.8: Substitution with Definite Integrals Let u = g ( x ) and let g ′ be continuous over an interval ⎡ ⎣ a , b ⎤
⎦ , and let f be continuous over the range of u = g ( x ).
Then,
⌠ ⌡ a b
g ( b )
⎞ ⎠ g ′( x ) dx = ∫
f ⎛
f ( u ) du .
⎝ g ( x )
g ( a )
Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for indefinite integrals, if F ( x ) is an antiderivative of f ( x ), we have ∫ f ⎛ ⎝ g ( x ) ⎞ ⎠ g ′( x ) dx = F ⎛ ⎝ g ( x ) ⎞ ⎠ + C . Then (5.20) ∫ a b f ⎡ ⎣ g ( x ) ⎤ ⎦ g ′( x ) dx = F ⎛ ⎝ g ( x ) ⎞ ⎠ | x = a x = b = F ⎛ ⎝ g ( b ) ⎞ ⎠ − F ⎛ ⎝ g ( a ) ⎞ ⎠
= F ( u ) | u = g ( a ) u = g ( b ) = ∫ g ( a ) g ( b ) f ( u ) du ,
and we have the desired result.
Example 5.34 Using Substitution to Evaluate a Definite Integral Use substitution to evaluate ⌠ ⌡ 0 1 x 2 ⎛ ⎝ 1+2 x 3 ⎞ ⎠ 5 dx .
Solution Let u =1+2 x 3 , so du =6 x 2 dx . Since the original function includes one factor of x 2 and du =6 x 2 dx , multiply both sides of the du equation by 1/6. Then, du = 6 x 2 dx
1 6
du = x 2 dx .
x =0, u =1+2(0)=1,
To adjust the limits of integration, note that when
and when
x =1, u =1+2(1)=3. Then
⌠ ⌡ 0 1
3
5
x 2 ⎛
⎝ 1+2 x 3 ⎞ ⎠
6 ∫
u 5 du .
dx = 1
1
Evaluating this expression, we get
Made with FlippingBook - professional solution for displaying marketing and sales documents online