http://www.ck12.org Chapter 7. Integration Techniques
Using Substitution on Definite Integrals
Example 6:
Evaluate∫ 13 √ 2 xx− 1 dx.
Solution:
Letu= 2 x− 1 .Thendu= 2 dx,ordx=du/ 2 .Before we substitute, we need to determine the new limits of
integration in terms of theuvariable. To do so, we simply substitute the limits of integration intou= 2 x−1:
Lower limit: Forx= 1 ,u= 2 ( 1 )− 1 = 1.
Upper limit: Forx= 3 ,u= 2 ( 3 )− 1 = 5.
We now substituteuand the associated limits into the integral:
∫ 5
1
√x
udu
2.As you may notice, the variablexis still hanging there. To write it in terms ofu,sinceu= 2 x− 1 ,solving forx,we
get,x= (u+ 1 )/( 2 ).Substituting back into the integral,
=
∫ 5
1u+ 1
2 √udu
2
=^14∫ 5
1u√+ 1
udu
=^14∫ 5
1 (u+^1 )u− 1 / (^2) du
=^14
∫ 5
1 (u
1 / (^2) +u− 1 / (^2) )du
=^14
[
2 u^3 /^2
3 +2 u^1 /^2
1] 5
1.
Applying the Fundamental Theorem of Calculus by inserting the limits of integration and calculating,
=^14
([
2 ( 5 )^3 /^2
3 +
2 ( 5 )^1 /^2
1
]
−
[
2 ( 1 )^3 /^2
3 +
2 ( 1 )^1 /^2
1
])
.
Calculating and simplifying, we get
=^4
√ 5 − 2
3.
We could have chosenu=√ 2 x−1 instead. You may want to try to solve the integral with this substitution. It might
be easier and less tedious.
Example 7: