Solutions to U-Substitution Practice Problems
Home > Review of First & Second Year Calculus > U-Substitution > PProb Solutions
Practice Problem Solutions: Problem 1 :: Problem 2 :: Problem 3
PROBLEM 1:
Problem: Evaluate
using u-substitution.
Solution: Let
. Then
and
. Thus we have the integral
. Proceeding with normal integration, we find
PROBLEM 2:
Problem: Evaluate
using u-substitution.
Solution: Let
. Then we have
and the integral
. Integrating, we find
.
PROBLEM 3:
Problem: Evaluate
using u-substitution.
Solution: Most students will see the answer to this problem intuitively, but it can carefully found using u-substitution. Since
, we have
. Let
, so
and
. Thus, we get
. Integration gives us
.
Note that the method used above made use of the exponent rule. One can use u-substitution to solve the problem as it is written without making use of this exponent rule. To do this, let
. Then is
also
, so we have
. The integral becomes
, which is the same answer we obtained using the first method.
This page last updated 12 January, 2008 11:26 AM
using u-substitution.
. Then
and
.
Thus we have the integral
.
Proceeding with normal integration, we find 
using u-substitution.
. Integrating, we find
.
using u-substitution.