Engineering Board Exam Reviewer · Definite Integrals · medium
Evaluate: 01 e2x,dx
- 3.19
- 6.39
- 1.59
- 2.00
Answer
A. 3.19
Solution
12≤ft[e2x ]01= 12(e2-1) 3.19.
Engineering Board Exam Reviewer · Definite Integrals · medium
A. 3.19
12≤ft[e2x ]01= 12(e2-1) 3.19.