Engineering Board Exam Reviewer · Integration by Parts · hard

Evaluate: x3 ex2,dx

1. ≤ft(x2 + 1 ) ex22+C
2. ≤ft(x2 - 1 ) ex2+C
3. ≤ft(x2 - 1 ) ex22+C
4. x4 ex24+C

Answer

C. ≤ft(x2 - 1 ) ex22+C

Solution

Write x3ex2=x2· xex2. Let u=x2, dv=xex2dx so v= 12ex2: x3ex2dx= 12x2ex2- xex2dx= 12x2ex2- 12ex2= (x2-1)ex22+C.

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