What is f(x) = int xe^(x-3)dx if f(0)=-2 ?

2 Answers
Jan 2, 2016

f(x) = xe^(x-3)-e^(x-3) + e^-3-2

Explanation:

f(x)=int xe^(x-3) dx

Use integration by parts

int u (dv) = uv - int v (du)

Selecting u and dv is the first step we should take.

Let u=x and dv = e^(x-3)dx

u=x
du = dx

dv = e^(x-3)dx

int dv = int e^(x-3)dx

v = e^(x-3)

Now our integration becomes

f(x)=int xe^(x-3) dx = xe^(x-3) - int e^(x-3) dx

int xe^(x-3) dx = xe^(x-3)-e^(x-3) + C

f(x) = xe^(x-3)-e^(x-3) + C
Given f(0) = -2
f(0) = 0e^(0-3)-e^(0-3)+C
-2=-e^-3+C
e^-3-2=C

xe^(x-3)-e^(x-3) + e^-3-2

Jan 2, 2016

f(x) = int_0^x te^{t-3} dt - 2
= xe^{x-3} - e^{x-3} + e^{-3} - 2

Explanation:

Instead of using indefinite integration (refer to the other answer by Karthik), one can also use definite integration for this question. The integration is mostly the same.