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

2 Answers
Karthik G
Jan 2, 2016

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

Explanation:

f(x)=int xe^(x-3) dxf(x)=∫xex−3dx

Use integration by parts

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

Selecting uu and dvdv is the first step we should take.

Let u=xu=x and dv = e^(x-3)dxdv=ex−3dx

u=xu=x
du = dxdu=dx

dv = e^(x-3)dxdv=ex−3dx

int dv = int e^(x-3)dx∫dv=∫ex−3dx

v = e^(x-3)v=ex−3

Now our integration becomes

f(x)=int xe^(x-3) dx = xe^(x-3) - int e^(x-3) dxf(x)=∫xex−3dx=xex−3−∫ex−3dx

int xe^(x-3) dx = xe^(x-3)-e^(x-3) + C∫xex−3dx=xex−3−ex−3+C

f(x) = xe^(x-3)-e^(x-3) + Cf(x)=xex−3−ex−3+C
Given f(0) = -2f(0)=−2
f(0) = 0e^(0-3)-e^(0-3)+Cf(0)=0e0−3−e0−3+C
-2=-e^-3+C−2=−e−3+C
e^-3-2=Ce−3−2=C

xe^(x-3)-e^(x-3) + e^-3-2xex−3−ex−3+e−3−2

Bio
Jan 2, 2016

f(x) = int_0^x te^{t-3} dt - 2f(x)=∫x0tet−3dt−2
= xe^{x-3} - e^{x-3} + e^{-3} - 2=xex−3−ex−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.

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