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.