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

2 Answers
Karthik G
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#

Bio
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.

Related questions
See all questions in Evaluating the Constant of Integration
Impact of this question
1990 views around the world
You can reuse this answer
Creative Commons License