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

1 Answer
Praveer Sharan
Nov 29, 2017

#F(x) = 1/2x^2 +xe^-x + e^-x + 1#

Explanation:

#F(x) = intx - xe^-xdx = 1/2x^2 - intxe^-xdx#
Finding #intxe^-xdx# is tricky, we'll have to use integration of parts.
#intxe^-xdx = -xe^-x - e^-x + C#
Therefore, #F(x) = 1/2x^2 +xe^-x + e^-x + C#.
Now we have to solve for #C#.
#F(0) = 1/2 * 0^2 + 0e^0 + e^0 + C#
#=1+C = 2 # (since #F(0) = 2#)
Therefore, #C = 1#.
#F(x) = 1/2x^2 +xe^-x + e^-x + 1#.

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