What is the net area between #f(x) = e^(2x)-xe^x # and the x-axis over #x in [2, 4 ]#?

1 Answer
Jul 7, 2016

#=1/2e^8-7/2e^4+e^2#

Explanation:

We can split the integral up:

#int_2^4 (e^(2x) - xe^x)dx = int_2^4 e^(2x)dx - int_2^4 xe^xdx#

The first integral is easy

#[1/2e^(2x)]_2^4 = 1/2(e^8-e^4)#

For the second, we need to use integration by parts:

#int u*v' = u*v - int u'*v#

Let #u(x) = x# and #v'(x) = e^x#
then #u'(x) = 1# and #v(x) = e^x#

#int xe^xdx = xe^x - int e^xdx = e^x(x-1)#

Therefore, we have

#1/2(e^8-e^4) - [e^x(x-1)]_2^4#

#1/2(e^8-e^4) -[3e^4 - e^2]#

#=1/2e^8-7/2e^4+e^2#