What is #int_(0)^(2) xe^(x^2 + 2)dx #?

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Answer 1 · 2015-11-14T00:34:27

#1/2(e^6-e^2)#

which can be factored into

#e^2/2(e^2+1)(e^2-1)#

Approximately #198.02#

Perform a u-substitution

Let #u=x^2+2#

#(du)/dx=2x#

#dx=(du)/(2x)#

When #x=0#, #u=0^2+2=2#

When #x=2#, #u=2^2+2=6#

Now make the substitution into the integral

#int_2^6xe^u(du)/(2x)#

#1/2int_2^6e^udu#

Now integrate,

#1/2e^u#

Now evaluate

#1/2e^6-1/2e^2#

#1/2e^2(e^4-1)#

#1/2e^2(e^2+1)(e^2-1)#

This is approximately #198.02#