What is #int_(0)^(2) x^3*e^(x^2)dx #?

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Answer 1 · 2015-11-24T19:08:20

#1/2(3e^4+ 1)#

First do substitution

#u = x^2#
#du = 2x dx#

Now we have

#1/2int_0^2u*e^u du#

Now by part

#v = u#
#v' = 1#
#w' = e^u#
#w = e^u#

if #x = 2 => u = 4#
if #x = 0 =>u = 0#

#1/2int_0^4u*e^u du = 1/2([u*e^u]_0^4-int_0^4e^udu)#

Which is

#1/2int_0^2u*e^u du = 1/2([u*e^u]_0^4-[e^u]_0^4)#

#1/2(3e^4+ 1)#