How do you integrate #int x^2e^(x^3/2)dx# from #[-2,3]#?

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Answer 1 · 2017-01-14T18:44:52

#(2e^(35/2)-2)/(3e^4)#

#I=int_(-2)^3x^2e^(x^3/2)dx#

Use the substitution:

#u=x^3/2" "=>" "du=3/2x^2dx#

We already have #x^2dx# in the integral, but are off by a factor of #3/2#. Also note that changing from #x# to #u# will change our bounds:

#x=3" "=>" "u=3^3/2=27/2#

#x=-2" "=>" "u=(-2)^3/2=-4#

Thus:

#I=2/3inte^(x^3/2)(3/2x^2dx)=2/3int_(-4)^(27/2)e^udu#

The integral of #e^u# is itself:

#=2/3[e^u]_(-4)^(27/2)=2/3(e^(27/2)-e^(-4))=2/3((e^(35/2)-1)/e^4)=(2e^(35/2)-2)/(3e^4)#