How do you integrate #int x^2e^(x^3/2)dx# from #[-2,3]#?
1 Answer
Jan 14, 2017
Explanation:
#I=int_(-2)^3x^2e^(x^3/2)dx#
Use the substitution:
#u=x^3/2" "=>" "du=3/2x^2dx#
We already have
#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
#=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)#