How do you evaluate the definite integral #(x^43)e^(-x^(44)) dx# for a=0, b=1?

1 Answer
Alberto M.
May 24, 2015

We have to look for a primitive of the function #f(x)=x^43e^{-x^44}#

For the chain rule,

#d/dx(e^{-x^44})=e^{-x^44}(-44x^43)#

So we have a primitive for #f# (we just have to divide by #-44#), and we have, for the fundamental theorem of integral calculus:

#int_0^1x^43e^{-x^44}=e^{-x^44}/(-44)|_0^1##=-1/(44e)+1/44#

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