How do you find the definite integral of #2/(sqrt(x) *e^-sqrt(x))# from #[1, 4]#?

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Answer 1 · 2016-10-10T21:01:50

#4e^2-4e#

#I=int_1^4 2/(sqrtx*e^(-sqrtx))dx#

Simplify the negative exponent by bringing it to the numerator.

#I=2int_1^4e^sqrtx/sqrtxdx#

We will use the substitution #u=sqrtx#. This implies that #du=1/(2sqrtx)dx#. Don't forget to plug the bounds of #1# and #4# into #sqrtx#.

#I=4int_1^4e^sqrtx/(2sqrtx)dx=4int_1^2e^udu#

The integral #inte^udu=e^u+C#:

#I=4[e^u]_1^2=4(e^2-e^1)=4e^2-4e#