Using the substitution #u=sqrtx#, how do you integrate #int e^sqrtx/sqrtx# from [1,4]?

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Answer 1 · 2017-01-11T11:18:44

#int_1^4 e^sqrt(x)/sqrt(x) dx = 2e(e-1)#

If we substitute:

#u = sqrt(x)#

we have:

#du = (dx) / (2sqrt(x))#

and of course:

#x=1 => u = 1#
#x=4 => u=2#

So:

#int_1^4 e^sqrt(x)/sqrt(x) dx = 2 int_1^2 e^udu = [e^u]_1^2 = 2(e^2-e)= 2e(e-1)#