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What is the antiderivative of #y=sqrt(x)# from 0 to 5?

1 Answer

Oct 15, 2016

#(2sqrt(125))/3#

Explanation:

#int_0^5sqrtxdx=int_0^5(x)^(1/2)dx->#We want to convert the function into index form so that we can apply simple power rules.

#int_0^5(x)^(1/2)dx=[(2x^(3/2))/3]_0^5#

#=[(2(5)^(3/2))/3]-[0]#

#=(2(5)^(3/2))/3=(2sqrt(125))/3#

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