What is the antiderivative of #y=sqrt(x)# from 0 to 5? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Henry W. 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# Answer link Related questions What is the difference between definite and indefinite integrals? What is the integral of #ln(7x)#? Is f(x)=x^3 the only possible antiderivative of f(x)=3x^2? If not, why not? How do you find the integral of #x^2-6x+5# from the interval [0,3]? What is a double integral? What is an iterated integral? How do you evaluate the integral #1/(sqrt(49-x^2))# from 0 to #7sqrt(3/2)#? How do you integrate #f(x)=intsin(e^t)dt# between 4 to #x^2#? How do you determine the indefinite integrals? How do you integrate #x^2sqrt(x^(4)+5)#? See all questions in Definite and indefinite integrals Impact of this question 1352 views around the world You can reuse this answer Creative Commons License