How do you evaluate the integral #int 1/sqrtxdx# from 0 to 1? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Alexander Jul 11, 2016 #int_0^1 (1)/(sqrt(x)) dx =[2sqrt(x)]_0^1 = (2sqrt(1)-2sqrt(0)) = 2# Explanation: We can rewrite #int_0^1 (1)/(sqrt(x)) dx# as #int_0^1 (x^(-1/2)) dx#, which is now easier to see and evaluate. #int_0^1 (x^(-1/2)) dx = [(x^(-1/2 + 1/1)) * (1/ (-1/2+1/1))]_0^1# Since #(-1/2) + (1/1) = 1/2#, we have #int_0^1 (1)/(sqrt(x)) dx =[2sqrt(x)]_0^1 = (2sqrt(1)-2sqrt(0)) = 2# 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 1381 views around the world You can reuse this answer Creative Commons License