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 1548 views around the world You can reuse this answer Creative Commons License