How do you integrate #int(sqrtx - x^2) dx#? Calculus Introduction to Integration Integrals of Polynomial functions 1 Answer Konstantinos Michailidis Sep 27, 2015 See explanation Explanation: It is #int(sqrtx - x^2) dx=int ((x^(3/2))/(3/2)-x^3/3)'dx=2/3*x^(2/3)-x^3/3+c# Answer link Related questions How do you evaluate the integral #intx^3+4x^2+5 dx#? How do you evaluate the integral #int(1+x)^2 dx#? How do you evaluate the integral #int8x+3 dx#? How do you evaluate the integral #intx^10-6x^5+2x^3 dx#? What is the integral of a constant? What is the antiderivative of the distance function? What is the integral of #|x|#? What is the integral of #3x#? What is the integral of #4x^3#? What is the integral of #sqrt(1-x^2)#? See all questions in Integrals of Polynomial functions Impact of this question 1185 views around the world You can reuse this answer Creative Commons License