Question #53b73 Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Jim H Nov 27, 2016 I get #1-sqrt2/3# Explanation: On #(0,1)#, #{x^2} = x^2# On #(1,sqrt2)#, #{x^2} = x^2-1# To evaluate #int_0^sqrt2 {x^2} dx#, evaluate both and add: #int_0^1 x^2 dx + int_1^sqrt2 (x^2-1) dx = (1/3) + (2/3-sqrt2/3)# 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 1203 views around the world You can reuse this answer Creative Commons License