How do you find the definite integral of #int (x^2 + x) dx# from #[1,3]#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Konstantinos Michailidis Nov 1, 2015 It is #int_1^3 (x^2+x)dx=int_1^3 d/dx[x^3/3+x^2/2]dx= [x^3/3+x^2/2]_1^3=9+9/2-(1/3+1/2)=38/3# 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 1200 views around the world You can reuse this answer Creative Commons License