How do you evaluate #int x^2 + x + 4dx# for [0, 2]? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Dharma R. Sep 21, 2015 #38/3# Explanation: #int_0^2x^2+x+4dx=int_0^2x^2dx+int_0^2xdx+int_0^24dx =[x^3/3]_0^2+[x^2/2]_0^2+4[x]_0^2=8/3+4/2+8-0=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 1522 views around the world You can reuse this answer Creative Commons License