How do you evaluate the definite integral #int (x^2-4)dx# from [2,4]? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Narad T. Feb 22, 2017 The answer is #=32/3# Explanation: We need #intx^ndx=x^(n+1)/(n+1)+C(n!=-1)# #int_2^4(x^2-4)dx# #=[x^3/3-4x]_2^4# #=(64/3-16)-(8/3-8)# #=64/3-8/3-16+8# #=56/3-8# #=32/3# 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 3931 views around the world You can reuse this answer Creative Commons License