How do you integrate #(1/2)x-1dx# for the interval (0,3)? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer GiĆ³ Aug 18, 2015 I found: #-3/4# Explanation: Given: #int_0^3(x/2-1)dx=# #[x^2/4-x]_0^3=# substituting the extrema: #=(9/4-3)-0=(9-12)/4=-3/4# 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 1345 views around the world You can reuse this answer Creative Commons License