How do you evaluate the definite integral #int (1/2x-1) dx# from #[0, 3]#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Narad T. Nov 2, 2016 The answer is #=-3/4# Explanation: Use #intx^ndx=x^(n+1)/(n+1)+C# #int_0^3(x/2-1)dx=int_0^3x/2dx-int_0^3 1*dx=(x^2/4-x)_0^3# #=(9/4-3)-0=-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 3289 views around the world You can reuse this answer Creative Commons License