How do you find the integral from 0 to 3 of #(1/2 x - 1) dx#? Calculus Introduction to Integration Integrals of Polynomial functions 1 Answer Tom Apr 13, 2015 #int_0^3(1/2x-1)dx = int_0^3(1/4x^2-x)dx# We have : #1/2[(1/2x^2-2x)]_0^3# Therefore : #1/2(9/2-6)# #=-3/4# Answer link Related questions How do you evaluate the integral #intx^3+4x^2+5 dx#? How do you evaluate the integral #int(1+x)^2 dx#? How do you evaluate the integral #int8x+3 dx#? How do you evaluate the integral #intx^10-6x^5+2x^3 dx#? What is the integral of a constant? What is the antiderivative of the distance function? What is the integral of #|x|#? What is the integral of #3x#? What is the integral of #4x^3#? What is the integral of #sqrt(1-x^2)#? See all questions in Integrals of Polynomial functions Impact of this question 2219 views around the world You can reuse this answer Creative Commons License