How do you evaluate the definite integral #int (x-2) dx# from [-1,0]? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Andrea S. May 26, 2017 #int_(-1)^0 (x-2)dx = - 5/2# Explanation: #int_(-1)^0 (x-2)dx = int_(-1)^0 (x-2)d(x-2)# #int_(-1)^0 (x-2)dx = [(x-2)^2/2]_(-1)^0 # #int_(-1)^0 (x-2)dx = (0-2)^2/2 - ((-1)-2)^2/2# #int_(-1)^0 (x-2)dx = 2 - 9/2# #int_(-1)^0 (x-2)dx = - 5/2# 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 1249 views around the world You can reuse this answer Creative Commons License