How do you find the definite integral of #int (1-2x-3x^2)dx# from #[0,2]#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer ali ergin Mar 11, 2016 #int_0^2(1-2x-3x^2)d x=-10# Explanation: #int_0^2(1-2x-3x^2)d x=|x-2*1/2*x^2-3*1/3*x^3|_0^2# #int_0^2(1-2x-3x^2)d x=|x-x^2-x^3|_0^2# #int_0^2(1-2x-3x^2)d x=2-2^2-2^3# #int_0^2(1-2x-3x^2)d x=2-4-8# #int_0^2(1-2x-3x^2)d x# #int_0^2(1-2x-3x^2)d x=-10# 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 3364 views around the world You can reuse this answer Creative Commons License