How do you find the definite integral for: #(3x^2+2x+3) dx# for the intervals #[0, 6]#? Calculus Introduction to Integration Formal Definition of the Definite Integral 1 Answer ali ergin Aug 3, 2016 #int_0^6(3x^2+2x+3)d x=?# #int_0^6(3x^2+2x+3)d x=|3*x^3/3+2*x^2/2+3x|_0^6# #int_0^6(3x^2+2x+3)d x=|x^3+x^2+3x|_0^6# #int_0^6(3x^2+2x+3)d x=[(6^3+6^2+3*6)-(0^3+0^2+3*0)]# #int_0^6(3x^2+2x+3)d x=[(216+36+18)-(0)]# #int_0^6(3x^2+2x+3)d x=270# Answer link Related questions What is the Formal Definition of the Definite Integral of the function #y=f(x)# over the... How do you use the definition of the definite integral? What is the integral of dy/dx? What is an improper integral? How do you calculate the double integral of #(xcos(x+y))dr# where r is the region: 0 less than... How do you apply the evaluation theorem to evaluate the integral #3t dt# over the interval [0,3]? What is the difference between an antiderivative and an integral? How do you integrate #3x^2-5x+9# from 0 to 7? Question #f27d5 How do you evaluate the definite integral #int sqrtt ln(t)dt# from 2 to 1? See all questions in Formal Definition of the Definite Integral Impact of this question 3968 views around the world You can reuse this answer Creative Commons License