How do you find the definite integral for: #(6x+3)dx# for the intervals #[3, 9]#? Calculus Introduction to Integration Formal Definition of the Definite Integral 1 Answer Henry W. Oct 12, 2016 234 Explanation: #int_3^9(6x+3)dx=[3x^2+3x]_3^9# #=[3(9)^2+3(9)]-[3(3)^2+3(3)]# #=270-36=234# 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 2922 views around the world You can reuse this answer Creative Commons License