How do you find the definite integral of #2x* (x-3)^5 dx# from #[-1, 2]#? Calculus Introduction to Integration Formal Definition of the Definite Integral 1 Answer Eddie Jul 2, 2016 # = 4101/7# Explanation: we'll use IBP #int u v' = uv - int u'v# so for #int_{-1}^2 dx qquad 2x (x-3)^5 # we can set #u = 2x, u' = 2# #v' = (x-3)^5, v = 1/6(x-3)^6# #implies [2x * 1/6(x-3)^6]\_(-1)^(2) - int_{-1}^2 dx qquad 2 * 1/6(x-3)^6 # # =[1/3 x (x-3)^6]\_(-1)^(2) - 1/3 int_{-1}^2 dx qquad (x-3)^6# # =[1/3 x (x-3)^6 - 1/3* 1/7(x-3)^7 ]\_(-1)^(2)# # =[1/3 x (x-3)^6 - 1/21(x-3)^7 ]\_(-1)^(2)# #= 2/3 + 1/21 + 4^6/3 - 4^7/21# # = 4101/7# 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 1797 views around the world You can reuse this answer Creative Commons License