How do you use the Fundamental Theorem of Calculus to find the derivative of #int (2t - 1)^3dt# from x^5 to x^6?

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Answer 1 · 2016-07-17T17:21:32

#= 6x^5 (2x^6 - 1)^3 - 5x^4(2x^5 - 1)^3 #

by FTC #d/(dx) int_a^x \ f(t) \ dt = f(x) qquad triangle#

by chain rule where #u = u(x)#

#d/(dx) int_a^(u(x)) \ f(t) \ dt = f(u)*(du)/dx qquad star#

in this case we have #d/dx int_(x^5)^(x^6) (2t - 1)^3dt#

.........and we need to rewrite this is the forms given in #triangle# and #star#

#d/dx int_(x^5)^(x^6) (2t - 1)^3dt#

#= d/dx int_(x^5)^(a) (2t - 1)^3dt + int_(a)^(x^6) (2t - 1)^3dt#

#= - d/dx int_(a)^(x^5) (2t - 1)^3dt + int_(a)^(x^6) (2t - 1)^3dt#

and applying FTC and the chain rule

#= - (2(x^5) - 1)^3 d/dx x^5 + (2(x^6) - 1)^3 d/dx x^6#

#= - (2x^5 - 1)^3 * 5x^4 + (2x^6 - 1)^3 * 6x^5#

#= 6x^5 (2x^6 - 1)^3 - 5x^4(2x^5 - 1)^3 #