How to find the derivative of the following function F(x)=∫ (2t−1)^3dt (a=x^3 & b=x^4) ?

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Answer 1 · 2015-09-08T01:51:47

#F'(x) = -3x^2(2x^3-1)^3 + 4x^3(2x^4-1)^3#

#F(x) = int_(x^3)^(x^4) (2t-1)^3 dt#.

To find #F'(x)# rewrite so we can use Fundamental Theorem of Calculus Part 1. And we also need the chain rule:

#F(x) = int_(x^3)^(x^4) (2t-1)^3 dt = int_(x^3)^0(2t-1)^3 dt+int_0^(x^4) (2t-1)^3 dt#

# = - int_0^(x^3)(2t-1)^3 dt+int_0^(x^4) (2t-1)^3 dt#.

Now use

#d/dx(- int_0^(x^3)(2t-1)^3 dt) = -(2x^3-1)^3(3x^2)# (FTC1 and chain)

and

#d/dx(int_0^(x^4) (2t-1)^3 dt#) = (2x^4-1)^3(4x^3)# (FTC1 and chain)

to get

#F'(x) = -3x^2(2x^3-1)^3 + 4x^3(2x^4-1)^3#