Use the fundamental theorem of calculus to find the derivative

Question

Use the Fundamental Theorem of Calculus to find the derivative of

 

f(x)=∫((1/3)t²-1)⁵dt with upper bound x² and lower bound 3

f'(x)=?

Answer 1

This is an example of the 2nd Fundamental Theorem of Calculus.

 

Only the first part of this f(b) - f(a) will have a derivative, since when we plug in 3 and derive, that is the derivative of a constant which is zero.

 

Essentially, taking the derivative of an integral cancel each other. But, due to taking the derivative, there is a chain rule with an extra term tacked on the end.

 

Taking: f(x)=∫((1/3)t²-1)⁵dt with upper bound x² and lower bound 3.

 

Finding f'(x), we simply plug in x² for every t in the integrand.  This yields: f'(x)=(1/3)(x²)²-1)⁵ but due to the chain rule when taking f'(x), we also need to multiply by the derivative of the upper limit.  The derviative of x² is 2x so we need to multiply by 2x, giving: f'(x)=(1/3)(x²)²-1)⁵ * 2x

 

Therefore, our final answer is: f'(x)=(1/3)(x⁴-1)⁵ * 2x