If f ( x ) = x∫ 0 ( t 3 + 2 t 2 + 6 ) d t then f ' ' ( x ) =

Question

If f ( x ) = x∫ 0 ( t 3 + 2 t 2 + 6 ) d t then f ' ' ( x ) =??

I am confused how to do this. How do I start this problem?

Answer 1

I am assuming that the problem is the integral from 0 to x of t^3 + 2t^2 + 6

as the upper limit is x, from the Fundamental Theorem of Calculus, f'(x) = x^3 + 2x^2 + 6 (t replaced by x)

Then, f'(x) = 3x^2 + 4x

Answer 2

Hi Bri!

This is a great question! If f ( x ) = x∫ 0 ( t 3 + 2 t 2 + 6 ) d t, we can simplify this equation by saying f ( x ) = x∫ 0 d t because anything multiplied by 0 is 0. Therefore, if we take the first derivative of f(x) we will find that f'(x)=0 because the derivative of a constant is 0. Therefore when we take the derivative of f'(x), we are once again taking the derivative of a constant and therefore get f''(x)=0.

Hope this helps!