Question

If f(x) = integral (3,x) sqr root (1+t^3) dt, find (f^1)'(0)

Answer 1

If \(\int\limits_{a}^{b}f(t)\;dt = F(b) - F(a)\)

Then it is reasonable to conclude that \(\int\limits_{a}^{x}f(t)\;dt = F(x) - F(a)\)

It is then also quite reasonable to conclude:

\(\dfrac{d}{dx}\left(\int\limits_{a}^{x}f(t)\;dt\right) = \dfrac{d}{dx}\left(F(x) - F(a)\right) = F'(x) - F'(a) = f(x) - 0 = f(x)\)

Answer 2

can you rewrite it using LaTex?!

Answer 3

This looks like FToC Part ii

Answer 4

I am assume that last expression in the original post was simple the first derivative. I suppose it may have means some sort of inverse.

Answer 5

yes that formula is \[1/((f')(f^-1)(a_\]f