Question

Got a Fundamental Theorem of Calculus Question. Screenshot in comments!

Answer 1

@freckles Think you could help me with this D:

Answer 2

The FTC states that
\[\frac{d}{dx}\int_c^{g(x)}f(t)\,dt=f(g(x))\,g'(x)\]
Identifying your \(f(t)\) and \(g(x)\) is half the battle (a pretty easy one).

Answer 3

So f(t) is (1/5t^2-1) and g(x) is x^2?

Answer 4

Would this be correct @SithsAndGiggles ?

Answer 5

\(g(x)\) is right, but \(f(t)\) is missing that exponent, 8.

Alright, so if \(g(x)=x^2\), what is \(g'(x)\)?

And if \(f(t)=\left(\dfrac{1}{5}t^2-1\right)^8\), what is \(f(g(x))\)?

Answer 6

so g'(x) is 2x, so, I put in 2x where the t is?

Answer 7

Hello @wio , would you be able to help me with the rest D:

Answer 8

\[\Large f(g(x))\,g'(x)=2x\,f(x^2)\]
Since \(f(t)=\left(\dfrac{1}{5}t^2-1\right)^8\), that means
\[\Large f(g(x))=f(x^2)=\left(\frac{1}{5}(x^2)^2-1\right)^8\]

Answer 9

But I thought g'(x) came out to 2x? Do I just disregard that?

Answer 10

\[\Large\begin{align*}
f(x)&=\int_5^{\color{red}{g(x)}=x^2}\underbrace{\left(\frac{1}{5}t^2-1\right)^8}_{\color{blue}{f(t)}}\,dt\\\\
\frac{d}{dx}f(x)&=\frac{d}{dx}\int_5^{x^2}\left(\frac{1}{5}t^2-1\right)^8\,dt\\\\
&\stackrel{\text{FTC}}{=}\color{blue}{f(}\color{red}{g(x)}\color{blue})\frac{d}{dx}[\color{red}{g(x)}]\\\\
&=2x\color{blue}{f(}x^2\color{blue})\\\\
&=2x\left(\frac{1}{5}(x^2)^2-1\right)^8\end{align*}\]

Answer 11

Oh ok, thanks!