# Fundamental Theorem and Integral Problem

$F'(1)$ given that

$$F(x) = \int_{5}^{x^9}\frac{1}{5+t^2}dt .$$

So far I have simplify the problem to

$$F'(x)=\frac{9x^8}{5+x^9} .$$

So what I'm wondering is do I replace x with 1 $F'(1)$ or do I replace x with the $\int_{5}^{x^9}$ and subtract them?

Can anybody please help me out and tell me if I'm on the right track and what is my next step.

• You should really take out the real-analysis and fundamental-groups tags. They do not belong in your question. Nov 21, 2013 at 20:53
• Your derivative is not correct, though at $1$ it doesn't matter. We have $F'(x)=\frac{1}{5+x^{18}}\cdot 9x^8$. Plug in $1$. Nov 21, 2013 at 20:55

You are on the correct path. You applied the fundamental theorem of calculus correctly and deduced (almost) correctly that

$$F'(x) = \frac{9x^8}{5+(x^9)^2} = \frac{9x^8}{5+x^{18}}.$$

All you have to do now is really do $x=1$. I do not understand your question of replacing $x$ with an integral. What you forgot was that when you apply the fundamental theorem of calculus what you have is

$$F(x) = \int_a^x f(t) \, dt \implies F'(x) = f(x).$$

In the general case, you have

$$F(x) = \int_a^{h(x)} f(t) \, dt \implies F'(x) = f(x) \cdot h'(x).$$

The Fundamental Theorem of Calculus(TFTC): $\int_{a}^{b}g(x)dx=G(b)-G(a)$

Replace $g(x)$ by $\frac{1}{5+t^2}$, and replace the boundaries as appropriate($a=5$ and $b=x^8$). Apply the fundamental theorem as follows, we know

$$F(x)=\int_{5}^{x^8}\frac{1}{5+t^2}dx$$ by TFTC this is the same as $$F(x)=\int_{5}^{x^8}\frac{1}{5+t^2}dx=G(x^8)-G(5)$$ Now take the derivative, $$F'(x)=\frac{d}{dx}\left[ \int_{5}^{x^8}\frac{1}{5+t^2}dx \right]$$ $$\; \; \; \qquad \;=\frac{d}{dx}\left[G(x^8)-G(5) \right]$$ $$\qquad \; \; \; \; \; \qquad =\frac{d}{dx}\left[G(x^8)\right]-\frac{d}{dx}\left[G(5) \right]$$ $$\qquad=\frac{d}{dx}\left[G(x^8)\right]+0$$ $$=8x^7g(x^8)$$ So $F'(1)=8(1)^7g(1)=8g(1)$.

If you work out your problems like this you will always understand what gets replace and why. And more importantly, what makes sense as an answer and what does not.