# Fundamental Theorem of Calculus Proof

Find $f'$ where is $f$ is defined on $[0, 1]$ as indicated: $$f(x) = \int_x^{\sqrt{x}} \frac 1{1+t^3}dt$$ I know that the fundamental theorem is going to be used in this proof, but I'm not really sure where to begin. So any help would be greatly appreciated. Thank you!

## 2 Answers

Let $g'(x)=\frac{1}{1+x^3}$, then by the FTC, $$f(x) = \int_x^{\sqrt{x}} g'(t)\,dt=g(\sqrt x)-g(x)$$ then $$f'(x)=\frac{g'(\sqrt x)}{2\sqrt x}-g'(x)=\frac{1}{2(\sqrt x+x^2)}-\frac{1}{1+x^3}$$

• Thank you, I was definitely over thinking this problem. Great solution man! Commented Nov 20, 2014 at 6:34

We have: $f(x) = \displaystyle \int_{0}^{\sqrt{x}} \dfrac{1}{1+t^3} dt - \displaystyle \int_{0}^x \dfrac{1}{1+t^3} dt$. Apply the fundamental theorem of calculus and the chainrule we arrive at..:

$f'(x) = \dfrac{1}{2\sqrt{x}\cdot (1+x\sqrt{x})} - \dfrac{1}{1+x^3} = \dfrac{1}{2\sqrt{x} + 2x^2} - \dfrac{1}{1+x^3}$.

• Thank you, yours and Pauly B's answers were both spot on. Thanks again! Commented Nov 20, 2014 at 6:35
• I have another question about a similar topic if you want to help me out? Commented Nov 20, 2014 at 6:55
• Should I post it in a comment or just open up a second question? It seems a little long to just post in a comment. Commented Nov 20, 2014 at 6:57
• You should post it as a second question, then you get more responses. Commented Nov 20, 2014 at 6:59
• Alright I posted it, feel free to check it out. Thanks! Commented Nov 20, 2014 at 7:05