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I just learned about the fundamental theorem of calculus and I am doing some applications of the theorem. I have been given this problem:

If $F(x) =$ $\int_{x}^{x^2}f$ what is $F'(x)$?

I what the theorem itself states that given an integral like the one above $F'(x) = f(x)$ and about primitive functions of F. I just really have no idea how to attack this problem and define what $F(x)$ is, because I have been given no examples of how the hell to apply this and I am completely lost. I don't even know where to start, and my intuition would just say that $F'(x) = f(x^2) - f(x)$, which I am pretty sure is too simplistic and incorrect. Help is greatly appreciated!

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Hint 1. Fix $a$ and write $F(x) = \int_a^{x^2} f(t) \, dt - \int_a^{x} f(t) \, dt$.

Hint 2. If $g(x) = \int_a^{x} f(t) \, dt$, can you express $F(x)$ using the function $g$?

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It is alas a bit too simple. The fundamental theorem of calculus, combined with the chain rule, says that if you have a function $F(x)=\int _c^{g(x)}f(x)dx$, where C is a constant, then $F'(x)=f(g(x))(g'(x))$.

So, the key here is that you need a constant on the bottom, a function on the top, and the chain rule.....so we rewrite your equation as $F(x)=\int _0 ^{x^2}f(x)dx-\int_0^xf(x)dx$.

Then, taking the derivative and using the chain rule, we have $F'(x)=f(x^2)\cdot 2x-f(x)$

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I apologize for the lack of rigor in this answer, as I do not have the conditions for the FTC memorized. However, this is how I usually teach this concept to a Calculus class. Consider the integral $$\int\limits_{\ell(x)}^{u(x)}f(t)\text{ d}t\text{.}$$ (I use $\ell$ for "lower" and $u$ for "upper.") By the FTC, we know that $$\int\limits_{\ell(x)}^{u(x)}f(t) \text{ d}t= F\left(u(x)\right)-F\left(\ell(x)\right)\text{,}$$ where $F$ is an antiderivative of $f$.

Assume $F$, $u$, and $\ell$ are differentiable, so that $$\dfrac{\text{d}}{\text{d}x}\left[\int\limits_{\ell(x)}^{u(x)}f(t) \text{ d}t\right] = \dfrac{\text{d}}{\text{d}x}\left[F\left(u(x)\right)-F\left(\ell(x)\right)\right] = F^{\prime}(u(x))u^{\prime}(x)-F^{\prime}(\ell(x))\ell^{\prime}(x)\text{.}$$ [Note the use of the chain rule above.] But as $F$ is an antiderivative of $f$, $F^{\prime} = f$, so that $$\dfrac{\text{d}}{\text{d}x}\left[\int\limits_{\ell(x)}^{u(x)}f(t) \text{ d}t\right] = f(u(x))u^{\prime}(x)-f(\ell(x))\ell^{\prime}(x)\text{.}$$ Take $u(x) = x^2$, $\ell(x) = x$, and their respective derivatives and you're done.

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  • $\begingroup$ It would be wise not to use $x$ for two different variables, it is very confusing. $\endgroup$ – yo' Oct 6 '14 at 9:44
  • $\begingroup$ Oops, thanks. Didn't notice. $\endgroup$ – Clarinetist Oct 6 '14 at 14:35
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The fundamental theorem of calculus only directly applies when you are differentiating something of the form $\int_a^x g(t)\,dt$, with $a$ constant. Your $F(x)$ isn't immediately in that form, but we can write $$ F(x)=\int_x^af(t)\,dt+\int_a^{x^2}f(t)\,dt\\ \qquad\,\,=-\int_a^xf(t)\,dt+\int_a^{x^2}f(t)\,dt $$ We can use FTOC to find the derivative of the first integral, $\int_a^xf(t)\,dt$. For the second one, you need to use the chain rule: letting $G(x) = \int_a^xf(t)\,dt$, the second integral is $G(x^2)$.

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