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$y'=\frac{\mathrm{d}}{\mathrm{d}x} \int_{a}^{\sqrt{x}}f(t)dt$

Here, $y$ is the area function given by $y = \int_{a}^{\sqrt{x}}f(t)dt$

Let us say that $y=F(x)$.

$$y'=F'(x)=\lim_{h\rightarrow 0}\frac{ \int_{a}^{\sqrt{x+h}}f(t)dt-\int_{a}^{\sqrt{x}}f(t)dt}{h}$$ On working it out we get, $$y'=\lim_{h\rightarrow 0}\frac{\int_{\sqrt{x} }^{\sqrt{x+h}}f(t)dt}{h}$$ We know that, $$\frac{\int_{\sqrt{x}}^{\sqrt{x+h}}f(t)dt}{h}$$ is the average of $f(t)$. We then have, $$y'= \lim_{h\rightarrow 0}f(c)$$ where $f(c)$ is the average of $f(t)$ in the interval. On taking the limit, $$y'=f(\sqrt{x})$$ which is incorrect.

Where could I have gone wrong?

Edit: My statements about "Linearizing" have been withdrawn.

R004
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3 Answers3

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Let $F$ be the antiderivative of $f$. Then

$$\int_a^{\sqrt x}f(t)\,dt=F(\sqrt x)-F(a)$$ and by the chain rule

$$\frac d{dx}\int_a^{\sqrt x}f(t)\,dt=\frac d{dx}F(\sqrt x)=\frac1{2\sqrt x}f(\sqrt x).$$

user254433
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  • I understand. I have just altered my statements. I wish to know where I might have made a mistake while working out the derivative. – R004 Jun 14 '17 at 08:47
  • @R004: user254433 explained it. –  Jun 14 '17 at 08:57
6

The statement "We know that $\frac{1}{h}\int_{\sqrt x}^{\sqrt{x+h}}f(t)dt$ is the average of $f$" is not correct, since the interval of integration does not have length $h$. It has length $\sqrt{x+h}-\sqrt{x}$, so the average is actually $$ A_hf(x):=\frac{1}{\sqrt{x+h}-\sqrt{x}}\int_{\sqrt{x}}^{\sqrt{x+h}}f(t)dt. $$

Like you suggested, this tends towards $f(\sqrt{x})$ as $h\to 0$, so your original expression becomes: $$ \frac{1}{h}\int_{\sqrt x}^{\sqrt{x+h}}f(t)dt=\frac{\sqrt{x+h}-\sqrt{x}}{h}A_hf(x)\to \frac{1}{2\sqrt{x}}f(\sqrt{x}), $$ since $\lim\frac{\sqrt{x+h}-\sqrt{x}}{h}$ is just the derivative.

user254433
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3

There is not enough information to determine what $f(x)$ is.

As far as determining the derivative of an integral, the Second Fundamental Theorem of Calculus tells us that ---

$\displaystyle \frac{d}{dx} \int_{a(x)}^{b(x)}f(t)\,dt=f(b(x))b'(x)-f(a(x))a'(x)$

This can be derived from $\displaystyle \frac{d}{dx} \int_{a(x)}^{b(x)}f(t)\,dt = \frac{d}{dx}\left(F(a(x)-F(b(x)\right)$, and applying Chain Rule, to get $f(b(x))b'(x)-f(a(x))a'(x)$.

Therefore, $\displaystyle y' = \frac{d}{dx}\int_a^{\sqrt{x}}f(t)\,dt=\frac{f(\sqrt{x})}{2\sqrt{x}}.$

Saketh Malyala
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