# Integral with a limit; integral and inequality

I am trying to solve the following problem. $$\lim_{h \to 0} \int_0^h\frac{\sqrt{t^2+9}}{h}\mathrm{d}t$$

My presumption is that I should just evaluate the function at $0$, but I can't justify why the answer is $3$ and where the $h$ goes. How does this work?

Also, I am asked to prove that $\sin{x} \le x$ in the interval $(0,1)$. Then I need to justify why the following is true.

$$\int_0^1(\sin{x})^{2012} \le \frac{1}{2013}$$

I'm sorry about the trouble. I haven't seen exercises like these two before and I am unsure of how I am supposed to solve them.

For the first one, I should have placed it in a form where it may be seen as a derivative.

$$\lim_{h \to 0} \frac{\int_0^h\sqrt{t^2+9}\space \mathrm{d}t - \int_0^0\sqrt{t^2+9}\space \mathrm{d}t}{h}$$

Where, as André suggested, we define

$$F(x) = \int_0^x\sqrt{9+t^2}\mathrm{d}t$$

So it basically is

$$\lim_{h \to 0}\frac{F(0+h)-F(0)}{h} = F'(0) = \sqrt{0^2+9}=3$$

This is my attempt at the second problem. Is it right?

$$\sin{x}\le x \\ (\sin{x})^{2012} \le x^{2012} \\ \int_0^1(\sin{x})^{2012} \le \int_0^1x^{2012} \\ \int_0^1(\sin{x})^{2012} \le \int_0^1x^{2012} = \frac{x^{2013}}{2013}\bigg|_0^1=\frac{1}{2013}$$

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For the inequality, show that $|\sin(t)| \leq |t|$ using the mean value theorem. – Amateur Dec 14 '13 at 3:14
Alternatively, use the Taylor expansion for $\sin x$ to see that $\sin x \leq x$ in $(0, 1)$. For the second question, use the fact that $\sin x \leq x$ in $(0, 1)$ to conclude $(\sin x)^{2012} \leq x^{2012}$ in $(0, 1)$, and use that to conclude $\int_{0}^{1} (\sin x)^{2012}dx \leq \int_{0}^{1} x^{2012}dx$. – Alex Wertheim Dec 14 '13 at 3:18

One way to do the first question is to let $$F(x)=\int_0^x \sqrt{t^2+9}\,dt.$$ Then our limit, by definition, is $F'(0)$. Evaluate $F'(0)$ using the Fundamental Theorem of Calculus.

In my opinion a better way goes as follows. For simplicity, suppose that $h\gt 0$. Then $$h\sqrt{0+9}\lt \int_0^h \sqrt{t^2+9}\,dt \lt h\sqrt{h^2+9}.$$ Divide through by $h$, and Squeeze. (Negative $h$ can be dealt with similarly, the inequalities reverse.)

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So (I think I'm supposed to use the first solution) I should do something to turn the expression into something that becomes a derivative, to then apply the Fundamental Theorem of Calculus? – OFRBG Dec 14 '13 at 3:43
The expression is a derivative. Define $F(x)$ as I did. Then $F(0)=0$. So we are asked to find $\lim_{h\to 0}\frac{F(0+h)-F(0)}{h}$. By definition, this is $F'(0)$. – André Nicolas Dec 14 '13 at 3:47
Just a small last question. The "intermediate" step in that reasoning is that $\lim_{h \to 0}\frac{F(h)}{h}$ has an integral above and the $h$ below may be taken inside, hence turning it into the original expression? – OFRBG Dec 14 '13 at 3:56
The expression $\frac{F(0+h)-F(0)}{h}$ is exactly the same as the given expression, since $F(0)=0$. No intermediate step. – André Nicolas Dec 14 '13 at 4:02
But if $F(h)=\int_0^h f(t)dt$, the $h$ of the denominator doesn't appear right away, does it? That's what I meant. i.e. $\frac{\int_0^h f(t)dt}{h}$. I ask because I am pretty much new to integrals formally. – OFRBG Dec 14 '13 at 4:10

Hint 1: $$\lim_{h \to 0} \int_0^h\frac{\sqrt{t^2+9}}{h}\mathrm{d}t=\lim_{h \to 0} \frac{\int_0^h\sqrt{t^2+9}\mathrm{d}t}{h}$$ This is of the form $\frac00$ so you can apply L'Hôpital's rule.

Hint 2: Notice that $\sin0=0$ and that $$\frac{d}{dx}\sin x=\cos x \le1=\frac{d}{dx}x$$ Hint 3: Use Hint 2 to show that $$\int_0^1(\sin{x})^{2012}dx \le\int_0^1x^{2012} dx$$

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Thanks. Got it, I think. – OFRBG Dec 14 '13 at 3:33

Hint: Note that,

$$\lim_{h \to 0} \int_0^h\frac{\sqrt{t^2+9}}{h}\mathrm{d}t = \lim_{h \to 0} \frac{F(h)}{h} .$$

Now, you can use L'hopital's rule.

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