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I found this exercise in an old exam but I don't know how to attack it because is a limit of an integration and I don't know if the limit affects the process of the integral or it makes it easier. The exercise is this:

The value of the limit $$ \lim_{h\to 0}\frac{1}{h}\int^{2+h}_2\sqrt{1+t^3}dt $$ Is:

a) $0$

b) $1$

c) $\sqrt{2}$

d) $2$

e) $3$

I want to know two things:

1) If there is a method of how to solve limit-integral problems.

2) An explained solution of this example to at least try to understand how to solve exercises of the same style.

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As $h \rightarrow 0$, we have $\displaystyle \int_2^{2+h} \sqrt{1+t^3} \ dt \rightarrow 0$. Thus, we arrive at a $0/0$ indeterminant form when trying to evaluate the limit, so we can apply L'Hopital's rule.

The hardest part about doing this is evaluating $\displaystyle \frac{d}{dh}\int_2^{2+h} \sqrt{1 + t^3} \ dt$. For a continuous function $f,$ the fundamental theorem tells us that $\displaystyle \frac{d}{dx} \int_a^x f(t) \ dt = f(x)$. To get our integral into this form, we can simply shift the function to the left $2$ units and adjust the bounds of integration accordingly: $$\displaystyle \frac{d}{dh}\int_2^{2+h} \sqrt{1 + t^3} \ dt = \frac{d}{dh} \int_0^h \sqrt{1 + (t+2)^3} \ dt = \sqrt{1+(h+2)^3}$$

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  • $\begingroup$ Yes, my answer as is in the exam was 3 because of that, but I don't know how affects L'Hopital to the integral. It lets the fuction inside the integrate just like it's now? (I mean $\sqrt{1+t^3}$) $\endgroup$ – MonsieurGalois Jun 24 '16 at 4:26
  • $\begingroup$ @MonsieurGalois, added some detail $\endgroup$ – Kaj Hansen Jun 24 '16 at 4:32
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Use L'Hopital's rule as follows: $$\lim_{h \to 0} \frac{\int_{2}^{2 + h} \sqrt{1 + t^3}dt}{h}$$ $$= \lim_{h \to 0} \frac{\sqrt{1 + (2 + h)^3}}{1}$$ $$= \boxed{3}.$$

And we are finished. Hope this helped!

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You should recognize that for well behaved $f(x)$ and fixed $x$: $$\frac{1}{h}\int^{x+h}_x f(t)dt=\frac{1}{h}\left[f(x)h+R_h(x) \right] $$ where $\lim_{h\to 0}R_h(x)/h=0$. So: $$ \lim_{h\to 0}\frac{1}{h}\int^{x+h}_x f(t)dt=f(x) $$

An alternate way of showing this is to observe that for a well behaved function on an interval containing $[x,x+h]$: $$ h \times \min_{t\in[x,x+h]}f(t)\le\int^{x+h}_x f(t)dt\le h \times \max_{t\in[x,x+h]}f(t) $$ and when $h\to 0$ we have $\min_{t\in[x,x+h]}f(t)\to \max_{t\in[x,x+h]}f(t)\to f(x)$.

Hence as $f(x)=\sqrt{1+t^3}$ is well behaved around $t=2$: $$ \lim_{h\to 0}\frac{1}{h}\int^{2+h}_2\sqrt{1+t^3}dt=\left.\sqrt{1+t^3}\right|_{t=2}=3 $$

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  • $\begingroup$ Why $R_h(x)$ tends to 0? $\endgroup$ – MonsieurGalois Jun 24 '16 at 4:47
  • $\begingroup$ There are a number of ways of showing this, as I have specified that $f(x)$ be well behaved we may take that to mean that $f(x)$ is differentiable on an interval containing $[x,x+h]$ when the given property follows from the Lagrange form of the remainder in the Taylor series up to the zeroth order term. $\endgroup$ – Conrad Turner Jun 24 '16 at 5:57
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By the Mean Value Theorem for Definite Integrals, https://en.wikipedia.org/wiki/Mean_value_theorem#First_Mean_Value_Theorem_for_Definite_Integrals,

$\displaystyle\int_2^{2+h}\sqrt{1+t^3}dt=\left(\sqrt{1+c^3}\right)h$ for some c between 2 and $2+h$,

so $\displaystyle\lim_{h\to 0}\frac{1}{h}\int^{2+h}_2\sqrt{1+t^3}dt=\lim_{c\to 2}\sqrt{1+c^3}=\sqrt{9}=3$.


$\textbf{Alternate solution:}$

Let $\displaystyle G(x)=\int_2^x \sqrt{1+t^3}dt,\;$ so $G^{\prime}(x)=\sqrt{1+x^3}$ by the Fundamental Theorem of Calculus.

Then $\displaystyle\lim_{h\to 0}\frac{1}{h}\int^{2+h}_2\sqrt{1+t^3}dt=\lim_{h\to 0}\frac{G(2+h)-G(2)}{h}=G^{\prime}(2)=\sqrt{1+2^3}=3$

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  • $\begingroup$ How you get the $\sqrt{1+c^3}h$? $\endgroup$ – MonsieurGalois Jun 24 '16 at 4:45
  • $\begingroup$ I have added some explanation to my answer. $\endgroup$ – user84413 Jun 24 '16 at 20:29
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You need to use Fundamental Theorem of Calculus. Let $$F(x) = \int_{2}^{x}\sqrt{1 + t^{3}}\,dt = \int_{2}^{x}f(t)\,dt$$ where $f(x) = \sqrt{1 + x^{3}}$ and we need to calculate the limit $$\lim_{h \to 0}\frac{F(2 + h)}{h}$$ Note that $F(2) = 0$ and hence $$\lim_{h \to 0}\frac{F(2 + h)}{h} = \lim_{h \to 0}\frac{F(2 + h) - F(2)}{h} = F'(2)$$ provided the limit exists.

By Fundamental Theorem of Calculus we know that $F'(x)$ exists if the function $f(x) = \sqrt{1 + x^{3}}$ is continuous at $x$ and then $F'(x) = f(x)$. Thus $F'(2) = f(2) = 3$.

The use of L'Hospital's Rule is unnecessary and is a roundabout way to evaluate this limit.

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