# Evaluate $\lim_{h\to 0}\frac{1}{h}\int^{x^2+h}_{x^2}\log(3+\arcsin t)\,dt$

I am concerned with how i might evaluate the statement in terms of $x$: $$\lim_{h\to0}[\frac{1}{h}\int^{x^2+h}_{x^2}\log[3+\arcsin({t})]\,dt$$

On first glance it appears to be just $\log[3+\arcsin({x^2})]$ But is this actually correct? Naturally, if $x$ was a parameter this statement would be true. But what if $x$ is variable?

It is fairly obvious, by first principles of differentiation that $\lim_{h\to0}[\frac{1}{h}\int^{(x+h)^2}_{x^2}\log[3+\arcsin({t})]\,dt$] is by the fundamental theorem of calculus just $2x\log[3+\arcsin({x^2})$. Is there a similar relationship that can be drawn between the problems?

Answer based on your first glance is correct. The problem is solved as a direct application of Fundamental Theorem of Calculus (FTC). Note that the variable of integration is $t$ and the variable of the limit operation is $h$ and hence the expression $x^{2}$ occurring in the integral is a constant and may well be denoted by $a$. So with $a = x^{2}$ we have to find the limit $$L = \lim_{h \to 0}\frac{1}{h}\int_{a}^{a + h}\log(3 + \arcsin t)\,dt\tag{1}$$ Now let's note that if $$F(x) = \int_{a}^{x}f(t)\,dt\tag{2}$$ then via FTC we have $F'(c) = f(c)$ and in particular $F'(a) = f(a)$. Note also by definition of derivative that $$f(a) = F'(a) = \lim_{h \to 0}\frac{F(a + h) - F(a)}{h} = \lim_{h \to 0}\frac{1}{h}\int_{a}^{a + h}f(t)\,dt\tag{3}$$ and now looking at $(1)$ and $(3)$ we can easily see that $$L = f(a) = \log(3 + \arcsin a) = \log(3 + \arcsin x^{2})$$

Consider $$A=\frac{\int^{x^2+h}_{x^2}\log\left(3+\sin^{-1}(t)\right))\,dt }h$$ which is $\frac 00$. Apply L'Hospital with $$u=\int^{x^2+h}_{x^2}\log\left(3+\sin^{-1}(t)\right))\,dt \qquad , \qquad v=h$$ This leads to $$\frac {u'}{v'}=\frac {d}{dh}\int^{x^2+h}_{x^2}\log\left(3+\sin^{-1}(t)\right))\,dt$$ Now, use the fundamental theorem of calculus which will give you Jacky Chong's result.

If you have had $$u'=\frac {d}{dh}\int^{x^2+b(h)}_{x^2+a(h)}\log\left(3+\sin^{-1}(t)\right))\,dt$$ the result would have been $$\frac {u'}{v'}=b'(h) \log \left(3+\sin ^{-1}\left(x^2+b(h)\right)\right)-a'(h) \log \left(3+\sin ^{-1}\left(x^2+a(h)\right)\right)$$ and since $a(h)\to 0$ and $b(h)\to 0$ then $$A \to (b'(h)-a'(h))\log \left(3+\sin ^{-1}\left(x^2\right)\right)$$

Edit

Since you edited the post while I was typing, for the most general case where $$u'=\frac {d}{dh}\int^{b(x,h)}_{a(x,h)}f(t)\,dt$$ the result would have been $$A \to b'_h(x,h) f(b(x,h))-a'_h(x,h) f(a(x,h))$$ For the second case in the post, then $$u'= 2 (h+x) f\left((h+x)^2\right)$$ making $$A \to 2x\,f(x^2)$$

• I don't think this is correct. Sep 22, 2016 at 5:06
• @JackyChong. Could you elaborate, please ? Sep 22, 2016 at 5:15
• I think the value of $A$ should be $f(x^{2})$ and not $2xf(x^{2})$. The reason is that the $x^{2}$ here is just acting as a constant and is independent of the variable of limit $h$ and variable of integration $t$. Sep 22, 2016 at 9:02
• @ParamanandSingh. May I disagree ? If we consider $$u'=\frac {d}{dh}\int^{(x+h)^k}_{x^k}f(t)\,dt$$ the FTC would lead to $$u'=k (x+h)^{k-1} f\left((x+h)^k\right)$$ and then $A=k\,x^{k-1}f(x^k)$. For sure I may be wrong ! Sep 22, 2016 at 9:23
• agree fully with your last comment. your answer works when limits of integration are $x^2$ and $(x +h)^{2}$. but if the limits are $x^2$ and $x^2 + h$ then you need to drop the factor $2x$. Sep 22, 2016 at 9:25

Hint:

\begin{align} \frac{1}{h} \int^{x^2+h}_{x^2} \log[3+\arcsin(t)]\ dt \approx \frac{\log[3+\arcsin(x^2)]}{h} \int^{x^2+h}_{x^2} \ dt= \log[3+\arcsin(x^2)]. \end{align}

• It would have better to use the FTC here rather than using symbol $\approx$. Your approach is better expressed by use of mean value theorem for integrals. Thus $$\frac{1}{h} \int^{x^2+h}_{x^2} \log[3+\arcsin(t)]\ dt = \log[3+\arcsin(c)]$$ where $c$ lies between $x^{2}$ and $x^{2} + h$. When $h \to 0$ then $c \to x^{2}$ and answer follows by continuity of $\log(3 + \arcsin t)$. Sep 22, 2016 at 9:08