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Solution of Term Exam 3

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Problem 1.

1. Compute $$\int_0^1\sqrt{x} dx$$.
2. Compute $$\int_0^\pi\sin x dx$$.
3. For $x\geq 0$, compute $$\frac{d}{dx}\int_{x^3}^{157}\sqrt{t} dt$$.

Solution. (Graded by Shay Fuchs)

1. To use the second fundamental theorem of calculus we are looking for a function $f$ for which $f'=\sqrt{x}=x^{1/2}$. The most obvious guess is $f(x)=x^{3/2}$, but this is off by a factor of $3/2$, for $(x^{3/2})'=\frac32=x^{1/2}$. So a good answer would be $f(x)=\frac23x^{3/2}$. Now
   $$\int_0^1\sqrt{x} dx =\int_0^1f'(x)dx =\left.f\right\vert _0^1 =f(1)-f(0) =\frac231^{3/2}-\frac230^{3/2} =\frac23.$$

2. Likewise choose $f(x)=-\cos x$ to get $f'(x)=\sin x$, and so using the second fundamental theorem of calculus,
   $$\int_0^\pi\sin x dx =\int_0^\pi f'(x)dx =f(\pi)-f(0) =-\cos\pi-(-\cos 0) =2.$$

3. Let $g(y)=\int_{157}^y\sqrt{t} dt$ and let $f(x)=x^3$. Using the first fundamental theorem of calculus, $g'(y)=\sqrt{y}$. So using the chain rule,
   

   $$\frac{d}{dx}\int_{x^3}^{157}\sqrt{t} dt = \frac{d}{dx}\left(-\int_{157}^{x^3}\sqrt{t} dt\right) = -(g\circ f)'$$

   $$= -g'(f(x))f'(x) = -\sqrt{x^3}3x^2 = -3x^{7/2}.$$

   
   

Problem 2.

1. Perhaps using L'Hôpital's law, compute $$\lim_{x\to 0}\frac{\sin x}{x}$$ and $$\lim_{x\to 0}\frac{1-\cos x}{x^2}$$.
2. Use these results to give educated guesses for the values of $\sin 0.1$ and $\cos 0.1$ (no calculators, please).

Solution. (Graded by Shay Fuchs)

1. $\sin x$ is differentiable at 0 and $\sin 0=0$. So
   $$\lim_{x\to 0}\frac{\sin x}{x} = \lim_{x\to 0}\frac{\sin x - \sin 0}{x} = \sin' 0 = \cos 0 = 1.$$
   (L'Hôpital's law also works and gives the same result).

   The second limit is of the form $\frac00$ so we can use L'Hôpital:

   $$\lim_{x\to 0}\frac{1-\cos x}{x^2} = \lim_{x\to 0}\frac{(1-\cos x)... ...im_{x\to 0}\frac{\sin x}{2x} = \frac12\lim_{x\to 0}\frac{\sin x}{x} = \frac12.$$

2. $0.1$ is close to 0, so $\frac{\sin 0.1}{0.1}\sim\lim_{x\to 0}\frac{\sin x}{x}=1$. Multiplying both sides by $0.1$ we get $\sin 0.1\sim 0.1$.

   Likewise, $\frac{1-\cos 0.1}{0.1^2}\sim\frac12$, so $1-\cos 0.1\sim\frac120.1^2=0.005$, so $\cos 0.1\sim0.995$.

Problem 3.

1. State the ``one partition for every $\epsilon$'' criterion of the integrability of a bounded function $f$ defined on an interval $[a,b]$.
2. Let $f$ be an increasing function on $[0,1]$ and let $P_n$ be the partition defined by $t_i=i/n$, for $i=0,1,\ldots,n$. Write simple formulas for $U(f,P_n)$ and for $L(f,P_n)$.
3. Under the same conditions, write a very simple formula for $U(f,P_n)-L(f,P_n)$.
4. Prove that an increasing function on $[0,1]$ is integrable.

Solution. (Graded by Derek Krepski)

1. A bounded function $f$ defined on an interval $[a,b]$ is integrable iff for every $\epsilon>0$ there is a partition $P$ of $[a,b]$ for which $U(f,P)-L(f,P)<\epsilon$.
2. As $f$ is increasing, $m^{P_n}_i=\inf_{[t_{i-1},t_i]}f(x)=f(t_{i-1})$ and $M^{P_n}_i=\sup_{[t_{i-1},t_i]}f(x)=f(t_i)$. Thus
   $$U(f,P_n) = \sum_{i=1}^n M^{P_n}_i(t_i-t_{i-1}) = \sum_{i=1}^n f(t_i)\frac1n = \frac1n\sum_{i=1}^n f(\frac{i}{n}).$$
   Likewise, $L(f,P_n)=\frac1n\sum_{i=1}^n f(t_{i-1})=\frac1n\sum_{i=1}^n f(\frac{i-1}{n})$.

3. $$U(f,P_n)-L(f,P_n) = \frac1n\sum_{i=1}^n f(t_i) - \frac1n\sum_{i=1}^n f(t_{i-1}) = \frac1n\sum_{i=1}^n (f(t_i)-f(t_{i-1}))$$
   using telescopic summation this is
   $$= \frac1n(f(t_n)-f(t_0)) = \frac1n(f(1)-f(0)).$$

4. Since $f$ is increasing, $f$ is bounded (with upper bound $f(1)$ and lower bound $f(0)$). So using the criterion of part 1, to show that $f$ is integrable it is enough to show that for every $\epsilon>0$ there is a partition $P$ of $[0,1]$ for which $U(f,P)-L(f,P)<\epsilon$. Indeed, let $\epsilon>0$ be given. Choose $n$ so big so that $\frac1n(f(1)-f(0))<\epsilon$, and then the partition $P=P_n$ of before satisfies $U(f,P_n)-L(f,P_n)=\frac1n(f(1)-f(0))<\epsilon$, as required.

Problem 4.

1. Show that the function $f(x)=3x-x^3$ is monotone on the interval $[-1,1]$.
2. Deduce that for every $c\in[-2,2]$ the equation $3x-x^3=c$ has a unique solution $x$ in the range $-1\leq x\leq 1$.
3. For $c\in[-2,2]$, let $g(c)$ be the unique $x$ in the range $-1\leq x\leq 1$ for which $3x-x^3=c$. Write a formula for $g'(c)$ and simplify it as much as you can. Your end result may still contain $g(c)$ in it, but not $f$, $f'$ or $g'$.

Solution. (Graded by Brian Pigott)

1. $f'(x)=3-3x^2=3(1-x^2)$. On $(-1,1)$ we know that $x^2<1$, so $f'(x)>0$. So $f$ is increasing on $[-1,1]$.
2. By the theorem about the existence of inverses of monotone functions, $f$ has an inverse on $[-1,1]$ and it is defined on $[f(-1),f(1)]=[-2,2]$. This precisely means that for $c\in[-2,2]$ the equation $3x-x^3=c$ (which defines $f^{-1}(c)$) has a unique solution with $x$ in the range $-1\leq x\leq 1$.
3. By the theorem about the derivative of an inverse function,
   $$g'(c)=\frac{1}{f'(g(c))}=\frac{1}{3(1-g(c)^2)}.$$

The results. 67 students took the exam; the average grade was 77.7 and the standard deviation was about 22.

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