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Solution of Term Exam 3

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Problem 1.

  1. Compute $ \displaystyle\int_0^1\sqrt{x} dx$.
  2. Compute $ \displaystyle\int_0^\pi\sin x dx$.
  3. For $ x\geq 0$, compute $ \displaystyle\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

    $\displaystyle \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,

    $\displaystyle \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,
    $ \displaystyle\frac{d}{dx}\int_{x^3}^{157}\sqrt{t} dt$ $\displaystyle =$ $\displaystyle \frac{d}{dx}\left(-\int_{157}^{x^3}\sqrt{t} dt\right)
= -(g\circ f)'$  
      $\displaystyle =$ $\displaystyle -g'(f(x))f'(x)
= -\sqrt{x^3}3x^2
= -3x^{7/2}.$  

Problem 2.

  1. Perhaps using L'Hôpital's law, compute $ \displaystyle\lim_{x\to 0}\frac{\sin x}{x}$ and $ \displaystyle\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

    $\displaystyle \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:

    $\displaystyle \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

    $\displaystyle 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. $\displaystyle 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

    $\displaystyle = \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,

    $\displaystyle 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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Dror Bar-Natan 2005-02-07