Midterm 1 Solutions

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Math 2300 – Calculus II – University of Colorado

Fall 2010 – Midterm I – Solutions

  1. (26 points) Consider the integral \displaystyle\int _{0}^{{\sqrt{2}}}\sqrt{2-x^{2}}\, dx

    1. (6 points) Draw a picture of the area that this integral represents. Evaluate this integral any way you like.

      Solution: The graph of the function f(x)=\sqrt{2-x^{2}} for x in [0,\sqrt{2}] represents the upper right quarter (first quadrant) of the circle centered at the origin and of radius r=\sqrt{2}. We now want to compute the integral:

      Solution 1: Since the integrand f is positive on [0,\sqrt{2}], the integral \int _{0}^{{\sqrt{2}}}\sqrt{2-x^{2}}\, dx represents the area under the graph of f and above the x-axis; this is exactly the area of a quarter of the disc, which is \frac{1}{4}\pi r^{2}=\frac{2\pi}{4}=\frac{\pi}{2}.

      Solution 2: We make the trig substitution, and write x in [-\sqrt{2},\sqrt{2}] as x=\sqrt{2}\sin t, with t in [\frac{-\pi}{2},\frac{\pi}{2}]. Then the radical becomes: \sqrt{2-x^{2}}=\sqrt{2-2\sin^{2}t}=\sqrt{2\cos^{2}t}=\sqrt{2}\cos t. The integration limits for t are t=\sin^{{-1}}(0)=0 and t=\sin^{{-1}}(1)=\frac{\pi}{2}, and dx=\sqrt{2}\cos t\, dt. So:

      \displaystyle\int _{0}^{{\sqrt{2}}}\sqrt{2-x^{2}}\, dx \displaystyle=\int _{{0}}^{{\pi/2}}{\sqrt{2}\cos t\sqrt{2}\cos t\, dt}
      \displaystyle=\int _{{0}}^{{\pi/2}}{2\cos^{2}t\, dt}
      \displaystyle=\int _{{0}}^{{\pi/2}}{[1+\cos(2t)]\, dt}
      \displaystyle=\left.\left(t+\frac{\sin(2t)}{2}\right)\right\rvert _{0}^{{\pi/2}}
      \displaystyle\approx 1.57
    2. (6 points) Compute Left(2) and Mid(2). Show your work. An answer that appears to have come from a calculator will receive 0 points.


      \displaystyle\textsc{Left}(2) \displaystyle=\Delta x\left[f(x_{0})+f(x_{1})\right]
      \displaystyle\approx 1.86
      \displaystyle\textsc{Mid}(2) \displaystyle=\Delta x\left[f(\frac{x_{0}+x_{1}}{2})+f(\frac{x_{1}+x_{2}}{2})\right]
      \displaystyle\approx 1.63
    3. (4 points) The formula for error in an approximation is given by

      \textsc{Error }=\textsc{Actual Value }-\textsc{ Approximation}.

      Compute the errors in Left(2) and Mid(2).


      \textsc{Error Left}(2)\approx 1.57-1.86=-0.29
      \textsc{Error Mid}(2)\approx 1.57-1.62=-0.05
    4. (6 points) Estimate the errors in Left(20) and Mid(20) without actually computing these values. Again, an answer that appears to have come from a calculator will receive 0 points.


      \textsc{Error Left}(20)\approx-0.29/10=-0.029
      \textsc{Error Mid}(20)\approx-0.05/100=-0.0005
    5. (4 points) Order the following approximations from least to greatest for any integer n\ge 2.



      The function f is decreasing, so \textsc{Right}(n)<\displaystyle\int _{0}^{{\sqrt{2}}}\sqrt{2-x^{2}}\, dx<\textsc{Left}(n).

      The function f is concave down, so \textsc{Trap}(n)<\displaystyle\int _{0}^{{\sqrt{2}}}\sqrt{2-x^{2}}\, dx<\textsc{Mid}(n).

      Since Mid(n) and Trap(n) are generally closer estimates than Left(n) and Right(n), we can order:

      \textsc{Right}(n)<\textsc{Trap}(n)<\displaystyle\int _{0}^{{\sqrt{2}}}\sqrt{2-x^{2}}\, dx<\textsc{Mid}(n)<\textsc{Left}(n)
  2. (12 points)

    1. (8 points) Prove that

      \int _{{0}}^{{\pi/2}}\sin^{n}x\, dx=\frac{n-1}{n}\int _{0}^{{\pi/2}}\sin^{{n-2}}x\, dx

      Solution: Start by rewriting:

      I_{n}=\int _{{0}}^{{\pi/2}}\sin^{n}x\, dx=\int _{{0}}^{{\pi/2}}\sin^{{n-1}}x\sin x\, dx

      Use an integration by parts with u=\sin^{{n-1}}x and dv=\sin x\, dx. Then du=(n-1)\sin^{{n-2}}x\cos x\, dx and v=-\cos x, so:

      \displaystyle I_{n} \displaystyle=\int _{{0}}^{{\pi/2}}\sin^{{n-1}}x\sin x\, dx
      \displaystyle=\left.\left[-\sin^{{n-1}}x\cos x\right]\right\rvert _{0}^{{\pi/2}}+(n-1)\int _{{0}}^{{\pi/2}}\sin^{{n-2}}x\cos^{2}x\, dx
      \displaystyle=0+(n-1)\int _{{0}}^{{\pi/2}}\sin^{{n-2}}x(1-\sin^{2}x)\, dx\text{\qquad(multiply out)}
      \displaystyle=(n-1)\int _{{0}}^{{\pi/2}}\sin^{{n-2}}x\, dx-(n-1)\int _{{0}}^{{\pi/2}}\sin^{n}x\, dx

      Bring \displaystyle-(n-1)\int _{{0}}^{{\pi/2}}\sin^{n}x\, dx to the left side and get:

      \int _{{0}}^{{\pi/2}}\sin^{n}x\, dx-(n-1)\int _{{0}}^{{\pi/2}}\sin^{n}x\, dx=(n-1)\int _{{0}}^{{\pi/2}}\sin^{{n-2}}x\, dx

      Adding the terms together on the left:

      n\int _{{0}}^{{\pi/2}}\sin^{n}x\, dx=(n-1)\int _{{0}}^{{\pi/2}}\sin^{{n-2}}x\, dx

      Dividing both sides by the coefficient n, it follows immediately that:

      \int _{{0}}^{{\pi/2}}\sin^{n}x\, dx=\frac{n-1}{n}\int _{{0}}^{{\pi/2}}\sin^{{n-2}}x\, dx
    2. (4 points) Use the formula in (a) to evaluate \displaystyle\int _{0}^{{\pi/2}}\sin^{3}x\, dx.

      Solution: We apply the formula at (a) for n=3:

      \displaystyle\int _{0}^{{\pi/2}}\sin^{3}x\, dx \displaystyle=\frac{2}{3}\int _{0}^{{\pi/2}}\sin x\, dx
      \displaystyle=\frac{2}{3}[-\cos x]|_{0}^{{\pi/2}}=\frac{2}{3}[0-(-1)]
  3. (12 points)

    1. (6 points) Show that

      \int _{0}^{\infty}x^{2}e^{{-x^{2}}}\, dx=\frac{1}{2}\int _{0}^{\infty}e^{{-x^{2}}}\, dx

      Solution: We start by setting up the limit for the improper integral:

      \int _{0}^{\infty}x^{2}e^{{-x^{2}}}\, dx=\lim _{{b\to\infty}}{\int _{0}^{b}x^{2}e^{{-x^{2}}}\, dx}

      We use integration by parts, with u=x and dv=xe^{{-x^{2}}}dx. Then du=dx and (using a substitution w=x^{2}) v=\int{xe^{{-x^{2}}}}=-\frac{1}{2}e^{{-x^{2}}}.

      \displaystyle\lim _{{b\to\infty}}{\int _{0}^{b}x^{2}e^{{-x^{2}}}\, dx} \displaystyle= \displaystyle\lim _{{b\to\infty}}\left.\left[-\frac{1}{2}xe^{{-x^{2}}}\right]\right\rvert _{0}^{b}+\frac{1}{2}\lim _{{b\to\infty}}{\int _{0}^{b}e^{{-x^{2}}}\, dx}
      \displaystyle= \displaystyle\lim _{{b\to\infty}}\left[-\frac{1}{2}be^{{-b^{2}}}\right]+\frac{1}{2}\lim _{{b\to\infty}}{\int _{0}^{b}e^{{-x^{2}}}\, dx}


      \lim _{{b\to\infty}}\left[-\frac{1}{2}be^{{-b^{2}}}\right]=-\frac{1}{2}\lim _{{b\to\infty}}\left[\frac{b}{e^{{b^{2}}}}\right]

      Since both top and bottom of the fraction go to \infty when b\to\infty, we can use L'Hôpital's rule. Calculate the limit for the ratio of the derivatives:

      \lim _{{b\to\infty}}\left[\frac{1}{2be^{{b^{2}}}}\right]=0

      Hence the original limit is zero, which means that:

      \lim _{{b\to\infty}}{\int _{0}^{b}x^{2}e^{{-x^{2}}}\, dx}=\frac{1}{2}\lim _{{b\to\infty}}{\int _{0}^{b}e^{{-x^{2}}}\, dx}
    2. (6 points) Show that

      \int _{0}^{2}e^{{w^{2}}}\, dw=\int _{0}^{1}2e^{{4x^{2}}}\, dx

    Solution: Start with the right side: make a substitution that will transform 4x^{2} into w^{2}. That is, w=2x. Then dw=2dx, and the new integration limits are w(0)=0 and w(1)=2.

    \int _{0}^{1}2e^{{4x^{2}}}\, dx=\int _{0}^{2}2e^{{w^{2}}}\,\frac{dw}{2}=\int _{0}^{2}e^{{w^{2}}}\, dw
  4. (12 points) 

    1. (6 points) Let f(x) be a continuous function. Prove that

      \int _{0}^{a}f(x)\, dx=\int _{0}^{a}f(a-x)\, dx

      Solution: Start with the right side, for example, and make the substitution w=a-x. Then dw=dx and the new integration limits are w(0)=a and w(a)=0:

      \displaystyle\int _{0}^{a}f(a-x)\, dx \displaystyle=\int _{a}^{0}f(w)\,(-dw)
      \displaystyle=-\int _{a}^{0}f(w)\, dw
      \displaystyle=-\int _{0}^{a}f(w)\, dw
      \displaystyle=-\int _{0}^{a}f(x)\, dx
    2. (6 points) Let f^{{\prime}}(x) be a continuous function and let \displaystyle\lim _{{x\rightarrow\infty}}f(x)=0. Prove that

      \int _{0}^{{\infty}}f^{{\prime}}(x)\, dx=-f(0)


      \displaystyle\int _{0}^{{\infty}}f^{{\prime}}(x)\, dx \displaystyle=\lim _{{b\to\infty}}{\int _{0}^{b}f^{{\prime}}(x)\, dx}
      \displaystyle=\lim _{{b\to\infty}}[f(x)]\rvert _{0}^{b}
      \displaystyle=\lim _{{b\to\infty}}[f(b)-f(0)]
      \displaystyle=\lim _{{b\to\infty}}f(b)-f(0)
  5. (12 points) Assume \displaystyle\int _{1}^{\infty}f(x)\, dx converges and that a is a positive constant.

    1. (6 points) Does \displaystyle\int _{1}^{\infty}f(a+x)\, dx converge? Explain in 3 sentences or less.

      Solution 1: Make the substitution w=a+x. Then:

      \int _{1}^{\infty}f(a+x)\, dx=\int _{{a+1}}^{{a+\infty}}f(w)\, dw=\int _{{a+1}}^{{\infty}}f(w)\, dw

      So \int _{{a+1}}^{{\infty}}f(w)\, dw converges, since it differs only by a finite value \int _{{1}}^{{a+1}}f(w)\, dw from the original integral \int _{1}^{\infty}f(w)\, dw, which is known to be convergent.

      Solution 2: For a>, the transformation from f(x) to f(a+x) is a shift to the left (notice that the interval [1,\infty] is still contained in the domain of the new function). This horizontal shift does not affect the convergence of the integral, only the finite limit value.

    2. (6 points) Does \displaystyle\int _{1}^{\infty}(a+f(x))\, dx converge as well? Explain in 3 sentences or less.

      Solution 1:

      \int _{1}^{\infty}(a+f(x))\, dx=\int _{1}^{\infty}a\, dx+\int _{1}^{\infty}f(x)\, dx

      But \displaystyle\int _{1}^{\infty}a\, dx=\lim _{{b\to\infty}}\int _{1}^{b}a\, dx=\lim _{{b\to\infty}}{ab}=\infty. Since we known that \int _{1}^{\infty}f(x)\, dx is convergent, their sum will be blow up to \infty, making \int _{1}^{\infty}(a+f(x))\, dx divergent.

      Solution 2: The function a+f(x) is a shift up by a units. This will prevent the proper decay of f(x), and thus introduces an additional positive infinite area between the x-axis and the graph of the integrand, preventing it from converging. So \int _{1}^{\infty}(a+f(x))\, dx diverges.

  6. (12 points) Some values of the continuous, differentiable function g(x) are given in the table below.

    x 1 5/4 3/2 7/4 2
    g(x) 2 3 4 7 10
    1. (6 points) Estimate the integral \displaystyle\int _{1}^{4}\frac{g(t)}{t}\, dt using these data.

      Solution: We can use, for example, the trapezoidal sum for n=4:

      \displaystyle\textsc{TRAP}(4) \displaystyle=\frac{1}{4}\left[\frac{g(1)}{1}+2\frac{g(5/4)}{5/4}+2\frac{g(3/2)}{3/2}+2\frac{g(7/4)}{7/4}+\frac{g(2)}{2}\right]
    2. (6 points) Estimate the integral \displaystyle\int _{1}^{4}g(t)\, dt using these data.


      \displaystyle\textsc{TRAP}(4) \displaystyle= \displaystyle\frac{1}{4}[g(1)+2g(5/4)+2g(3/2)+2g(7/4)+g(2)]
      \displaystyle= \displaystyle\frac{1}{4}[2+2\cdot 3+2\cdot 4+2\cdot 7+10]
  7. (14 points) The Law of Mass Action tells us that the time, T, taken by a chemical to create a quantity p of the product (in molecules) is given by

    T=\int _{0}^{{p}}\frac{k\, dx}{(a-x)(b-x)}

    where a and b are initial quantities of the two ingredients used to make the product, and k is a positive constant. Suppose 0<a<b.

    Let's first calculate \displaystyle\int _{0}^{p}\frac{k\, dx}{(a-x)(b-x)}=\int _{0}^{p}\frac{k\, dx}{(x-a)(x-b)} using partial fractions:


    We want to find expressions for A and B (that will depend on the constants k, a and b) so that:


    Plug in x=b and obtain B(b-a)=k, i.e. B=\frac{k}{b-a}.

    Plug in x=a and obtain A(a-b)=k, i.e. A=\frac{k}{a-b}=-\frac{k}{b-a}.


    \displaystyle\int\frac{k\, dx}{(x-a)(x-b)} \displaystyle=-\frac{k}{b-a}\int\frac{dx}{x-a}+\frac{k}{b-a}\int\frac{dx}{x-b}
    \displaystyle=-\frac{k}{b-a}\ln\lvert x-a\rvert+\frac{k}{b-a}\ln\lvert x-b\rvert
    \displaystyle=\frac{k}{b-a}\big(\ln\lvert x-b\rvert-\ln\lvert x-a\rvert\big)
    \displaystyle\int\frac{k\, dx}{(x-a)(x-b)} \displaystyle=\frac{k}{b-a}\left.\left[\ln\left\lvert\frac{x-b}{x-a}\right\rvert\right]\right\rvert _{0}^{p}
    1. (8 points) Find the time taken to make a quantity p=a/2 of the product.


      \displaystyle T(a/2) \displaystyle= \displaystyle\frac{k}{b-a}\ln\left(\frac{a}{b}\left\lvert\frac{a/2-b}{a/2-a}\right\rvert\right)=\frac{k}{b-a}\ln\left(\frac{2b-a}{b}\right)
    2. (6 points) What happens to T as p\rightarrow a?


      \displaystyle\lim _{{p\to a}}{T(p)} \displaystyle= \displaystyle\lim _{{p\to a}}{\frac{k}{b-a}\ln\left(\frac{a}{b}\left\lvert\frac{p-b}{p-a}\right\rvert\right)}

      Since \displaystyle\lim _{{p\to a}}{\frac{p-b}{p-a}}=\pm\infty, it means that \displaystyle\lim _{{p\to a}}{\ln\left(\frac{a}{b}\left\lvert\frac{p-b}{p-a}\right\rvert\right)}=\infty. So: \displaystyle\lim _{{p\to a}}{T(p)}=\infty. In other words, a is the equilibrium quantity.

© 2011 Jason B. Hill. All Rights Reserved.