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Midterm 2 Solutions

Table of Contents

Question 1

True or false:

  1. The convergent sequences {an}\set{a_n} and {bn}\set{b_n} differ in their first 100100 terms, but an=bna_n = b_n for n>100n > 100. It follows that lim⁡n→∞an=lim⁡n→∞bn\displaystyle \lim_{n\to\infty} a_n = \lim_{n\to\infty} b_n.
  2. If the sequence {an}\set{a_n} converges, then the sequence {(−1)nan}\set{\p{-1}^n a_n} converges.
Solution.
(1) is true and (2) is false.
  1. If two convergent sequences differ for only finitely many terms, then they have the same limit.
  2. If an=1a_n = 1, then ana_n converges to 11, but (−1)nan=(−1)n\p{-1}^na_n = \p{-1}^n diverges.

Question 2

  1. If an infinite series converges conditionally, then the value of the series depends on the order of the summation. Give one example of such an infinite series and explain.
  2. Give an example of an antiderivative that can be found using a trigonometric reduction formula.
Solution.
  1. ∑n=1∞(−1)nn\displaystyle \sum_{n=1}^\infty \frac{\p{-1}^n}{n} converges by the alternating series test, but ∑n=1∞1n\displaystyle \sum_{n=1}^\infty \frac{1}{n} diverges because of the pp-series theorem.
  2. A reduction formula can be used for integrals involving powers of sin⁡x\sin{x} and cos⁡x\cos{x}, e.g., ∫sin⁡2xcos⁡4x dx\displaystyle \int \sin^2{x}\cos^{4}{x} \,\diff{x}.

Question 3

Evaluate the integrals using the specified technique.

  1. ∫x2+16x4 dx\displaystyle \int \frac{\sqrt{x^2 + 16}}{x^4} \,\diff{x} (trig substitution)
  2. ∫excos⁡x dx\displaystyle \int e^x\cos{x} \,\diff{x} (integration by parts)
Solution.

  1. Set x=4tan⁡θ  ⟹  dx=4sec⁡2θ dθx = 4\tan\theta \implies \diff{x} = 4\sec^2\theta \,\diff\theta. Then

    ∫x2+16x4 dx=∫4sec⁡θ256tan⁡4θ⋅4sec⁡2θ dθ=116∫sec⁡3θtan⁡4θ dθ=116∫1cos⁡3θ⋅cos⁡4θsin⁡4θ dθ=116∫cos⁡θsin⁡4θ dθ.\begin{aligned} \int \frac{\sqrt{x^2 + 16}}{x^4} \,\diff{x} &= \int \frac{4\sec\theta}{256\tan^4\theta} \cdot 4\sec^2\theta \,\diff\theta \\ &= \frac{1}{16} \int \frac{\sec^3\theta}{\tan^4\theta} \,\diff\theta \\ &= \frac{1}{16} \int \frac{1}{\cos^3\theta} \cdot \frac{\cos^4\theta}{\sin^4\theta} \,\diff\theta \\ &= \frac{1}{16} \int \frac{\cos\theta}{\sin^4\theta} \,\diff\theta. \end{aligned}

    If we set v=sin⁡θv = \sin\theta, then dv=cos⁡θ dθ\diff{v} = \cos\theta \,\diff\theta and we get

    116∫cos⁡θsin⁡4θ dθ=116∫dvv4=−1481v3=−1481sin⁡3θ+C.\begin{aligned} \frac{1}{16} \int \frac{\cos\theta}{\sin^4\theta} \,\diff\theta &= \frac{1}{16} \int \frac{\diff{v}}{v^4} \\ &= -\frac{1}{48} \frac{1}{v^3} \\ &= -\frac{1}{48} \frac{1}{\sin^3\theta} + C. \end{aligned}

    To undo the substitution,

    x2+16=16tan⁡2θ+16=16sec⁡2θcos⁡2θ=16x2+161−sin⁡2θ=16x2+16sin⁡2θ=x2x2+16sin⁡3θ=(x2x2+16)3/2=x3(x2+16)3/2.\begin{aligned} x^2 + 16 &= 16\tan^2\theta + 16 \\ &= 16\sec^2\theta \\ \cos^2\theta &= \frac{16}{x^2 + 16} \\ 1 - \sin^2\theta &= \frac{16}{x^2 + 16} \\ \sin^2\theta &= \frac{x^2}{x^2 + 16} \\ \sin^3\theta &= \p{\frac{x^2}{x^2 + 16}}^{3/2} \\ &= \frac{x^3}{\p{x^2 + 16}^{3/2}}. \end{aligned}

    So in the end, we get

    −(x2+16)3/248x3+C.\boxed{-\frac{\p{x^2 + 16}^{3/2}}{48x^3} + C}.

  2. Integrate by parts with u=exu = e^x and dv=cos⁡x dx\diff{v} = \cos{x} \,\diff{x}, which gives

    du=ex dxandv=sin⁡x.\diff{u} = e^x \,\diff{x} \quad\text{and}\quad v = \sin{x}.

    So, the integral becomes

    ∫excos⁡x dx=exsin⁡x−∫exsin⁡x dx.\int e^x\cos{x} \,\diff{x} = e^x\sin{x} - \int e^x\sin{x} \,\diff{x}.

    Integrate by parts again with u=exu = e^x and dv=sin⁡x dx\diff{v} = \sin{x} \,\diff{x}, so

    du=ex dxandv=−cos⁡x,\diff{u} = e^x \,\diff{x} \quad\text{and}\quad v = -\cos{x},

    which turns the integral into

    ∫excos⁡x dx=exsin⁡x−(−excos⁡x+∫excos⁡x dx)=exsin⁡x+excos⁡x−∫excos⁡x dx  ⟹  2∫excos⁡x dx=exsin⁡x+excos⁡x+C\begin{aligned} \int e^x\cos{x} \,\diff{x} &= e^x\sin{x} - \p{-e^x\cos{x} + \int e^x\cos{x} \,\diff{x}} \\ &= e^x\sin{x} + e^x\cos{x} - \int e^x\cos{x} \,\diff{x} \\ \implies 2\int e^x\cos{x} \,\diff{x} &= e^x\sin{x} + e^x\cos{x} + C \end{aligned}

    by solving for the integral. Dividing by 22 and absorbing constants into CC, we get

    ∫excos⁡x dx=ex2(sin⁡x+cos⁡x)+C.\int e^x\cos{x} \,\diff{x} = \boxed{\frac{e^x}{2}\p{\sin{x} + \cos{x}} + C}.

Question 4

Determine whether each integral is convergent or divergent. Evaluate those that are convergent.

  1. ∫1∞xx3+1 dx\displaystyle \int_1^\infty \frac{x}{x^3 + 1} \,\diff{x}
  2. ∫01dx1−x2\displaystyle \int_0^1 \frac{\diff{x}}{\sqrt{1 - x^2}}
Solution.

  1. We can factor sums of cubes, so x3+1=(x+1)(x2−x+1)x^3 + 1 = \p{x + 1}\p{x^2 - x + 1}. We can then do partial fractions:

    x(x+1)(x2−x+1)=Ax+1+Bx+Cx2−x+1  ⟹  x=A(x2−x+1)+(Bx+C)(x+1)  ⟹  0=Ax2−Ax+A+Bx2+Bx+Cx+C−x  ⟹  0=(A+B)x2+(−A+B+C−1)x+A+C\begin{aligned} \frac{x}{\p{x + 1}\p{x^2 - x + 1}} &= \frac{A}{x + 1} + \frac{Bx + C}{x^2 - x + 1} \\ \implies x &= A\p{x^2 - x + 1} + \p{Bx + C}\p{x + 1} \\ \implies 0 &= Ax^2 - Ax + A + Bx^2 + Bx + Cx + C - x \\ \implies 0 &= \p{A + B}x^2 + \p{-A + B + C - 1}x + A + C \end{aligned}

    For the right-hand side to be 00 for every xx, all the coefficients need to be 00. So, we get the system

    {A+B=0−A+B+C−1=0A+C=0.\begin{cases} A + B &= 0 \\ -A + B + C - 1 &= 0 \\ A + C &= 0. \end{cases}

    If we subtract the first equation from the second equation, we get −2A+C−1=0-2A + C - 1 = 0, so if we combine that with the third equation,

    3C−1=0  ⟹  C=13.3C - 1 = 0 \implies C = \frac{1}{3}.

    Then

    A+C=0  ⟹  A=−13andA+B=0  ⟹  B=13,A + C = 0 \implies A = -\frac{1}{3} \quad\text{and}\quad A + B = 0 \implies B = \frac{1}{3},

    so the decomposition is

    xx3+1=13(−1x+1+x+1x2−x+1).\frac{x}{x^3 + 1} = \frac{1}{3} \p{-\frac{1}{x + 1} + \frac{x + 1}{x^2 - x + 1}}.

    The integral of the first term in the parentheses is just −ln⁡∣x+1∣+C-\ln\abs{x + 1} + C, so we need to worry about the second term. We want to use the substitution u=x2−x+1  ⟹  du=2x−1 dxu = x^2 - x + 1 \implies \diff{u} = 2x - 1 \,\diff{x}, so we need to separate the second term based on that:

    x+1x2−x+1⋅22=122x+2x2−x+1=122x−1+3x2−x+1=122x−1x2−x+1+321x2−x+1.\begin{aligned} \frac{x + 1}{x^2 - x + 1} \cdot \frac{2}{2} = \frac{1}{2} \frac{2x + 2}{x^2 - x + 1} &= \frac{1}{2} \frac{2x - 1 + 3}{x^2 - x + 1} \\ &= \frac{1}{2} \frac{2x - 1}{x^2 - x + 1} + \frac{3}{2} \frac{1}{x^2 - x + 1}. \end{aligned}

    Using the uu-substitution, the integral of the first term here is 12ln⁡∣x2−x+1∣+C\frac{1}{2}\ln\abs{x^2 - x + 1} + C. So, we just need to work on the last term. If we complete the square, x2−x+1=(x−12)2+34x^2 - x + 1 = \p{x - \frac{1}{2}}^2 + \frac{3}{4}, so the substitution x−12=32tan⁡θx - \frac{1}{2} = \frac{\sqrt{3}}{2}\tan\theta works:

    32∫dxx2−x+1=32∫dx(x−12)2+34=32∫32sec⁡2θ34sec⁡2θ dθ=3∫ dθ=3θ+C=3arctan⁡(x−1232)+C=3arctan⁡(2x−13)+C.\begin{aligned} \frac{3}{2} \int \frac{\diff{x}}{x^2 - x + 1} &= \frac{3}{2} \int \frac{\diff{x}}{\p{x - \frac{1}{2}}^2 + \frac{3}{4}} \\ &= \frac{3}{2} \int \frac{\frac{\sqrt{3}}{2}\sec^2\theta}{\frac{3}{4}\sec^2\theta} \,\diff\theta \\ &= \sqrt{3} \int \,\diff\theta \\ &= \sqrt{3} \theta + C \\ &= \sqrt{3} \arctan\p{\frac{x - \frac{1}{2}}{\frac{\sqrt{3}}{2}}} + C \\ &= \sqrt{3} \arctan\p{\frac{2x - 1}{\sqrt{3}}} + C. \end{aligned}

    Putting everything together,

    ∫xx3+1 dx=13(−ln⁡∣x+1∣+12ln⁡∣x2−x+1∣+3arctan⁡(2x−13))+C.\int \frac{x}{x^3 + 1} \,\diff{x} = \frac{1}{3}\p{-\ln\abs{x + 1} + \frac{1}{2}\ln\abs{x^2 - x + 1} + \sqrt{3} \arctan\p{\frac{2x - 1}{\sqrt{3}}}} + C.

    To finish the problem, we need to evaluate the integral at the bounds. The lower bound is given at x=1x = 1:

    13(−ln⁡2+3arctan⁡(13))=−ln⁡23+33π6=π3−6ln⁡218.\frac{1}{3}\p{-\ln{2} + \sqrt{3}\arctan\p{\frac{1}{\sqrt{3}}}} = -\frac{\ln{2}}{3} + \frac{\sqrt{3}}{3} \frac{\pi}{6} = \frac{\pi\sqrt{3} - 6\ln{2}}{18}.

    For the upper bound, we need to take a limit:

    lim⁡R→∞13(−ln⁡∣R+1∣+12ln⁡∣R2−R+1∣+3arctan⁡(2R−13))=lim⁡R→∞13(ln⁡∣R2−R+1R+1∣+3arctan⁡(2R−13))=13(ln⁡1+3π2)=π36.\begin{aligned} &\lim_{R\to\infty} \frac{1}{3}\p{-\ln\abs{R + 1} + \frac{1}{2}\ln\abs{R^2 - R + 1} + \sqrt{3} \arctan\p{\frac{2R - 1}{\sqrt{3}}}} \\ ={}& \lim_{R\to\infty} \frac{1}{3}\p{\ln\abs{\frac{\sqrt{R^2 - R + 1}}{R + 1}} + \sqrt{3} \arctan\p{\frac{2R - 1}{\sqrt{3}}}} \\ ={}& \frac{1}{3}\p{\ln{1} + \sqrt{3} \frac{\pi}{2}} \\ ={}& \frac{\pi\sqrt{3}}{6}. \end{aligned}

    So the final answer is

    ∫1∞xx3+1 dx=π36−π3−6ln⁡218=π3+3ln⁡29.\int_1^\infty \frac{x}{x^3 + 1} \,\diff{x} = \frac{\pi\sqrt{3}}{6} - \frac{\pi\sqrt{3} - 6\ln{2}}{18} = \boxed{\frac{\pi\sqrt{3} + 3\ln{2}}{9}}.

  2. We can integrate directly:

    ∫01dx1−x2=lim⁡t→1−∫0tdx1−x2=lim⁡t→1−arcsin⁡x∣0t=lim⁡t→1−arcsin⁡t=arcsin⁡1=π2.\begin{aligned} \int_0^1 \frac{\diff{x}}{\sqrt{1 - x^2}} = \lim_{t\to1^-} \int_0^t \frac{\diff{x}}{\sqrt{1 - x^2}} &= \lim_{t\to1^-} \Bigl. \arcsin{x} \Bigr\rvert_0^t \\ &= \lim_{t\to1^-} \arcsin{t} \\ &= \arcsin{1} \\ &= \boxed{\frac{\pi}{2}}. \end{aligned}

Question 5

Assume the function ff has an inverse on its domain.

  1. Let y=f−1(x)y = f^{-1}\p{x} and show that ∫f−1(x) dx=∫yf′(y) dy\displaystyle \int f^{-1}\p{x} \,\diff{x} = \int yf'\p{y} \,\diff{y}.
  2. Use (i) to prove that ∫f−1(x) dx=yf(y)−∫f(y) dy\displaystyle \int f^{-1}\p{x} \,\diff{x} = yf\p{y} - \int f\p{y} \,\diff{y}.
  3. Use (ii) to evaluate ∫ln⁡x dx\displaystyle \int \ln{x} \,\diff{x}.
Solution.

  1. If y=f−1(x)y = f^{-1}\p{x}, then x=f(y)  ⟹  dx=f′(y) dyx = f\p{y} \implies \diff{x} = f'\p{y} \,\diff{y}, so

    ∫f−1(x) dx=∫yf′(y) dy.\int f^{-1}\p{x} \,\diff{x} = \int yf'\p{y} \,\diff{y}.

  2. We can integrate by parts with u=yu = y and dv=f′(y) dy\diff{v} = f'\p{y} \,\diff{y}, which gives

    du=dyandv=f(y),\diff{u} = \diff{y} \quad\text{and}\quad v = f\p{y},

    so we get

    ∫f−1(x) dx=yf(y)−∫f(y) dy.\int f^{-1}\p{x} \,\diff{x} = yf\p{y} - \int f\p{y} \,\diff{y}.

  3. If y=f−1(x)=ln⁡xy = f^{-1}\p{x} = \ln{x}, then f(y)=eyf\p{y} = e^y, so

    ∫ln⁡x dx=yf(y)−∫f(y) dy=ln⁡x⋅eln⁡x−∫ey dy=xln⁡x−ey+C=xln⁡x−eln⁡x+C=xln⁡x−x+C.\begin{aligned} \int \ln{x} \,\diff{x} &= yf\p{y} - \int f\p{y} \,\diff{y} \\ &= \ln{x} \cdot e^{\ln{x}} - \int e^y \,\diff{y} \\ &= x\ln{x} - e^y + C \\ &= x\ln{x} - e^{\ln{x}} + C \\ &= \boxed{x\ln{x} - x + C}. \end{aligned}

Question 6

Determine whether the series is convergent or divergent. If it is convergent, find its sum (if possible).

  1. ∑n=1∞3n+14−n\displaystyle \sum_{n=1}^\infty 3^{n+1} 4^{-n}
  2. ∑n=2∞1n(ln⁡n)2\displaystyle \sum_{n=2}^\infty \frac{1}{n\p{\ln{n}}^2}
Solution.

  1. This is a geometric series:

    ∑n=1∞3n+14−n=∑n=1∞3⋅3n4n=3∑n=1∞(34)n=3⋅341−34=9.\sum_{n=1}^\infty 3^{n+1} 4^{-n} = \sum_{n=1}^\infty 3 \cdot \frac{3^n}{4^n} = 3 \sum_{n=1}^\infty \p{\frac{3}{4}}^n = 3 \cdot \frac{\frac{3}{4}}{1 - \frac{3}{4}} = \boxed{9}.

  2. If f(x)=1x(ln⁡x)2f\p{x} = \frac{1}{x\p{\ln{x}}^2}, then if x>e2x > e^2,

    f′(x)=−(ln⁡x)2−2xln⁡xx(x(ln⁡x)2)2=−(ln⁡x−2)ln⁡x(x(ln⁡x)2)2<0,f'\p{x} = -\frac{\p{\ln{x}}^2 - \frac{2x\ln{x}}{x}}{\p{x\p{\ln{x}}^2}^2} = -\frac{\p{\ln{x} - 2}\ln{x}}{\p{x\p{\ln{x}}^2}^2} < 0,

    so ff is continuous, non-negative, and decreasing on x>e2x > e^2. So, we can apply the integral test: let u=ln⁡x  ⟹  du=dxxu = \ln{x} \implies \diff{u} = \frac{\diff{x}}{x}, and we get

    ∫2∞dxx(ln⁡x)2=∫ln⁡2∞duu2.\int_2^\infty \frac{\diff{x}}{x\p{\ln{x}}^2} = \int_{\ln{2}}^\infty \frac{\diff{u}}{u^2}.

    Since ln⁡2>0\ln{2} > 0, this converges because this is a pp-integral with p>1p > 1, so by the integral test, ∑n=2∞1n(ln⁡n)2\displaystyle \sum_{n=2}^\infty \frac{1}{n\p{\ln{n}}^2} converges, but we can't calculate its sum.

Question 7

Find the sum of the series, if it converges.

  1. ∑n=3∞1(n−2)(n−1)\displaystyle \sum_{n=3}^\infty \frac{1}{\p{n-2}\p{n-1}}
  2. ∑n=1∞ln⁡(n+1n)\displaystyle \sum_{n=1}^\infty \ln\p{\frac{n+1}{n}}
Solution.

  1. By partial fractions,

    1(n−2)(n−1)=1n−2−1n−1,\frac{1}{\p{n-2}\p{n-1}} = \frac{1}{n - 2} - \frac{1}{n - 1},

    so we have a telescoping series:

    ∑n=3N(1n−2−1n−1)=∑n=3N1n−2−∑n=3N1n−1=(11+12+13+⋯+1N−2)−(12+13+⋯+1N−2+1N−1)=1−1N−1.\begin{aligned} \sum_{n=3}^N \p{\frac{1}{n - 2} - \frac{1}{n - 1}} &= \sum_{n=3}^N \frac{1}{n - 2} - \sum_{n=3}^N \frac{1}{n - 1} \\ &= \p{\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{N - 2}} \\ &\qquad - \p{\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{N - 2} + \frac{1}{N - 1}} \\ &= 1 - \frac{1}{N - 1}. \end{aligned}

    So,

    ∑n=3∞1(n−2)(n−1)=lim⁡N→∞∑n=3N1(n−2)(n−1)=lim⁡N→∞(1−1N−1)=1.\begin{aligned} \sum_{n=3}^\infty \frac{1}{\p{n-2}\p{n-1}} &= \lim_{N\to\infty} \sum_{n=3}^N \frac{1}{\p{n-2}\p{n-1}} \\ &= \lim_{N\to\infty} \p{1 - \frac{1}{N - 1}} \\ &= \boxed{1}. \end{aligned}

  2. This series diverges. Here are two ways to prove this:

    ∑n=1Nln⁡(n+1n)=∑n=1N(ln⁡(n+1)−ln⁡n)=ln⁡(N+1),\sum_{n=1}^N \ln\p{\frac{n+1}{n}} = \sum_{n=1}^N \p{\ln\p{n+1} - \ln{n}} = \ln\p{N+1},

    which diverges as N→∞N \to \infty. Alternatively, you can use the limit test with 1n\frac{1}{n}:

    lim⁡n→∞ln⁡(n+1n)1n=lim⁡n→∞ln⁡(1+1n)n=ln⁡(lim⁡n→∞(1+1n)n)=ln⁡e=1>0.\begin{aligned} \lim_{n\to\infty} \frac{\ln\p{\frac{n+1}{n}}}{\frac{1}{n}} &= \lim_{n\to\infty} \ln\p{1 + \frac{1}{n}}^n \\ &= \ln\p{\lim_{n\to\infty} \p{1 + \frac{1}{n}}^n} \\ &= \ln{e} \\ &= 1 > 0. \end{aligned}

    By the limit test, because ∑n=1∞1n\displaystyle \sum_{n=1}^\infty \frac{1}{n} diverges, so does ∑n=1∞ln⁡(n+1n)\displaystyle \sum_{n=1}^\infty \ln\p{\frac{n+1}{n}}.