Math 161 — Test 2 Practice Questions (with Solutions)

CCU  |  Dr. Dahal

Integrals using a Trigonometric Substitution

Problem 1. Evaluate the following integrals using a trig substitution. No credit for any other methods.

  1. (a)
    \(\displaystyle\int \frac{x^3}{\sqrt{x^2+25}}\, dx\)
  2. (b)
    \(\displaystyle\int \frac{x}{\sqrt{4-x^2}}\, dx\)
  3. (c)
    \(\displaystyle\int \frac{\sqrt{x^2-4}}{x}\, dx\)
  4. (d)
    \(\displaystyle\int \frac{x^2}{\sqrt{9-x^2}}\, dx\)
  5. (e)
    \(\displaystyle\int \frac{1}{\sqrt{4+x^2}}\, dx\)
  6. (f)
    \(\displaystyle\int \frac{x}{\sqrt{x^2-9}}\, dx\)
  7. (g)
    \(\displaystyle\int_0^5 \frac{dt}{\sqrt{25+t^2}}\)
Solution: https://ww2.coastal.edu/rdahal/math161/trig_sub/

Partial Fraction Decomposition (PFD)

Problem 2. Evaluate the following integrals using partial fraction decomposition.

(a)   \(\displaystyle\int \frac{2x+19}{(x-3)(x+2)}\, dx\)

Solution:

Let \(\dfrac{2x+19}{(x-3)(x+2)} = \dfrac{A}{x-3} + \dfrac{B}{x+2}\), so \(2x+19 = A(x+2)+B(x-3)\).

Let \(x=3\): \(25 = 5A \implies A=5\).
Let \(x=-2\): \(15 = -5B \implies B=-3\).

\[\int \frac{2x+19}{(x-3)(x+2)}\, dx = 5\ln|x-3| - 3\ln|x+2| + C.\]

(b)   \(\displaystyle\int \frac{10x+2}{(3x+1)(x-1)}\, dx\)

Solution:

Let \(\dfrac{10x+2}{(3x+1)(x-1)} = \dfrac{A}{3x+1} + \dfrac{B}{x-1}\), so \(10x+2 = A(x-1)+B(3x+1)\).

Let \(x=1\): \(12 = 4B \implies B=3\).
Let \(x=0\): \(2 = -A+B \implies A=1\).

\[\int \frac{10x+2}{(3x+1)(x-1)}\, dx = \frac{1}{3}\ln|3x+1| + 3\ln|x-1| + C.\]

(c)   \(\displaystyle\int_0^1 \frac{x-6}{x^2-6x+8}\, dx\)

Solution:

The denominator factors as \(x^2-6x+8=(x-4)(x-2)\). Let \(\dfrac{x-6}{(x-4)(x-2)} = \dfrac{A}{x-4}+\dfrac{B}{x-2}\), so \(x-6=A(x-2)+B(x-4)\).

Let \(x=2\): \(-4=-2B \implies B=2\).
Let \(x=4\): \(-2=2A \implies A=-1\).

So \(\displaystyle\int \frac{x-6}{x^2-6x+8}\,dx = -\ln|x-4|+2\ln|x-2|+C\). Evaluating:

\[ \int_0^1 \frac{x-6}{x^2-6x+8}\,dx = \Bigl[-\ln|x-4|+2\ln|x-2|\Bigr]_0^1 = [-\ln 3+0]-[-\ln 4+2\ln 2] = -\ln 3. \]

(d)   \(\displaystyle\int \frac{5x^2+2x-5}{x^3-x}\, dx\)

Solution:

The denominator factors as \(x^3-x=x(x-1)(x+1)\). Let \(\dfrac{5x^2+2x-5}{x(x-1)(x+1)} = \dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{C}{x+1}\), so \(5x^2+2x-5=A(x-1)(x+1)+Bx(x+1)+Cx(x-1)\).

Let \(x=0\): \(-5=-A \implies A=5\).
Let \(x=1\): \(2=2B \implies B=1\).
Let \(x=-1\): \(-2=2C \implies C=-1\).

\[\int \frac{5x^2+2x-5}{x^3-x}\,dx = 5\ln|x|+\ln|x-1|-\ln|x+1|+C.\]

(e)   \(\displaystyle\int \frac{x+2}{x^3-2x^2}\, dx\)

(f)   \(\displaystyle\int \frac{x^2+1}{(x-2)(x-5)^2}\, dx\)

(g)   \(\displaystyle\int \frac{2x}{(x+1)(x^2+1)}\, dx\)

Solutions (e)–(g): https://tinyurl.com/ydw7t8ep (course webpage)

(h)   \(\displaystyle\int_1^5 \frac{x^2+5}{6x-x^2}\, dx\)

Solution:

The integrand is improper, so long division gives \(\dfrac{x^2+5}{6x-x^2} = -1 + \dfrac{6x+5}{6x-x^2}\). For the partial fractions: \(\dfrac{6x+5}{x(6-x)} = \dfrac{A}{x}+\dfrac{B}{6-x}\), so \(6x+5=A(6-x)+Bx\).

Let \(x=0\): \(5=6A \implies A=\tfrac{5}{6}\).
Let \(x=6\): \(41=6B \implies B=\tfrac{41}{6}\).

\[ \int_1^5 \frac{x^2+5}{6x-x^2}\,dx = \left[-x+\tfrac{5}{6}\ln|x|-\tfrac{41}{6}\ln|6-x|\right]_1^5 = -4 + \frac{23}{3}\ln 5. \]

(i)   \(\displaystyle\int_1^2 \frac{4y^2-7y-12}{y(y+2)(y-3)}\, dy\)

(j)   \(\displaystyle\int_3^4 \frac{2x^2-4}{x^3-2x^2}\, dx\)

Solutions (i)–(j): https://tinyurl.com/ydw7t8ep (course webpage)

Improper Integrals

Problem 3. Determine whether each improper integral converges or diverges (show your work). If it converges, find its value.

  1. (a)
    \(\displaystyle\int_5^{\infty} \frac{4}{(x-2)^3}\, dx\)
  2. (b)
    \(\displaystyle\int_{0}^{\infty} \frac{1}{\sqrt[3]{x+2}}\, dx\)
  3. (c)
    \(\displaystyle\int_{3}^{\infty} 12e^{-2x}\, dx\)
  4. (d)
    \(\displaystyle\int_0^{\infty} \frac{1}{(2x+5)^3}\, dx\)
  5. (e)
    \(\displaystyle\int_0^{\infty} \frac{1}{1+x^2}\, dx\)
  6. (f)
    \(\displaystyle\int_0^{\infty} \frac{x^2}{5+x^3}\, dx\)
  7. (g)
    \(\displaystyle\int_0^{\infty} \frac{x^2}{\sqrt{6+x^3}}\, dx\)
  8. (h)
    \(\displaystyle\int_2^{\infty} \frac{e^{-1/x}}{x^2}\, dx\)
  9. (i)
    \(\displaystyle\int_1^{4} \frac{1}{\sqrt{x-1}}\, dx\)
Solutions: https://ww2.coastal.edu/rdahal/math161/improper_integral/

Arc Length

Problem 4. Find the arc length of the curve \(y = 3x-5\) on the interval \(1 \leq x \leq 3\).

Solution:

Here \(y'=3\), so \(1+(y')^2=10\). The arc length is

\[ L = \int_1^3 \sqrt{10}\, dx = \sqrt{10}\,x\Big]_1^3 = 3\sqrt{10}-\sqrt{10} = 2\sqrt{10}. \]

Problem 5. Find the arc length of the curve \(y = 4x^{3/2}+5\) on the interval \(0 \leq x \leq 1\).

Solution:

Here \(y'=6\sqrt{x}\), so \(1+(y')^2=1+36x\). Let \(u=1+36x\), \(du=36\,dx\). When \(x=0\), \(u=1\); when \(x=1\), \(u=37\).

\[ L = \int_0^1\sqrt{1+36x}\,dx = \frac{1}{36}\int_1^{37}u^{1/2}\,du = \frac{1}{36}\cdot\frac{2}{3}\left[u^{3/2}\right]_1^{37} = \frac{1}{54}\left(37^{3/2}-1\right). \]

Problem 6. Find the arc length of the curve \(y = \ln(\sec x)\) on the interval \(0 \leq x \leq \dfrac{\pi}{4}\).

Solution: https://ww2.coastal.edu/rdahal/math161/arc_length/

Differential Equations

Problem 7. Find general solutions of the following differential equations. Write your solution as \(y=f(x)\).

  1. (a)
    \(\dfrac{dy}{dx}=\dfrac{1}{xy}\)
  2. (b)
    \(\dfrac{dy}{dx}=\dfrac{1}{1+x^2}\)
  3. (c)
    \(\dfrac{dy}{dx}=ye^x\)
  4. (d)
    \(e^{-2x}\,\dfrac{dy}{dx}=\dfrac{1}{y}\)
Solution: https://youtu.be/VYW-F2lDDOY

Problem 8. Find the solution satisfying the given initial condition. Write your solution as \(y=f(x)\).

  1. (a)
    \(\dfrac{dy}{dx}=4x,\quad y(1)=5\)
  2. (b)
    \(\dfrac{dy}{dx}=\dfrac{1}{\sec^2 y},\quad y(2)=0\)
  3. (c)
    \(xy^2\,\dfrac{dy}{dx}=x+1,\quad y(1)=3\)
  4. (d)
    \(\dfrac{dy}{dx}=y^2 e^x,\quad y(0)=1\)
  5. (e)
    \(\dfrac{dy}{dx}=\dfrac{6x^2}{y},\quad y(0)=-2\)
  6. (f)
    \(\dfrac{dy}{dx}=\dfrac{\sin(x)}{2y},\quad y(0)=-1\)
Solutions (a)–(d): https://youtu.be/1rZw2-Hsb7c
Solution (e):

Separating variables: \(y\,dy = 6x^2\,dx\). Integrating: \(\dfrac{y^2}{2}=2x^3+C\).

Using \(y(0)=-2\): \(2=C\). So \(y^2=4x^3+4\).

Since \(y(0)=-2< 0\), we take the negative root: \(y = -\sqrt{4x^3+4}\).

Solution (f):

Separating variables: \(2y\,dy = \sin(x)\,dx\). Integrating: \(y^2=-\cos(x)+C\).

Using \(y(0)=-1\): \(1=-1+C \implies C=2\). So \(y^2=2-\cos(x)\).

Since \(y(0)=-1< 0\), we take the negative root: \(y = -\sqrt{2-\cos(x)}\).

Sequences

Problem 9. Determine whether the following sequences converge or diverge. If it converges, find its limit. You do not need to show work.

(a)   \(\displaystyle a_n = \frac{5n^2+2n+20}{3n^5-3n^2-5}\)

\(\lim_{n\to\infty} a_n = 0\).   Converges to \(0\).

(b)   \(\displaystyle a_n = \frac{3n^4-2n}{8n^4+100n^2+1000}\)

\(\lim_{n\to\infty} a_n = \dfrac{3}{8}\).   Converges to \(\dfrac{3}{8}\).

(c)   \(\displaystyle a_n = \frac{2n^3+1}{1500n^2+100n+3000}\)

\(\lim_{n\to\infty} a_n = \infty\).   Diverges.

(d)   \(\displaystyle a_n = 2^{-n}\)

\(\lim_{n\to\infty} 2^{-n} = 0\).   Converges to \(0\).

(e)   \(\displaystyle a_n = 3^n \cdot 2^{1-n}\)

\(3^n \cdot 2^{1-n} = 3^n \cdot 2^1 \cdot 2^{-n} =2\!\left(\tfrac{3}{2}\right)^n\). Geometric with \(r=\tfrac{3}{2}\). Since \(|r|=\tfrac{3}{2}\gt 1\).   Diverges.

(f)   \(\displaystyle a_n = 4^n\)

Geometric with \(r=4\). Since \(|r| =4 \gt 1\).   Diverges.

(g)   \(\displaystyle a_n = 2^{\frac{2n}{n+1}}\)

\(\lim_{n\to\infty}\dfrac{2n}{n+1}=2\), so \(\lim a_n = 2^2=4\).   Converges to \(4\).

(h)   \(\displaystyle a_n = 3^{\frac{n}{n^2+1}}\)

\(\lim_{n\to\infty}\dfrac{n}{n^2+1}=0\), so \(\lim a_n = 3^0=1\).   Converges to \(1\).

(i)   \(\{-4,\ 12,\ -36,\ 108,\ -324,\ \ldots\}\)

Geometric with \(r=-3\), so \(|r|=3\gt 1\).   Diverges.