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PAGE 1

Integration by Parts: Example 1

Compute the value of the integral \[ \int_{1}^{e} 2x \ln x \, dx \]

  1. A. \( \frac{1}{2} \)
  2. B. \( \frac{1 - e^2}{2} \)
  3. C. \( \frac{1 + e^2}{2} \) (Correct)
  4. D. \( \frac{3e^2 - 1}{2} \)
  5. E. \( \frac{e^2}{2} \)

Method Selection

  • Direct: ✘
  • Substitution: ✘
  • By parts: ✔ (different kinds of functions)
  • Trig subs: ✘

LIATE Rule

Step-by-Step Solution

Let: \[ u = \ln x \quad dv = 2x \, dx \] \[ du = \frac{1}{x} \, dx \quad v = x^2 \]

Using the formula \( \int u \, dv = uv - \int v \, du \):

\[ uv \Big|_1^e - \int_{1}^{e} v \, du \]\[ = x^2 \ln x \Big|_1^e - \int_{1}^{e} x^2 \cdot \frac{1}{x} \, dx \]\[ = x^2 \ln x \Big|_1^e - \int_{1}^{e} x \, dx \]\[ = x^2 \ln x \Big|_1^e - \frac{1}{2} x^2 \Big|_1^e = \left( x^2 \ln x - \frac{1}{2} x^2 \right) \Big|_1^e \]\[ = e^2 \ln e - \frac{1}{2} e^2 - (1 · \ln 1 + \frac{1}{2}) \]\[ = e^2 - \frac{1}{2} e^2 + \frac{1}{2} = \frac{1}{2} e^2 + \frac{1}{2} \]
PAGE 2

Trigonometric Integrals: Example 2

Evaluate \[ \int_{0}^{\pi/2} \sin^3 x \cos^4 x \, dx \]

  1. A. \( \frac{5}{12} \)
  2. B. \( \frac{1}{7} \)
  3. C. \( \frac{2}{35} \) (Correct)
  4. D. \( \frac{3}{28} \)
  5. E. \( \frac{5}{16} \)

Strategy: \( \sin x \) and \( \cos x \)

Option 1: \( u = \cos x \), \( du = -\sin x \, dx \)

OR

Option 2: \( u = \sin x \), \( du = \cos x \, dx \)

Then use identity: \( \sin^2 x + \cos^2 x = 1 \)

Solution using \( u = \cos x \)

Split a factor of \( \sin x \). For the remaining \( \sin^2 x \), use \( \sin^2 x = 1 - \cos^2 x \) to change into \( \cos x \).

Substitution details:

\[ u = \cos x \]\[ du = -\sin x \, dx \]\[ x = \pi/2 \rightarrow u = \cos(\pi/2) = 0 \]\[ x = 0 \rightarrow u = \cos(0) = 1 \]\[ \int_{0}^{\pi/2} \sin^2 x \cos^4 x \sin x \, dx \]\[ = \int_{1}^{0} (1 - u^2) u^4 (-du) = - \int_{1}^{0} (u^4 - u^6) \, du \]\[ = - \left( \frac{u^5}{5} - \frac{u^7}{7} \right) \Big|_1^0 \]\[ = - \left[ (0) - \left( \frac{1}{5} - \frac{1}{7} \right) \right] = \frac{2}{35} \]
PAGE 3

Integration of Trigonometric Functions

Compute

\[ \int 7 \sec^4 x \, dx \]
  1. A. \( \frac{7}{3} \tan^3 x + C \)
  2. B. \( -\frac{7}{3} \tan^3 x + C \)
  3. C. \( 7(\sec x + \tan x)^5 + C \)
  4. D. \( \frac{7}{3} \tan x + 7 \tan^3 x + C \)
  5. E. \( 7 \tan x + \frac{7}{3} \tan^3 x + C \)

Substitution Options:

  • \( u = \sec x \implies du = \sec x \tan x \, dx \)
  • \( u = \tan x \implies du = \sec^2 x \, dx \)

Identity:

\( \sin^2 x + \cos^2 x = 1 \)

Divide by \( \cos^2 x \):

\( \frac{\sin^2 x}{\cos^2 x} + \frac{\cos^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} \)

\( \tan^2 x + 1 = \sec^2 x \)

Step-by-Step Solution

\[ 7 \int \sec^2 x \cdot \sec^2 x \, dx \]

Using the substitution \( u = \tan x \), we have \( du = \sec^2 x \, dx \). The first \( \sec^2 x \) term becomes \( \tan^2 x + 1 \), which is \( u^2 + 1 \).

\[ 7 \int (u^2 + 1) \, du \]\[ = 7 \left( \frac{u^3}{3} + u \right) + C \]\[ = \frac{7}{3} u^3 + 7u + C \]\[ = \frac{7}{3} \tan^3 x + 7 \tan x + C \]
PAGE 4

Trigonometric Substitution and Limits

After a proper trigonometric substitution is used to transform \( \int_{1}^{4} \frac{dt}{t^2 - 2t + 10} \) into \( \int_{a}^{b} f(\theta) \, d\theta \), what is the new upper integration limit \( b \)?

  1. A. \( \frac{\pi}{6} \)
  2. B. \( \frac{\pi}{4} \)
  3. C. \( \frac{\pi}{2} \)
  4. D. \( \frac{\pi}{3} \)
  5. E. \( \pi \)

Trig Subs: Difference or Sum of Squares

\( t^2 - 2t + 10 = t^2 - 2t + 1 + 10 - 1 \)

\( = (t - 1)^2 + 9 \)

Setting up the Substitution

\[ \int_{1}^{4} \frac{dt}{t^2 - 2t + 10} = \int_{1}^{4} \frac{dt}{((t-1)^2 + 9)} \]

Triangle with sides: \( \sqrt{(t-1)^2 + 3^2} \), \( t-1 \), and \( 3 \).

Hypotenuse: \( \sqrt{(t-1)^2 + 3^2} \) because its square is the sum of the squares of the other two sides.

Right triangle with angle theta, opposite side t-1, adjacent side 3, and hypotenuse sqrt((t-1)^2 + 3^2).

\( \tan \theta = \frac{t-1}{3} \)

\( 3 \tan \theta = t - 1 \)

\( t = 3 \tan \theta + 1 \)

\( dt = 3 \sec^2 \theta \, d\theta \)

PAGE 5

Definite Integral with Trigonometric Substitution

\[ \int_{1}^{4} \frac{dt}{(t-1)^2 + 9} = \int \frac{3 \sec^2 \theta \, d\theta}{9 \tan^2 \theta + 9} = \int \frac{\sec^2 \theta \, d\theta}{3(\tan^2 \theta + 1)} \]
\[ = \int \frac{\sec^2 \theta \, d\theta}{3 \sec^2 \theta} = \int \frac{1}{3} \, d\theta = \int_{0}^{\pi/4} \frac{1}{3} \, d\theta \]

Substitution Details

Let \( \tan \theta = \frac{t-1}{3} \)

Change of limits:

  • When \( t = 4 \rightarrow \tan \theta = 1 \rightarrow \theta = \pi/4 \)
  • When \( t = 1 \rightarrow \tan \theta = 0 \rightarrow \theta = 0 \)
PAGE 6

Partial Fraction Expansion via Integration

Use the fact that

\[ \int \frac{2x^4 + 15x^3 + 9x^2 + 11x + 3}{x^5 + x^4 - x - 1} \, dx = 5 \ln |x - 1| - 3 \ln |x + 1| - \frac{3}{x + 1} + 2 \tan^{-1}(x) + C \]

to find the partial fraction expansion of

\[ \frac{2x^4 + 15x^3 + 9x^2 + 11x + 3}{x^5 + x^4 - x - 1} = \text{expansion?} \]

Reasoning

If we know the expansion, then

\[ \int \text{expansion} = 5 \ln |x - 1| - 3 \ln |x + 1| - \frac{3}{x + 1} + 2 \tan^{-1}(x) + C \]

So expansion = derivative of

\[ \frac{5}{x-1} - \frac{3}{x+1} + \frac{3}{(x+1)^2} + \frac{2}{x^2+1} \]

Multiple Choice Options

  • A. \( \frac{5}{x-1} + \frac{-3}{x+1} + \frac{3}{(x+1)^2} + \frac{2}{x^2+1} \) (Selected)
  • B. \( \frac{5}{x-1} + \frac{-3}{x+1} + \frac{-3x}{(x+1)^2} + \frac{2}{x^2+1} \)
  • C. \( \frac{5}{x-1} + \frac{-3}{x+1} + \frac{-3}{(x+1)^2} + \frac{2x}{x^2+1} \)
  • D. \( \frac{-5}{x-1} + \frac{3}{x+1} + \frac{3}{(x+1)^2} + \frac{-2}{x^2+1} \)
  • E. \( \frac{-5}{x-1} + \frac{3}{x+1} + \frac{3x}{(x+1)^2} + \frac{2x}{x^2+1} \)
PAGE 7

Improper Integrals: Example 1

Compute \[ \int_{1}^{2} \frac{dx}{\sqrt{2-x}} \]

  1. A 2
  2. B \( 2\sqrt{2} - 1 \)
  3. C \( \sqrt{2} - 1 \)
  4. D \( \sqrt{2} \)
  5. E 1

improper because integrand is undefined somewhere between or at the two bounds

here, \( x=2 \) is problem

Solution

\[ \lim_{a \to 2^-} \int_{1}^{a} \frac{dx}{\sqrt{2-x}} = \lim_{a \to 2^-} \int_{1}^{a} (2-x)^{-1/2} dx \]
\[ = \lim_{a \to 2^-} \int_{1}^{2-a} -u^{-1/2} du \]\[ = \lim_{a \to 2^-} \left. -2u^{1/2} \right|_{1}^{2-a} = \lim_{a \to 2^-} (-2(2-a)^{1/2} + 2) \]\[ = 2 \]
\[ u = 2-x \]\[ du = -dx \]
PAGE 8

Improper Integrals: Example 2

\[ \int_{1}^{3} \frac{dx}{\sqrt{2-x}} \]
\( x=2 \) is the problem

Since the discontinuity occurs at \( x=2 \), which is inside the interval \( [1, 3] \), we must split the integral:

\[ = \int_{1}^{2} \frac{dx}{\sqrt{2-x}} + \int_{2}^{3} \frac{dx}{\sqrt{2-x}} \]
\[ = \lim_{a \to 2^-} \int_{1}^{a} \frac{dx}{\sqrt{2-x}} + \lim_{b \to 2^+} \int_{b}^{3} \frac{dx}{\sqrt{2-x}} \]
PAGE 9

Improper Integral Evaluation

Evaluate

\[ \int_{0}^{\infty} x^2 e^{-x^3} dx \]

Multiple Choice Options:

  • (A) \( \frac{1}{2} \)
  • (B) 1
  • (C) the integral diverges
  • (D) \( \frac{1}{3} \)
  • (E) \( \frac{1}{6} \)

Solution Steps:

\[ \lim_{a \to \infty} \int_{0}^{a} x^2 e^{-x^3} dx \]

Substitution:

\[ u = x^3 \]

\[ du = 3x^2 dx \]

\[ = \lim_{a \to \infty} \int_{0}^{a^3} \frac{1}{3} e^{-u} du \]

\[ = \lim_{a \to \infty} -\frac{1}{3} e^{-u} \Big|_0^{a^3} = \lim_{a \to \infty} \left( -\frac{1}{3} e^{-a^3} + \frac{1}{3} \right) = \frac{1}{3} \]

PAGE 10

Convergence and Divergence of Sequences

Determine whether the following sequences are convergent or divergent.

  • (1) \( \{a_n = 2n/(3n + 1)\} \) C
  • (2) \( \{a_n = \cos n\pi\} \) D
  • (3) \( \{a_n = n \sin(1/n)\} \) C

Definitions:

Sequence: convergent if \( \lim_{n \to \infty} a_n \) exists

divergent if \( \lim_{n \to \infty} a_n \) DNE

Options:

  • A. (1) convergent (2) convergent (3) convergent
  • B. (1) divergent (2) convergent (3) convergent
  • C. (1) convergent (2) divergent (3) convergent
  • D. (1) convergent (2) convergent (3) divergent
  • E. (1) convergent (2) divergent (3) divergent

(1) Analysis

\[ \lim_{n \to \infty} \frac{2n}{3n+1} \to \frac{\infty}{\infty} \]

Using l'Hospital's: \[ \lim_{n \to \infty} \frac{2}{3} = \frac{2}{3} \] conv

(2) Analysis

\[ \{ \cos n\pi \}_{n=1}^{\infty} = \{-1, 1, -1, 1, -1, 1, -1, 1, -1, \dots \} \]

not settling anywhere, so no limit, so div

(3) Analysis

\[ \lim_{n \to \infty} n \sin\left(\frac{1}{n}\right) = \lim_{n \to \infty} \frac{\sin(1/n)}{1/n} \to \frac{0}{0} \]

l'Hospital's ok:

\[ = \lim_{n \to \infty} \frac{\cos(1/n) \cdot -\frac{1}{n^2}}{-\frac{1}{n^2}} = \lim_{n \to \infty} \cos\left(\frac{1}{n}\right) = \cos(0) = 1 \] conv