PAGE 1

Work Required to Pump Water from a Cylindrical Tank

A cylindrical tank of height 4 feet and radius 1 foot is filled with water. How much work is required to pump all but 1 foot of water out of the tank. (Density = $62.5 \text{ lbs.}/\text{ft}^3$)

Note: Density in lbs./ft³ is force, no gravity to multiply. kg is a mass so mult. by $g = 9.8$.

Figure: Cylindrical tank diagram with height 4, radius 1, and a water slice at height y with thickness dy.

A $9\pi(62.5) \text{ ft-lbs.}$
B $3\pi(62.5) \text{ ft-lbs.}$
C $\frac{9\pi}{2}(62.5) \text{ ft-lbs.}$
D $18\pi(62.5) \text{ ft-lbs.}$
E $6\pi(62.5) \text{ ft-lbs.}$

Solution Steps

work = force $\cdot$ distance

Work to move one slice of water then integrate over water to move:

$\text{work of slice} = (\text{weight}) (\text{distance to go})$

$= (62.5)(\pi)(1)^2(dy) \cdot (4-y)$

$= 62.5\pi (4-y)dy$

\[ \int_{1}^{4} 62.5\pi (4-y)dy \]\[ = 62.5\pi \left( 4y - \frac{1}{2}y^2 \right) \Big|_1^4 \]\[ = 62.5\pi \left[ (16-8) - (4 - \frac{1}{2}) \right] = 62.5\pi \left( 8 - \frac{7}{2} \right) \]\[ = 62.5\pi \left( \frac{9}{2} \right) \]

PAGE 2

Integration by Parts Evaluation

Evaluate \[ \int_{0}^{1} xe^{3x} dx \]

A. $\frac{2e^3}{9}$
B $\frac{1}{9} + \frac{2e^3}{9}$
C. 1

D. $\frac{1}{9}$
E. $\frac{e^3}{9} - 1$

Method: Integration by Parts

By parts: $uv - \int v du$

Order to pick $u$: LIATE

L: Logarithmic
I: Inverse trig
A: Algebraic (here, $x$)
T: Trig
E: Exponential (here, $e^{3x}$)

Here, $x$ is Algebraic and $e^{3x}$ is Exponential, so:

$u = x$

$du = dx$

$dv = e^{3x} dx$

$v = \frac{1}{3}e^{3x}$

\[ uv \Big|_0^1 - \int_{0}^{1} v du = \frac{1}{3}xe^{3x} \Big|_0^1 - \int_{0}^{1} \frac{1}{3}e^{3x} dx \]\[ = \frac{1}{3}xe^{3x} \Big|_0^1 - \frac{1}{9}e^{3x} \Big|_0^1 \]\[ = \frac{1}{3}e^3 - \frac{1}{9}e^3 + \frac{1}{9} = \frac{2}{9}e^3 + \frac{1}{9} \]

PAGE 3

Basic Trig Integrals

Basic trig integrals: $ \cos x, \sin x \rightarrow u = \cos x $ or $ u = \sin x $. Move powers around to make things fit.

\[ \int_{0}^{\pi/2} \sin^3 x dx = \]

A. 2/3

B. 4/3

C. 0

D. 1/4

E. 1/3

Solution Strategy

\[ \int_{0}^{\pi/2} \sin^2 x \cdot \sin x dx \]

Using identity: $ \sin^2 x = 1 - \cos^2 x $

Substitution:

If $ u = \cos x $, then $ du = -\sin x dx $

We have this part in the integral.

\[ = \int_{0}^{\pi/2} (1 - \cos^2 x) \sin x dx \]

$ u = \cos x $

$ du = -\sin x dx $

Definite Integral Calculation

\[ = \int_{1}^{0} -(1 - u^2) du = -u + \frac{1}{3}u^3 \Big|_1^0 = 0 - (-1 + \frac{1}{3}) = \frac{2}{3} \]

or back-substituting:

\[ -u + \frac{1}{3}u^3 \Big|_{x=0}^{x=\pi/2} = -\cos x + \frac{1}{3}\cos^3 x \Big|_0^{\pi/2} = \dots = \frac{2}{3} \]

PAGE 4

\[ \int_{0}^{\pi/4} \sec^4 x \tan x dx = \]

A. 1

B. 1/3

C. 4/3

D. 3/4

E. 2/9

Substitution Options

If $ u = \tan x $ then $ du = \sec^2 x dx $
If $ u = \sec x $ then $ du = \sec x \tan x dx $

Step-by-Step Solution

\[ \int_{0}^{\pi/4} \sec^4 x \tan x dx = \int_{0}^{\pi/4} \sec^2 x \cdot \tan x \cdot \sec^2 x dx \]

$ \sec^2 x = \tan^2 x + 1 $

$ = u^2 + 1 $

$ \tan x = u $

$ \sec^2 x dx = du $

if $ u = \tan x $

Changing limits: when $ x = \pi/4, u = \tan(\pi/4) = 1 $. When $ x = 0, u = 0 $.

\[ = \int_{0}^{1} (u^2 + 1)(u) du = \int_{0}^{1} (u^3 + u) du \]

\[ = \frac{u^4}{4} + \frac{u^2}{2} \Big|_0^1 = \frac{1}{4} + \frac{1}{2} = \frac{3}{4} \]

PAGE 5

Definite Integral with Substitution

Evaluate the following definite integral using $u$-substitution:

\[ \int_{0}^{\pi/4} \sec^4 x \tan x \, dx \]

Substitution:

If $ u = \sec x $,

then $ du = \sec x \tan x \, dx $

Rewrite the integral to isolate the $ du $ term:

\[ = \int_{0}^{\pi/4} \sec^3 x (\sec x \tan x) \, dx \]

Change the limits of integration and substitute $ u $:

\[ = \int_{1}^{\sqrt{2}} u^3 \, du = \dots \]

PAGE 6

Trigonometric Substitution

\[ \int \frac{dx}{\sqrt{9 - 4x^2}} = \]

A. $ \sec^{-1} (\frac{3x}{2}) + C $
B. $ \frac{1}{2} \sin^{-1} (\frac{2x}{3}) + C $
C. $ \tan^{-1} (\frac{2x}{3}) + C $
D. $ \frac{1}{3} \sin^{-1} (\frac{3x}{2}) + C $
E. $ \sqrt{9 - 4x^2} + \tan^{-1} (\frac{2x}{3}) + C $

Trigonometric Substitution Method

Trig subs: $ \sqrt{9 - 4x^2} $

Triangle with sides: $ \sqrt{9 - 4x^2} $, $ 3 $, $ 2x $

Hypotenuse: $ 3 $

Adjacent: $ \sqrt{9 - 4x^2} $ (contains constant)

\[ \sin \theta = \frac{2x}{3} \implies x = \frac{3}{2} \sin \theta \]

\[ dx = \frac{3}{2} \cos \theta \, d\theta \]

Substituting into the integral:

\[ \int \frac{dx}{\sqrt{9 - 4x^2}} = \int \frac{\frac{3}{2} \cos \theta \, d\theta}{\sqrt{9 - 4(\frac{9}{4} \sin^2 \theta)}} \]

PAGE 7

Integration by Trigonometric Substitution

\[ = \int \frac{\frac{3}{2} \cos \theta}{\sqrt{9(1 - \sin^2 \theta)}} d\theta = \int \frac{\frac{3}{2} \cos \theta}{3 \cos \theta} d\theta \]

Note: In the denominator, $ 1 - \sin^2 \theta $ simplifies to $ \cos^2 \theta $.

\[ = \int \frac{1}{2} d\theta = \frac{1}{2} \theta + C = \frac{1}{2} \sin^{-1} \left( \frac{2x}{3} \right) + C \]

Back Substitution

To return to the original variable $ x $:

\[ \text{back to } \sin \theta = \frac{2}{3} x \]\[ \theta = \sin^{-1} \left( \frac{2}{3} x \right) \]

PAGE 8

Integration by Partial Fractions

\[ \int \frac{x + 1}{x^3 - 2x^2 + x} dx = \]

A. $ \ln |x| + \ln |x - 1| + C $

B. $ \ln |x| - \ln |x - 1| + C $

C. $ \ln |x| - \frac{2}{x - 1} + C $

D. $ \ln |x - 1| - \frac{2}{x - 1} + C $

E. $ \ln |x| - \ln |x - 1| - \frac{2}{x - 1} + C $

Partial Fraction Decomposition

\[ \frac{x + 1}{x(x^2 - 2x + 1)} = \frac{x + 1}{(x)(x - 1)(x - 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{(x - 1)^2} \]

The factors are linear ($ x $) and repeated linear ($ (x - 1)^2 $).

Solving for Coefficients

Multiply by $ (x)(x - 1)(x - 1) $:

\[ x + 1 = A(x - 1)^2 + B(x)(x - 1) + C(x) \]\[ = A(x^2 - 2x + 1) + B(x^2 - x) + Cx \]\[ 0x^2 + 1x + 1 = (A + B)x^2 + (-2A - B + C)x + A \]

PAGE 9

So,

$ A + B = 0 $

$ -2A - B + C = 1 $

$ A = 1 $

$ B = -1 $

Substituting these values into the second equation:

\[ -2(1) - (-1) + C = 1 \]\[ -2 + 1 + C = 1 \]\[ C = 2 \]

Integration by Partial Fractions

\[ \int \left( \frac{1}{x} - \frac{1}{x-1} + \frac{2}{(x-1)^2} \right) dx \]\[ = \int \frac{1}{x} dx - \int \frac{1}{x-1} dx + 2 \int \frac{1}{(x-1)^2} dx \]

Let $ u = x - 1 $

$ du = dx $

\[ = \int \frac{1}{x} dx - \int \frac{1}{x-1} dx + 2 \int \frac{1}{u^2} du \]\[ = \ln|x| - \ln|x-1| - \frac{2}{u} + C \]\[ = \ln|x| - \ln|x-1| - \frac{2}{x-1} + C \]

PAGE 10

Indicate convergence or divergence for each of the following improper integrals:

$ (I) \int_{2}^{\infty} \frac{1}{(x-1)^2} dx $
conv.

$ (II) \int_{0}^{2} \frac{1}{(x-1)^2} dx $
div.

$ (III) \int_{0}^{1} \frac{\ln x}{x} dx $
div.

A. I converges, II and III diverge.

B. II converges, I and III diverge.

C. I and III converge, II diverges.

D. I and II converge, III diverges.

E. I, II and III diverge.

I. Evaluation of Integral I

\[ \lim_{a \to \infty} \int_{2}^{a} \frac{1}{(x-1)^2} dx = \lim_{a \to \infty} \left. -\frac{1}{(x-1)} \right|_{2}^{a} \]\[ = \lim_{a \to \infty} \left( -\frac{1}{a-1} \right) + \frac{1}{1} = 1 \]

Note: $ -\frac{1}{a-1} \to 0 $ as $ a \to \infty $.

II. Evaluation of Integral II

trouble at $ x = 1 $

\[ \int_{0}^{1} \frac{1}{(x-1)^2} dx + \int_{1}^{2} \frac{1}{(x-1)^2} dx \]\[ = \lim_{a \to 1^-} \int_{0}^{a} \frac{1}{(x-1)^2} dx + \lim_{b \to 1^+} \int_{b}^{2} \frac{1}{(x-1)^2} dx \]

PAGE 11

Improper Integrals and Divergence

\[= \lim_{a \to 1^-} \left. -\frac{1}{x-1} \right|_0^a + \lim_{b \to 1^+} \left. -\frac{1}{x-1} \right|_b^2\]

\[= \lim_{a \to 1^-} \left( -\frac{1}{a-1} - 1 \right) + \lim_{b \to 1^+} \left( -1 + \frac{1}{b-1} \right)\]

As $ a \to 1^- $:

\[ \frac{1}{a-1} = \frac{1}{\text{small negative}} = -\infty \]

\[= \infty + \text{whatever}\]

One part diverges, so the whole thing diverges.

Example IV

\[ \int_0^1 \frac{\ln x}{x} dx = \lim_{a \to 0^+} \int_a^1 \frac{\ln x}{x} dx \]

Substitution:

$ u = \ln x $

$ du = \frac{1}{x} dx $

\[ = \lim_{a \to 0^+} \int_{\ln a}^0 u \, du = \lim_{a \to 0^+} \left. \frac{1}{2} u^2 \right|_{\ln a}^0 \]

\[ = \lim_{a \to 0^+} 0 - \frac{1}{2} (\ln a)^2 \]

Note: As $ a \to 0^+ $, $ \ln a \to -\infty $.