Math 101 Final Answers

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1. Short-Answer Questions. Put your answers in the boxes provided but show your workalso. Each question is worth 3 marks, but not all questions are of equal difficulty. At mostone mark will be given for an incorrect answer. Unless otherwise stated, simplify youranswers as much as possible.Marks

(a) Evaluate

3 + x5

x

dx[3]

3 + x5

x dx=

3x1/2 + x9/2 dx= 6

x +

2

11x11/2 + C.

(b) What integral is defined by the following expression? limn

ni=1

4ntani

4n

Do not evaluate the integral.[3]

If one divides [0, /4] into n equal subintervals, then x = 4n and xi = ix =

i4n .

Therefore,

limn

ni=1

4ntan

i

4n = lim

n

ni=1

tan(xi)x=

/40

tan xdx.

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April 24, 2009 Math 101 Page 2 of 13 pages

(c) Evaluate

1

0

(y+ 1)

2y+ y2dy[3]

Let u = 2y+y2. Then du = (2 + 2y) dy = 2(y+ 1) dy. When y = 0, u = 0 and wheny = 1, u = 3. Thus,

1

0

(y+ 1)

2y+ y2 dy= 1

2

3

0

u du=

1

3u3/2

3

0

= 1

333/2 =

3.

(d) Evaluate

x2 ln x dx[3]

Integrate by parts: let u = ln x and dv = x2 dx so that du = dx/x and v = x3/3.Therefore,

x2 ln x dx=x3

3 ln x

x2

3 dx=

x3

3 ln x x

3

9 + C.

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[3](e) Find the length of the curve x = 100 + 2y3/2, 0 y 11.

dxdy = 3

y so the length of the curve L is

L= 11

01 + 9y dy= 2

27(1 + 9y)3/211

0

= 2

27

(1000

1) = 74.

(f) What is the average value of| sin cos | over the interval 0 /2?[3]

Notice that cos sin for 0 /4 and sin cos for/4 /2. Thereforethe average value is

2

/2

0

| sin cos | d = 2

/4

0

(cos sin ) d+ /2

/4(sin cos ) d

= 2

[sin + cos ]

/40

+ [ cos sin ]/2/4

=

2

(

2 1) + (1 +

2)

= 4(

2 1)

.

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(g) Give the first three nonzero terms of the Maclaurin series (power series in x) for[3]

ex2 1x

dx

Recall that

ex = 1 + x +x2

2 +

x3

6 + . . .

Therefore,

ex2

= 1 x2 + x4

2 x

6

6 + . . . ,

andex

2 1x

= x + x3

2 x

5

6 + . . . .

Finally,

ex2 1x dx= x2

2 +x4

8 x6

36+ . . .

[3](h) For what values ofr does the function y= xr satisfy the following differential equation

(for x >0)?x2y + 4xy + 2y = 0

Ify = xr theny =rxr1 and y =r(r 1)xr2. Therefore,x2y + 4xy + 2y= r(r 1)xr + 4rxr + 2xr = (r2 + 3r+ 2)xr = (r+ 1)(r+ 2)xr.

This will be zero when r = 1 or r= 2.

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(i) Calculate the derivative f(x) iff(x) =

exx

cos t dt.[3]

ex

x

cos t dt=

ex

0

cos t dt

x

0

cos tdt.

Therefore, by the chain rule and the first fundamental theorem of calculus,

f(x) =ex

cos ex cos x.

(j) Give the Simpsons rule approximation to

2

0

sin(ex)dx using 4 equal subintervals (do

not simplify your answer).[3]

Here x= 1/2, x0= 0, x1= 1/2, x2= 1, x3= 3/2, x4= 2. Therefore,

S4 = 1

6

sin(1) + 4 sin(e1/2) + 2 sin(e) + 4 sin(e3/2) + sin(e2)

.

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Full-Solution Problems. In the remaining questions, justify your answers and show all yourwork. If a box is provided, write your final answer there. If you need more space, use theback of the previous page. Unless otherwise indicated, simplification of answers isnot required.

2. For what values ofp does

e

dxx(ln x)p

converge?[6]

e

dx

x(ln x)p = lim

t

te

dx

x(ln x)p.

Letu = ln x so that du = dx/x. Whenx = e, u = 1 and when x = t, u = ln t. Therefore,

limt

t

e

dx

x(ln x)p = lim

t ln t

1

du

up.

Ifp = 1, then

limt

ln t

1

du

up = lim

t

ln t

1

du

u

= limt

[ln u]ln t1

= limt

ln(ln t) = .

Therefore ifp = 1, the integral does not converge.

Ifp

= 1, then

limt

ln t

1

du

up = lim

t

u1p

1pln t1

= limt

(ln t)1p

1p + 1

p 1=

ifp 1

Therefore, the integral converges if and only ifp >1.

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3. Let R be the finite region in the xy-plane bounded by x = 0, y = 0 and y = cos x.[10]

(a) Calculate the centroid ofR.

The area A ofR is

A=

/2

0

cos x dx= [sin x]/20

= 1.

The x-coordinate of the centroid x is

x=

/20

x cos x dx= [x sin x + cos x]/20

=

2 1.

We used integration by parts to evaluate the integral.

The y -coordinate of the centroid y is

y =

1

2 /20 cos

2

x dx

= 1

2

/20

1

2+

cos2x

2

dx

= 1

4

x +

sin2x

2

/20

=

8.

Therefore, the centroid ofR is (/2 1, /8).(b) Express the volume of the solid obtained by rotating R about the line x =2 as an

integral. Do not evaluate the integral.

V = 2

/20

(x + 2) cos xdx.

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4. Calculate the integrals:

(a)

(4 x2)3/2 dx[6]

Letx= 2 sin u so that dx= 2 cos u duand 4 x2

= 4 cos2

u. Thus, (4 x2)3/2 dx =

2cos u du

8cos3 u

= 1

4

sec2 u du

= 1

4tan u + C

= x

4

4 x2 + C.

(b) 1 + ex dx[6]

Letu=

1 + ex, i.e. u2 = 1 + ex. Then 2u du= ex dx= (u2 1) dx. Therefore,1 + ex dx=

2u2

u2 1du =

(2 + 2

u2 1 ) du.

By a partial fraction decomposition,

2

u2 1= 2

(u 1)(u + 1) = 1

u 1 1

u + 1.

Thus, (2 +

2

u2 1 ) du = 2u +

du

u 1

du

u + 1

= 2u + ln |u 1| ln |u + 1| + C= 2u + ln

u 1u + 1 + C

= 2

1 + ex + ln

1 + ex 11 + ex + 1

+ C.Alternatively: - which is really the same u-substitution as above, broken down into

two stepsLetu= 1 + ex, then ex =u 1, du = ex dx, so du

u 1=dx. Therefore,1 + ex dx=

u

u 1 du

Now let t =

u, sodt =1

2u

1

2 = du

2

u=

du

2t, i.e. 2t dt= du. Therefore

u

u 1 du=

t

t2 1 2t dt= 2

t2

t2 1 dt= 2

t2 1 + 1t2 1 dt= 2

1 +

1

t2 1 dt

Etc. - like above.8


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[10] 5. Let Xbe a random variable with probability density function

f(x) =

kx(1 x4) if 0 x 1,0 ifx 1.

(a) Find the value ofk .

1

0

x(1 x4) dx=

x2

2x

6

6

10

=1

3.

Therefore, we need to choose k = 3 in order for1

0 f(x) dx= 1.

(b) Find the mean.

=

1

0

xf(x) dx= 3

1

0

x2(1 x4) dx= 3

x3

3 x

7

7

10

=4

7.

(c) Find an algebraic equation satisfied by the median m.

The mean m is such that

1

2= 3

m0

x(1 x4) dx= 3

x2

2 x

6

6

m0

=3

2m2 m

6

2

orm6 3m2 + 1 = 0.

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6. Solve these differential equations:

(a) y =xy2 with initial conditionsy(0) = 1.[6]

This is a separable equation:

dy

dx = xy2

dy

y2 = x dx

dy

y2 =

x dx

1y

= x2

2 + C.

The initial conditiony(0) = 1 implies that C=

1. Therefore,

y= 1

1 x2/2 .

(b) y 2y = 4 + e3t. (Give the general solution)[6]

This is a linear equation. The integrating factor is I(t) = e 2 dt = e2t. Multiplying

both sides of the equation by I(t) yields

e2ty 2e2ty = 4e2t + et

(e

2t

y)

= 4e

2t

+ e

t

e2ty =

(4e2t + et) dt

e2ty = 2e2t + et + Cy = 2 + e3t + Ce2t.

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[10] 7. An open metal tank has two ends which are isosceles triangles with vertex at the bottom,two sides which are rectangular, and an open top. The tank is 1 metre wide, 2 metres deep.10 metres long and full of water (density =1000kg/m3)

(a) What is the hydrostatic force on each triangular end?

We choose a vertical x-axis with origin at the bottom of the tank. We divide the interval[0, 2] into n equal subintervals of length x. The i-th horizontal strip is approximatelya rectangle of height x and width wi where by similar triangles

wixi

=1

2 or wi=

xi2

.

Therefore the area of the i-th strip is approximately xi/2 x and the hydrostatic forceon the i-th strip is

Fi 9800(2 xi)xi/2 x.

Therefore the total hydrostatic force is

F = limn

ni=1

4900(2 xi)xix

= 4900

2

0

(2 x)x dx

= 4900

x2 x

3

3

20

= 19600

3 N .

(b) Express the work required to pump the water out of the tank, if it is pumped out overthe top edge, as an integral. (Do not evaluate the integral.)

With the same setup as in (a), the volume of the i-th slab is 10 xi/2 x = 5xix.Therefore the work Wi to pump the i-th slab is 9800 5(2 xi)xixand the total workdone is

W= 49000

2

0

(2 x)xdx.

(c) Express the hydrostatic force on each of the rectangular sides as an integral. (Do not

evaluate the integral.)

Note that the i-th strip is a rectangle with height (

17/4)x and width 10. Thereforethe force is

2450

17

2

0

xdx.

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[10] 8. Consider the chemical reactionA + B C + D.

Suppose at time t = 0 sec the concentration of chemical A is 0.1 mol/L, the concentrationof chemical B is 0.2 mol/L, and the concentrations of chemicals C and D are both 0. Fort 0, letx(t) be the concentration of chemical D in mol/L. It can be shown that x(t) is thesolution to the initial-value problem

dx

dt =k(0.1 x)(0.2 x), x(0) = 0,

wherek is a positive constant whose value can be determined by experiment.

(a) Solve the initial-value problem to find x(t) explicitly.

This is a separable equation.

dx

(0.1 x)(0.2 x) = k dt

dx

(0.1 x)(0.2 x) =

k dt

To evaluate the left hand side, we use partial fractions:

1

(0.1 x)(0.2 x) = 10

0.1 x 10

0.2 x ,

and therefore,

10

0.1

x 10

0.2

x

dx =

k dt

ln |0.2 x| ln |0.1 x| = kt + Cln

0.2 x0.1 x = kt + C

0.2 x0.1 x = Ae

kt

The initial conditionx(0) = 0 implies that A = 2. Therefore,

0.2 x0.1 x = 2e

kt

1 + 0.1

0.1

x = 2ekt

0.1

0.1 x = 2ekt 1

0.1 x = 120ekt 10

x = 0.1 120ekt 10

(b) What value does the concentration of chemical D approach ast approaches?

limt

x(t) = 0.1

(we couldve guessed this based on the physical interpretation).12

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Math 101 Final Answers