SOLUTIONS - Northwestern University

MATH 214-2 - Fall 2001 - Final Exam (solutions)

SOLUTIONS

1. (Numerical Integration) Find the Trapezoidal (T4), Midpoint (M4) and Simpson's (S4) approximations with 4 subintervals to the following integral:

2

x2 dx .

-2

Solution: Note that in all cases x = 2 - (-2) = 1. 4

1. Trapezoidal approximation:

T4

=

x 2

?

{y02

+

2y12

+ 2y22

+ 2y32

+ y42}

= 1 ? {(-2)2 + 2 ? (-1)2 + 2 ? 02 + 2 ? 12 + 22} = 6 2

2. Midpoint approximation:

M4 = x ? {y12/2 + y32/2 + y52/2 + y72/2} = 1 ? {(-1.5)2 + (-0.5)2 + 0.52 + 1.52} = 5

3. Simpson's approximation:

S4

=

x 3

? {y02

+

4y12

+

2y22

+

4y32

+ y42}

= 1 ? {(-2)2 + 4 ? (-1)2 + 2 ? 02 + 4 ? 12 + 22} = 16

3

3

1

2. (Volumes of solids) Find the volume of the solid obtained by rotating around the y-axis the area under y = sin x from x = 0 to x = .

Solution: We use the method of cylindrical shells:

V = 2xy dx = 2 x sin x dx .

0

0

The integral can be evaluated by parts, using u = x, v = - cos x:

x sin x dx = -x cos x + cos x dx = -x cos x + sin x + C ,

Hence:

V = 2 - x cos x + sin x = 2(- cos ) = 22

0

2

3. (Surface Areas) Find the area of the surface obtained by revolving the curve

y

=

2 3

x3/2,

3

x

8,

around

the

x-axis--just

setup

the

integral,

do

not

try

to evaluate it.

Solution:

A=

x=8

2y ds =

8

2y

1 + (y )2 dx = 4

8 x3/2 1 + x dx

x=3

3

33

3

4. (Separable Differential Equations) Solve the following initial value problem:

dx

dt

= x (1 - x)

x(0)

=

1 2

Solution: Separating variables we get: dx = dt . x (1 - x)

The left hand side can be integrated in the following way:

dx =

x (1 - x)

11

+

dx = ln x - ln (1 - x) + C ,

x 1-x

hence:

ln x - ln (1 - x) = t + C .

According to the initial condition we have:

ln (1/2) - ln (1/2) = C C = 0 .

So the solution is

ln x - ln (1 - x) = t

x = et .

1-x

Solving for x we get:

et x(t) = 1 + et

4

5. (Logarithmic Differentiation) Use logarithmic differentiation to find the deriva-

tive of

1 + x2 4 1 + x4

y=

3 1 + x3 5 1 + x5

Solution: First we take logarithms and simplify:

ln y = 1 ln (1 + x2) - 1 ln (1 + x3) + 1 ln (1 + x4) - 1 ln (1 + x5) .

2

3

4

5

Next we differentiate:

y

x

x2

x3

x4

y = 1 + x2 - 1 + x3 + 1 + x4 - 1 + x5 .

Hence:

1 + x2 4 1 + x4

y =

3 1 + x3 5 1 + x5

x

x2

x3

x4

1 + x2 - 1 + x3 + 1 + x4 - 1 + x5

5

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