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Midterm Exam 3 with Solution - Business Calculus | MATH 135, Exams of Calculus

Material Type: Exam; Professor: Annin; Class: Business Calculus; Subject: Mathematics; University: California State University - Fullerton; Term: Spring 2009;

Typology: Exams

Pre 2010

Uploaded on 08/16/2009

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May 4, 2009 Midterm III Name:
Math 135
SOLUTIONS
Problem 1. (6 points each) Evaluate each indefinite integral below:
(a):
Z(x24x3)dx
SOLUTION: We have
Z(x24x3)dx =1
3x32x23x+C.
2
(b):
Zt
(t2+ 1)2dt
SOLUTION: We need to use a u-substitution with u=t2+ 1 and du = 2t dt. Thus,
Zt
(t2+ 1)2dt =1
2Z2t
(t2+ 1)2dt
=1
2Z1
u2du
=1
2Zu2du
=1
2·u1
1+C
=1
2u+C
=1
2(t2+ 1) +C.
2
(c):
Z4
x6
x+x7.2dx.
pf3
pf4
pf5

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May 4, 2009 Midterm III Name:

Math 135

SOLUTIONS

Problem 1. (6 points each) Evaluate each indefinite integral below:

(a): (^) ∫

(x^2 − 4 x − 3)dx

SOLUTION: We have ∫ (x^2 − 4 x − 3)dx =

x^3 − 2 x^2 − 3 x

+ C.

(b): (^) ∫ t (t^2 + 1)^2

dt

SOLUTION: We need to use a u-substitution with u = t^2 + 1 and du = 2t dt. Thus, ∫ t (t^2 + 1)^2 dt =

2 t (t^2 + 1)^2 dt

u^2

du

u−^2 du

u−^1 − 1

+ C

2 u

+ C

2(t^2 + 1)

+ C.

(c): (^) ∫ ( 4 √ x

x

  • x^7.^2

dx.

SOLUTION: We have ∫ (^ 4 √ x

x

  • x^7.^2

dx =

4 x−^1 /^2 −

x

  • x^7.^2

dx

4 x^1 /^2 1 / 2

− 6 ln x + x^8.^2

  1. 2

+ C

x − 6 ln x +

x^8.^2

+ C.

Problem 2. (6 points each) Evaluate each definite integral below. You may use a calculator, but you must show all of your work as well.

(a): ∫ (^3)

− 2

(6x^2 − 5)dx.

SOLUTION: We have ∫ (^3)

− 2

(6x^2 − 5)dx = (2x^3 − 5 x)

∣^3 − 2

= (2 · 33 − 5 · 3) − (2 · (−2)^3 − 5(−2))

(b): ∫ (^1)

0

x^3 ex

4 dx.

SOLUTION: We use u-substitution with u = x^4 and du = 4x^3 dx. Then for the indefinite integral, we get (^) ∫

x^3 ex

4 dx =

4 x^3 ex

4 dx

eudu

=

eu^ + C

=

ex

4

  • C.

Now we evaluate: (^) ( 1 4

ex

4

1

0

(e^1 − e^0 ) = e − 1 4

to three decimal places.

SOLUTION: We have a = 3 and b = 13, so b−na = 13 − 5 3 = 2. So the approximation is given by

2 [f (3) + f (5) + f (7) + f (9) + f (11)] ≈ 2[0.910 + 0.621 + 0.514 + 0.455 + .417] = 5. 834.

2

(b): (4 points) Is the estimate in part (b) an under-estimate or an over-estimate of the exact value of the integral? Circle your answer and provide a brief justification.

SOLUTION: OVERESTIMATE. In a decreasing function, the height of each rectangle used in the approximation of the area exceeds the height of the function as each sub-interval is computed. That is, the rectangles sit above the level of the curve f (x) itself, thereby giving an area that is too large, compared with the actual area. 2

Problem 5.

(a): (9 points) The polonium isotope Po^210 has a half-life of 140 days. If a sample weighs 20 mg initially and decays exponentially, how much will be left after two weeks (14 days)?

SOLUTION: The exponential decay model is

P (t) = P 0 ekt,

where P (t) is the amount of polonium remaining at time t (measured in days). We are given that P (0) = 20 (mg), and we are given that P (140) = 12 P 0 , so we can solve for k:

1 2

P 0 = P (140) = P 0 e^140 k.

Thus, 1 2

= e^140 k.

Taking the natural logarithm of each side, we obtain

ln

= 140k,

so that

k =

ln

So the complete formula is P (t) = 20e−^0.^004951 t.

After two weeks, there will be

P (14) = 20e−^0.^004951 ·^14 ≈ 18 .66 (mg).

2

(b): (12 points) A hot plate ( 150 ◦F) is placed on a countertop in a room kept at 70 ◦F. If the plate cools to 125 ◦F in 10 minutes, when does the plate reach 100 ◦F?

SOLUTION: Newton’s Law of Cooling dictates that the temperature of the object at time t (minutes, in this case) is given by T (t) = ae−kt^ + c,

where c is the room temperature (70◦F), and a and k are solved for by using the given information. We have 150 = T (0) = a + c = a + 70,

so a = 80. So far, we have T (t) = 80e−kt^ + 70.

Next, 125 = T (10) = 80e−^10 k^ + 70,

so

e−^10 k^ =

Taking the natural logarithm of both sides, we obtain

− 10 k = ln

Therefore, k = 0. 0374693.

We conclude that the complete formula is

T (t) = 80e−^0.^0374693 t^ + 70.

The plate reaches 100◦F at time t such that

100 = T (t) = 80e−^0.^0374693 t^ + 70.

That is,

e−^0.^0374693 t^ =

Taking the natural logarithm again, we have

− 0. 0374693 t = ln

Solving for t, we conclude that the desired time is

t =

≈ 26 .18 minutes.

2

Problem 6. (7 points) The time to failure, t, in hours, of a machine is exponentially distributed with a probability density function

f (t) = 0. 01 e−^0.^01 t