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Material Type: Exam; Professor: Annin; Class: Business Calculus; Subject: Mathematics; University: California State University - Fullerton; Term: Spring 2009;
Typology: Exams
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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
(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
2 u
2(t^2 + 1)
(c): (^) ∫ ( 4 √ x
x
dx.
SOLUTION: We have ∫ (^ 4 √ x
x
dx =
4 x−^1 /^2 −
x
dx
4 x^1 /^2 1 / 2
− 6 ln x + x^8.^2
x − 6 ln x +
x^8.^2
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)
(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
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