Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
The solutions to problem 1-7 of a math 135 midterm exam. The problems involve evaluating indefinite and definite integrals using substitution and calculus. The solutions also include the use of newton's law of cooling and the concept of elasticity of demand.
Typology: Exams
1 / 7
This page cannot be seen from the preview
Don't miss anything!
Problem 1. (6 points each) Evaluate each indefinite integral below.
(a): (^) ∫ ( 3 √ x
x
dx
SOLUTION: We have ∫ (^ 3 √ x
x
dx =
3 x−^1 /^2 −
x
dx
3 x^1 /^2 1 / 2
− 5 ln x + 8 ·
e^2 x^ + C
= 6
x − 5 ln x + 4e^2 x^ + C.
2
(b): (^) ∫ t^5 4 t^6 + 3
dt
SOLUTION: We need to use u-substitution here. Let u = 4t^6 + 3. Then du = 24t^5 dt. Thus,
∫ t^5 4 t^6 + 3
dt =
24 t^5 4 t^6 + 3
dt
du u
=
ln u + C
=
ln(4t^6 + 3) + C.
2
(c): (^) ∫
x^7 (− 5 x^8 + 3)^11 dx
SOLUTION: We need to use u-substitution. Let u = − 5 x^8 + 3. Then du = − 40 x^7 dx. Thus, ∫ x^7 (− 5 x^8 + 3)^11 dx = −
− 40 x^7 (− 5 x^8 + 3)^11 dx
u^11 du
u^12 12
(− 5 x^8 + 3)^12 + C.
2
Problem 2. (6 points each) Evaluate each definite integral below. You may use a calculator, but you must show your work as well.
(a): ∫ (^) − 1
− 2
e−^4 tdt
SOLUTION: We have
∫ (^) − 1
− 2
e−^4 tdt = −
e−^4 t
− 1
− 2
e^4 +
e^8 =
(e^8 − e^4 ) ≈ 731. 59.
(b): (^) ∫ 5
e
x
dx
SOLUTION: We have
∫ (^5)
e
x
dx =
3 ln x + 5x −
x^5 /^2
5
e
3 ln 5 + 25 − 2 · 53 /^2
3 ln e + 5e −
e^5 /^2
(c): (^) ∫ 3
0
9 − x^2 dx
SOLUTION: The equation y =
9 − x^2 can be rearranged to read x^2 + y^2 = 9, which is a circle of radius 3, centered at the origin. We are only considered the one-fourth of the circle in the first quadrant (since x is in the interval [0, 3] and y =
9 − x^2 is positive). Since the integral of a positive
and taking the natural logarithm of both sides once more, we find that
− 0. 0223 t = ln
We immediately solve for the half-life t:
t = −
ln
≈ 31 .063 minutes.
Problem 4. (10 points) At 2 p.m. on a cool (34◦^ F) afternoon in March, Sherlock Holmes measured the temperature of a dead body to be 38◦F. One hour later, the temperature was 36◦F. After a quick calculation using Newton’s law of cooling, and taking the normal temperature of a living body to be 98◦F, Holmes concluded that the time of death was 10 a.m. Was Holmes right?
SOLUTION: YES. Newton’s law of cooling leads to the general equation
T (t) = ae−kt^ + C,
where a, k, and C are constants to be determined and T (t) is the function representing the body temperature as a function of time t. Let us set t = 0 at 2 p.m. and measure t in hours. We have
38 = T (0) = a + C.
Recall that C represents the ambient air temperature, which is 34. Hence,
38 = a + 34,
so that a = 4. Now 36 = T (1) = 4e−k^ + 34.
Hence, 2 = 4e−k.
Therefore, we have
e−k^ =
Taking the natural logarithm of both sides, we find that −k = ln
2
, so that k ≈ 0 .693. We conclude that the temperature function is
T (t) = 4e−^0.^693 t^ + 34.
Let us determine the time at which T (t) = 98. We have
98 = 4e−^0.^693 t^ + 34.
Hence, 16 = e−^0.^693 t.
Taking the logarithm of both sides, this becomes
− 0. 693 t = ln 16,
so that
t =
ln 16 − 0. 693
This means the time of death was at t = −4, which occurs 4 hours before 2 p.m., at 10 a.m., exactly as Sherlock predicted!
Problem 5. Let D(x) = (x − 6)^2 be the price (in dollars per unit) that consumers are willing to pay for x units of an item, and let S(x) = x^2 + 12 be the price (in dollars per unit) that producers are willing to accept for x units of the item.
(a): (4 points) Find the equilibrium point.
SOLUTION: We set D(x) = S(x):
(x − 6)^2 = x^2 + 12 x^2 − 12 x + 36 = x^2 + 12 12 x = 24.
Therefore, x = 2. The y-value is obtained from D(2) = S(2) = 16. Hence, the equilibrium occurs at a quantity Q = 2 and price P = 16. 2
(b): (4 points) Find the consumer surplus at the equilibrium point.
SOLUTION: The consumer surplus is ∫ (^) Q
0
D(x)dx − P Q =
0
(x − 6)^2 dx − (16)(2)
0
(x^2 − 12 x + 36)dx − 32
x^3 − 6 x^2 + 36x
2
0
dollars.
2
(c): (4 points) Find the producer surplus at the equilibrium point.
SOLUTION: The producer surplus is
0
S(x)dx = (16)(2) −
0
(x^2 + 12)dx
x^3 + 12x
2
0 = 32 −
dollars.
(a): (8 points) Use rectangles to estimate the value of
∫ (^11)
1
xex^ dx
by dividing the interval [1, 11] into 5 subintervals.
SOLUTION: We have a = 1, b = 11, and n = 5, so the width of each rectangle is b−na = 2. We have x 1 = 1, x 2 = 3, x 3 = 5, x 4 = 7, and x 5 = 9, so our approximation of area is
2
1 e^1 + 3e^3 + 5e^5 + 7e^7 + 9e^9
(b): (4 points) Using the fact that xex^ is increasing on the interval [1, 11], does your estimate in part (a) overestimate or underestimate the exact value of the integral? Circle your answer and explain your answer carefully.
SOLUTION: UNDERESTIMATE. If you draw a graph of an increasing function, you can see that the rectangles fit inside the region under the curve. This means that the rectangles come up short in estimating the actual area bounded beneath the curve. That is, our estimate in (a) is an UNDERESTIMATE. 2
