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Calculus and Analytic Geometry II: Exam I
Math 164 Section 51
February 27, 2003
Name: Signature:
- Questions 1–3 do not require any explanations. However, show all of your work. Partial credit will be very limited.
- Questions 4 and 5 are essay questions. The marks which you earn on these questions will depend upon the quality of your writing. Answers which are unclear or which contain grammatical errors will not earn full credit.
- (15 pts.) To the right is a graph of the function q.
(a) Suppose that Q′(s) = q(s) and that Q(6) = 12.0. Find Q(1).
By the Fundmantal Theorem of Calculus,
Q(6) − Q(1) =
1
q(s) ds
Q(1) = Q(6) −
1
q(s) ds.
We already know that Q(6) = 12. 0. So all that remains is to evaluate the definite integral on the right-hand side. We can estimate the definite integral by eigher constructing a Riemann sum or by interpreting the definite integral as an area; examining the graph, the signed area is approximately 6. 5 Hence,
Q(1) ≈ 12. 0 − 6 .5 = 5. 5.
(b) Evaluate ∫ (^3)
0
q′(2s − 1) ds.
y = q(s)
We can substitute for u = 2s − 1. With this substitution, ds = du/ 2 , and ∫ (^3)
0
q′(2s − 1) ds =
− 1
q′(u) du =
− 1
q′(u) du =
(q(5) − q(−1)).
(Notice the change in the limits of the definite integral.) The values of q(5) and q(−1) can be estimated from the graph. We hence arrive at ∫ (^3)
0
q′(2s − 1) ds ≈
- (10 pts.) Let
g′(t) =
t^2 and suppose that g(2) = 1. Find a formula for g.
The general antiderivative for g′(t) is ∫ 1 t^2
dt =
t−^2 dt = −t−^1 + C = −
t
+ C.
Hence, we know that g(t) = −
t
+ C
where C is some constant. We can solve for C by using the one data point given:
g(2) = −
+ C = 1
C =
Thus, g(t) =
t
One can also find a formula for g with a clever application of the Second Fundamental Theorem:
g(t) = 1 +
∫ (^) t
2
s^2
ds = 1 +
s
s=t
s=
t
t
- (10 pts.) In the year 2000, the United States produced carbon dioxide emissions at the rate of 1.57 billion metric tons each year. During the latter part of the nineties, carbon dioxide emissions increased by appoximately 1.8% each year. Assuming these trends hold, what will be the total amount of carbon dioxide emissions produced in the United States over the next 50 years? [Data from the International Energy Annual produced by the U.S. Department of Energy.]
The rate of carbon dioxide production is an exponential function with respect to time. If we let t be the number of years since the beginning of 2001 , then then number of billions of metric tons of carbon dioxide produced each year is
1 .57(1.018)t.
The amount of carbon dioxide that will be produced during the next fifty years is then ∫ (^53)
3
1 .57(1.018)t^ dt =
1 .57(1.018)t ln 1. 018
53
3
≈ 133. 7 billion metric tons.
Note that the definite integral could also be evaluated through numerical means. Also note that by integrating in the interval from t = 3 to t = 53, we are finding carbon dioxide emissions between 2003 and 2053. Integraint from t = 0 to t = 50 finds the carbon dioxide emissions between 2000 and 2050. Also, we are assuming that the 1 .8% is non-continuous growth rate. If we were to assume that carbon dioxide emissions increase continuously by 1 .8%, then the rate of carbon dioxide emissions would be
- 57 e^0.^018 t. The definite integral then would change accordingly.
In order to get three places of accuracy after the decimal point, we need to ensure that any errors are less than 0. 0005. First, let’s ensure the tail is small enough by choosing c large enough. Notice how we used the comparison function introduced earlier: ∫ (^) ∞
c
e(−t+sin^ t)^ dt ≤
c
e(−t+1)^ dt = lim b→∞
−e(−t+1)
b
c
= e(−c+1).
If we choose c such that e(−c+1)^ ≤ 0. 0005 , then c is sufficiently large. This will require c ≥ 1 − ln(0.0005) ≈ 8. 6. Let’s pick c = 9. Then the we can estimate the head using numerical methods. Using Simpson’s method with an increasing number of subdivisions n to estimate ∫ (^9)
0
e(−t+sin^ s)^ dt.
From the table on the right, we can be fairly confident to say that the improper integral evaluates to 1. 764.
n SIMP(n) 10 1. 76418 20 1. 76396 50 1. 76394
- (20 pts.) Consider the following table of data:
t 0. 0 0. 5 1. 0 1. 5 2. 0 2. 5 3. 0 P (t) 30. 8778 29. 9215 28. 1176 25. 6637 22. 8100 19. 8181 16. 7766
Estimate (^) ∫ 3
0
P (t) dt
using the right-hand sum, the left-hand sum, the midpoint rule, and the trapezoid rule. (You can determine the number of subdivisions which is reasonable.) What do you believe is the actual value of the integral? Explain in a well-organized essay using complete sentences. How much additional data would be required to compute the integral to an accuracy of four places after the decimal point? Again, explain using complete sentences. Use graphs, tables, and formulas where appropriate. If you need more space, use the back of the previous page.
The left-hand sum and the right-hand (with varying number of subdivisions) are shown below:
LHS(3) = (P (0.0) + P (1.0) + P (2.0))(1) = 81. 8054 , LHS(6) = (P (0.0) + P (0.5) + P (1.0) + P (1.5) + P (2.0) + P (2.5))(0.5) = 78. 60435 , RHS(3) = (P (1.0) + P (2.0) + P (3.0))(1) = 67. 7042 , RHS(6) = (P (0.5) + P (1.0) + P (1.5) + P (2.0) + P (2.5) + P (3.0))(0.5) = 71. 55375.
The trapezoid method can be arrived at by averaging the left-hand and right-hand sums.
TRAP(3) =
LHS(3) + RHS(3)
TRAP(6) =
LHS(6) + RHS(6)
We can also calculate the midpoint rule with 3 subdivisions. Notice that it is not possible to estimate the definite integral using the midpoint rule with 6 subdivisions. Also, notice that the midpoint values are not the average of the left-hand and right-hand values.
MID(3) = (P (0.5) + P (1.5) + (2.5))(1) = 75. 4033.
Analyzing the data, we notice that P appears to be decreasing and concave down. Because P is decreasing, the left-hand sums are overestimates while the right-hand sums are underestimates. Because P is concave down, the trapezoid rule yields an underestimate while the midpoint rule yields an overestimate. Using the trapezoid rule with 6 subdivisions and the midpoint rule with 3 subdivisions, we can be fairly certain that
0
P (t) dt ≤ 75. 4033.
Based on the above inequality, we can tell the the errors of the left-hand and the right-hand sums with 6 subdivisions are both less than 4. In order to get 4 digits of accuracy after the decimal point, we need to reduce the error to less than
- 00005 , or down by a factor of 4 / 0 .00005 = 80, 0000. So we need to increase the number of subdivisions by a factor of 80 , 000. Thus, we can estimate that it will take approximately 6 × 80 , 0000 = 480, 000 subdivisions in order to reduce the error to something low enough.
The midpoint and trapezoid rules with 3 subdivisions have errors less than 0. 65. So with these methods, we need to reduce the error by a factor of
- 65 / 0 .00005 = 13, 000. Since the midpoint and trapezoid rules are second-order approximations, the error is proportion is 1 /n^2 , and we only need to raise the number of subdivisions by a factor of 13 , 000 (1/2)^ ≈ 115. Hence, to get the required accuracy with the midpoint or the trapezoid methods will require at most 3 × 115 , or something on the order of 350 , subdivisions. This is a lot better than the left-hand and right-hand rules.
We could also try to estimate the definite integral using Simpson’s method, which is a weighted average of the trapezoid and midpoint rules. Simpson’s method with 3 subdivisions gives
SIMP(3) =
TRAP(3) + 2MID(3)
(Note that we need to use the trapezoid and the midpoint rules with equal number of subdivisions.) Based on our earlier upper and lower bounds for the definite integral, we can tell that the error for Simpson’s method with 3 subdivisions is at most 0. 25. To get an accuracy of within 4 places after the decimal point, we need to reduce the error by a factor of 0. 25 / 0 .00005 = 5000. Simpson’s method is a fourth- order approximation with an error proportional to 1 /n^4 , so we need to increase the number of subdivisions only by a factor of 5000 (1/4)^ ≈ 8. 4. So to get the required accuracy using Simpson’s method would require only about 3 × 8. 4 or around 30 subdivisions!
