**Unformatted text preview: **Math for Econ II, Written Assignment 6 (30 points)
Due Friday, November 6th
Please write neat solutions for the problems below. Show all your work. If you only write the answer with no work,
you will not be given any credit.
• Write your name and recitation section number.
• Staple your homework if you have multiple pages!
5-4h43 1. (4 pts) Using the graph of f below, arrange the following quantities in increasing order, from least to greatest.
1
0 (iii) In Problems ??, evaluate the expression, if possible, or
say what additional information is needed, given that
2
1
2
3
1
4
2
ii) 1 f (x)(x) dx iii) 2 f (x) dx g(x) dx 0 f (x) dx v) 2 f (x) dx vi) 1 f (x) dx
iv) = 12.
(ii)
f dx
−4
1 1 i) (x) dx dx
f 0 f (x)
2 0 (iv) f (x) dx 3 f (x) dx 2 4 20 viii) The 0 5-4h47 47.
(v) − vii) (x) dx
f The number (vi) The numbernumber −10
1 (vii) 2 The number 20 (viii) 1
5-4h43fig ins5-4h49-52 2 −10 5-4h49 7 49. 50. f (x) dx
0 44. (a) Using Figures ?? and ??, ﬁnd the average value on
Solution:
5
7
0 ≤ x ≤ 2 of
5-4h51 51. 3
5-4h52
1
2
f (x + 2) dx be the 52.
(f areas dx
dx g(x)
f (x) dx = A2
f
(positive)(x) + 2)of the three pieces cut out
(i) Let 0 f (x) (ii) = A1 , − 1(iii) f (x)·g(x) and 2 −2(x) dx = A3
f (x)
0
by the graph of y = true? and the x-axis.
(b) Is the following statement f (x) Explain your an- By visual inspection we see that 0 < A1 < A2 < A3 . Also, since |f (x)|
is
swer. smaller than 10 from [0, 2] and smaller than 20 on [2, 3], we see that A2 < 10 and A3 < 30.
2
5-4h53 53.
of (x) √
√
In terms of· the Ai , the integrals are equal to (a) Sketch a graph√ f√ = sin(x ) and mark on it
Average(f ) Average(g) = Average(f · g)
the points x = π, 2π, 3π, 4π.
(b) Use your graph to decide which of the 1
four numbers
1
2
1
1
f (x)
g(x) f (x) dx = A1 , (ii) = 0 5-4h44figa x 5-4h44figb 1 2 1 1 (iii) = 2 sin(x ) dx f (x) dx = A2 , 2 n = 1, 2, 3, 4 3 0 2 f (x) dx = A1 − A2 , (iv) = f (x) dx nπ −A2 ,
√= x 2 Figure 5.72 (v) = 1 f (x) dx = A3 , (vi) = f (x) dx = 0. is largest. Which is smallest? How many of the num2
1
bers are positive? 0
Figure 5.73 We need to arrange the eight numbers A1 , −A2 , A2 , A1 − A2 , A3 , 0, -10 and 20 in increasing order, knowing that 45. (a) Without A1 < A2 < A3integrals, explain< 10.the For Problems ??, assuming F = f , mark the quantity on a
why The result is
0 < computing any < 20 and A2 ins5-4h54-56
average value of f (x) = sin x on [0, π] must be be- copy of Figure ??.
tween 0.5 and 1.
−10 < −A2 < A1 − A2 < 0 < A1 < A2 < A3 < 20,
(b) Compute this average.
F (x)
which corresponds normal distribution from
46. Figure ?? shows the standard to
statistics, which is given by
(viii) < (ii) < (iv) < (vi) < (i) < (iii) < (v) < (vii).
1 −x2 /2
−1
4
4
√
.
2. (2 pts) Supposeef is even, −2 f (x) dx = 3, and 2 f (x) dx = 5. Either ﬁnd 1 f (x) dx, or show there is not
2π
4
enough information to ﬁnd as the followStatistics books often contain tables such 1 f (x) dx.
ing, which show the area under the curve from 0 to b for
x
ins5-4h54-56fig
a
b
various Solution: Because f is even, −1 f (x) dx = 2 f (x) dx = 3, so
values of b.
−2 ' Area = √1 2π 1 b 4−x2 /2
e
0 2 f (x) dx
dx = 1 5-4h46fig 3.5 5-4h50 f (x) dx
0 (i) = 5-4h46 g(−x) dx
−4 0 !
Figure 5.71 48. In Problems ??, evaluate the expression if possible, or say
7
what extra information is needed, given 0 f (x) dx = 25. x 3 4 5-4h48 g(x) dx The number −10 f (x) 10 5-4h45 311 43. Using the graph of f in Figure ??, arrange the ins5-4h47-48
following
Explain your reasoning!
quantities in increasing order, from least to greatest.
(i) 5-4h44 5.4 THEOREMS ABOUT DEFINITE INTEGRALS enough information to ﬁnd
Figure 5.74
b √1
2π b −x2 /2
e
0 1 0.3413 2 g(x) dx. 5-4h56 0.4987 4 5-4h55 0.5000 5-4h57 Use the information given in the table and the symmetry 2 54. A slope representing f (a). g(x) dx = 2, and
−2 0.4772 3 dx 6
2 5-4h54 f (x) dx = 3 + 5 = 8 . f (x) dx +
1 x 4 3. (2 pts) Suppose that g is odd,
0
b Figure 5.75 4 −4
−6 b
g(x) dx = 3. Either ﬁnd 55. A length representing f (x) dx. 6
2 g(x) dx, or show there is not a 56. A slope representing 1
b−a b f (x) dx.
a 57. In Chapter ??, the average velocity over the time interval a ≤ t ≤ b was deﬁned to be (s(b) − s(a))/(b − a),
where s(t) is the position function. Use the Fundamental
Theorem of Calculus to show that the average value of
the velocity function v(t), on the interval a ≤ t ≤ b, is
also (s(b) − s(a))/(b − a). Solution: Because g is odd, 2
−2 g(x) dx = 0, so
2 4 4 6
4 g(x) dx = − −4
−6 −2 −2 2 Also, g(x) dx = 2 − 0 = 2. g(x) dx − g(x) dx =
g(x) dx = −3. Hence
6 4 g(x) dx = 6 g(x) dx = 2 − 3 = −1. g(x) dx + 2 2 4 4. (5 pts) Evaluate the following integrals:
1
(f (x)
0 (a) (3 pts) If 1
(2f (x)
0 − 2g(x)) dx = 6 and − g(x)) dx = 9, ﬁnd 1
(f (x)
0 − g(x)) dx. 1 (x17 − x5 + x3 − x) dx (b) (2 pts)
−1 Solution:
1 1 (a) Denote A = 0 f (x) dx and B = 0 g(x) dx. We are asked to ﬁnd A − B given that A − 2B = 6 and
2A − B = 9. Adding these equations gives 3A − 3B = 15, and dividing by 3 gives A − B = 5.
1 (x17 − x5 + x3 − x) dx = 0 because f (x) = x17 − x5 + x3 − x is an odd function. (b)
−1 (check f is odd: f (−x) = (−x)17 −(−x)5 +(−x)3 −(−x) = −x17 +x5 −x3 +x = −(x17 −x5 +x3 −x) = −f (x)). 5. (2 pts) A honeybee population starts with 100 bees and increases at a rate of n (t) bees per week. What does
15
100 + 0 n (t) dt represent?
15 Solution: 0 n (t) dt is the net change in honeybee population from t = 0 to t = 15 weeks. Therefore,
15
100 + 0 n (t) dt is the number of bees in the colony 15 weeks after the start.
6. (2 pts) Explain what is wrong with the following calculation of the area under the curve
1
−1 1
x2 from x = −1 to x = 1: 1 1
1
dx = −
2
x
x −1 = −1 − 1 = −2. 1
Solution: We cannot apply the evaluation theorem because x2 is not continuous on [−1, 1] (it has an inﬁnite
discontinuity when x = 0). Whatever the integral is (if it even exists!), it can’t be negative, because 1/x2 is a
strictly positive function with plenty of positive area. 7. (3 pts) Suppose the area under the curve ex from x = 0 to x = a is twenty times the area under the curve 2e2x
from x = 0 to x = b. Solve for a in terms of b. (in other words, write a =(some formula involving b))
Solution: We use the Evaluation Theorem.
a b ex dx = 20
0
a ex dx = ex
0 a 2e2x dx.
0 b = ea − 1, 0 2e2x dx = 20 e2x 20
0 b = 20e2b − 20. 0 Thus
ea − 1 = 20e2b − 20, ea = 20e2b − 19, a = ln(20e2b − 19) . 2 1
dx = ln(2). Now ﬁnd a fraction which approximates ln(2), by
x
1
2
1
using M4 (midpoint sum with 4 rectangles) to approximate
dx.
1 x
(The actual value of ln(5) is 0.6931 . . . For fun, plug your approximation into a calculator and compare) 8. (4 pts) Use the Evaluation Theorem to show 2 Solution: By the Evaluation Theorem,
1 1
dx = ln(x)
x 2 = ln(2) − ln(1) = ln(2).
1 (the formulas for ln |x| and ln(x) are identical since x > 0 on the interval [1, 2]).
M4 = 1 · f 3
5
7
9
2 2 2 2
496
+1·f
+1·f
+1·f
= + + + =
2
2
2
2
3 5 7 9
315 How good is our approximation? ln(5) = 1.6094.... Our approximating fraction is 496
315 = 1.5746... 6300
and the supply curve is given by P = Q + 20. Find
Q + 40
the equilibrium price and quantity, and compute the consumer and producer surplus. 9. (6 pts) Suppose the demand curve is given by P = Solution: To ﬁnd Q∗ we set both values of P equal:
6300
= Q∗ + 20,
Q∗ + 40 6300 = (Q∗ + 40)(Q∗ + 20), (Q∗ )2 + 60Q∗ − 5500 = 0, 6300 = (Q∗ )2 + 60Q∗ + 800, (Q∗ − 50)(Q∗ + 110) = 0, Q∗ = 50 , Q∗ =-110. For practical purposes we ignored the negative equilibrium quantity Q∗ = −110 above.
Thus P ∗ = Q∗ + 20 = 50 + 20 = 70, so P ∗ = 70 .
Q∗ 50 (f (Q) − P ∗ ) dQ = CS =
0 0 6300
− 70
Q + 40 50 dQ = [6300 ln(Q + 40) − 70Q]
0 = (6300 ln(90) − 70(50)) − 6300 ln(40) = 6300 ln(9/4) − 3500 .
Also,
Q∗ 50 (P ∗ − g(Q)) dQ = PS =
0 50 (70 − (Q + 20)) dQ =
0 (50 − Q) dQ = [50Q −
0 Q2
2 50 =
0 2500
= 1250 .
2 Some extra practice (not to be handed in)
1. Suppose h is a function such that h(1) = −2, h (1) = 3, h (1) = 4, h(2) = 6, h (2) = 5, h (2) = 13, and h is
2
continuous everywhere. Find 1 h (u) du. 2. Use the ﬁgure below to ﬁnd the values of
(a)
(b) b
c
c
a f (x) dx
|f (x)| dx 5-2h29 29. Use Figure ?? to ﬁnd the values of
(a)
(c) b
a
c
a f (x) dx
f (x) dx 5-2h33
c (b)
(d) b
c
a 33. (a) Graph f (x) = x(x + 2)(x − 1).
(b) Find the total area between the graph and the x-axis
between x = −2 and x = 1.
1
(c) Find −2 f (x) dx and interpret it in terms of areas. 5-2h34 34. Compute the deﬁnite integral
the result in terms of areas. f (x) dx
|f (x)| dx f (x)
Area = 13 © 4
0 cos √ x dx and interpret 5-2h35 a b 5-2h29fig 35. Without computation, decide if 0 e−x sin x dx is positive or negative. [Hint: Sketch e−x sin x.] 5-2h36 36. Estimate x c sArea = 2 ! (a) Figure 5.36
5-2h30 30. Given 2 −2 (b) Left-hand sum Right-hand sum 5-2h38 38. (a) Draw the rectangles that give the left-hand sum apπ
proximation to 0 sin x dx with n = 2. f (x)dx f (x)
2
x
−2 2
−2 0 Figure 5.37 5-2h31fig 2 e−x dx using n = 5 rectangles to form a 37. (a) On a sketch of y = ln x, represent the left Riemann
2
sum with n = 2 approximating 1 ln x dx. Write
out the terms in the sum, but do not evaluate it.
(b) On another sketch, represent the right Riemann sum
2
with n = 2 approximating 1 ln x dx. Write out the
terms in the sum, but do not evaluate it.
(c) Which sum is an overestimate? Which sum is an underestimate? f (x)dx = 4 and Figure ??, estimate:
−2 2 5-2h31 1
0 5-2h37 0 f (x)dx
(b)
(a)
0
(c) The total shaded area. 5-2h30fig 2π 0 (b) Repeat part (a) for −π sin x dx.
(c) From your answers to parts (a) and (b), what is
the value of the left-hand sum approximation to
π
sin x dx with n = 4?
−π 31. (a) Using Figure ??, ﬁnd −3 f (x) dx.
(b) If the area of the shaded region is A, estimate
5-2h39 39. (a) Use a calculator or computer to ﬁnd 6 (x2 + 1) dx.
4
0
f (x) dx.
−3
Represent this value as the area under a curve.
6
(b) Estimate 0 (x2 + 1) dx using a left-hand sum with
1
f (x)
n = 3. Represent this sum graphically on a sketch
4
x
of f (x) = x2 + 1. Is this sum an overestimate or
−4 −3 −2 −1
1
2
3
5
underestimate of the true value found in part (a)?
−1
6
(c) Estimate 0 (x2 +1) dx using a right-hand sum with
n = 3. Represent this sum on your sketch. Is this
Figure 5.38
sum an overestimate or underestimate? ...

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