Download answers, in pdf - People @ EECS at UC Berkeley

016A Homework 14 Solution
Jae-young Park∗
December 6, 2008
• 6.2 #4 Determine ∆x and the midpoints of the subintervals for 3 ≤
x ≤ 5, n = 5
2
∆x
Solution ∆x = 5−3
5 = 5 = .4. The first midpoint is 3 + 2 = 3.2
The other four midpoints are found by adding ∆x = .4 successively.
So we get
3.2 3.6 4 4.4 4.8
• 6.2 #6 Estimate the area under the graph of f (x) = x2 for −2 ≤ x ≤ 2
using the Riemann sum for n = 4. with midpoints of subintervals.
Solution ∆x =
2−(−2)
4
= 1. So the midpoints are given by
−1.5
− .5
.5
1.5
So the Riemann sum is
∆x[(−1.5)2 + (−.5)2 + (.5)2 + (1.5)2 ] = 5
• 6.2 #12 3 ≤ x ≤ 7, n = 4, left endpoints.
Solution The length of the subintervals are ∆x = 1. Reading
four values at the left end points, we get f (3) = 8, f (4) = 7, f (5) =
6, f (6) = 4. So the Riemann sum is
∆x[f (3) + f (4) + f (5) + f (6)] = 8 + 7 + 6 + 4 = 25
To draw approximating rectangles, sketch 4 rectangles with width 1;
The first rests on the interval [3,4] with height 8, the second rests on
the interval [4,5] with height 7, the 3rd is on the interval [5,6] with
height 6 and the last rectangle is on the interval [6,7] with height 4.
∗
jaypark at m a t h . b e r k e l e y . e d u. GSI for 16A 101,104,105
1
• 6.3 #10 Calculate
R4
1
√
x2 xdx.
√
Solution
Let F (x) be an antiderivative of x2 x. Then F (x) =
R
√
x2 x = x5/2 = 72 x7/2 + C where C is an integration constant. By
the fundamental theorem of calculus, we have
Z 4
√
2
2
28 − 2
254
x2 x = F (4) − F (1) = [ 47/2 + C] − [ 17/2 + C] =
=
7
7
7
7
1
R
• 6.3 #12 Calculate
R1
0
(4x3 − 1)dx.
Solution Let F (x) be an antiderivative of 4x3 − 1. Then F (x) =
4x3 − 1 = x4 − x + C where C is an integration constant. By the
fundamental theorem of calculus, we have
Z 1
4x3 − 1 = F (1) − F (0) = [14 − 1 + C] − [0 − 0 + C] = 0
R
0
• 6.3 #22 Find
R2
1
( x2 +
2
x
+
1
)dx.
2x2
2
−1
Solution The antiderivative of ( x2 + x2 + 2x12 ) is x4 + 2 ln x + x−2 .
So we have
Z 2
x 2
1
22
1
( + + 2 ) = [ +2 ln 2−1/4+C]−[ +2 ln 1−1/2+C] = 2 ln 2+3/2
4
4
1 2 x 2x
• 6.3 #24 Calculate
R1
−1
1−x−e−x
dx.
2
Solution Let F (x) be an antiderivative of
R
1−x−e−x
2
x−(1/2)x2 +e−x
2
1−x−e−x
.
2
Then F (x) =
=
+ C where C is an integration constant. By
the fundamental theorem of calculus, we have
Z 1
1 − x − e−x
1 − (1/2)12 + e−1
−1 − (1/2)(−1)2 + e−(−1)
=[
+C]−[
+C]
2
2
2
−1
R1
−x
−1
= 1 + e 2 − 2e
So, −1 1−x−e
2
R1
• 6.3 #26 Calculate 0 (ex/3 − 2x
5 )dx.
Solution Let F (x) be an antiderivative of (ex/3 − 2x
5 ). Then F (x) =
2x
x2
x/3
x/3
(e
− 5 ) = (3e
− 5 ) + C where C is an integration constant.
By the fundamental theorem of calculus, we have
Z 1
2x
12
02
16
(ex/3 − ) = F (1)−F (0) = [(3e1/3 − )+C]−[(3e0/3 − )+C] = 3e1/3 −
5
5
5
5
0
R
2
• 6.3 #28
R9
3
4 t−2 dt
Solution
Z
4
9
3
dt = [3 ln(t − 2)]94 = 3 ln 7 − 3 ln 2
t−2
• 6.3 #34 Find the area under the curve y = e3x ; x = −1/3 to x = 0
The area under the curve is given by the definite integral
R 0 Solution
3x dx So,
e
−1/3
Z
0
e3x dx = [
−1/3
• 6.3 #38 Interpret
R7
5
1
1
e3x 0
]−1/3 = −
3
3 3e
p(t)dt
Solution It is the pollutants discharged into a lake from 1995 to
1997.
• 6.3 #42 (Profit) Suppose that the marginal profit function for a company is P 0 (x) = 100 + 50x − 3x2 at production level x.
(a) Find the extra profit earned from the sale of 3 additional units
if 5 units are currently being produced.
Solution We want to find P (8) − P (5). By the fundamental theorem of calculus, we have
Z 8
Z 8
0
P (8) − P (5) =
P (x)dx =
100 + 50x − 3x2 dx
5
So, the extra profits are
888
R8
5
5
100 + 50x − 3x2 dx = [100x + 25x2 − x3 ]85 =
(b) Describe the answer to part (a) as an area. (Do not make a
sketch)
Solution The area under the marginal profit curve y = P 0 (x) from
x = 5 to x = 8.
Rb
• 6.3 #48 (a) Compute 1 1t dt, where b > 1.
Solution Note that ln t is an antiderivative of 1t . So by the fundaRb
mental theorem of calculus, we get 1 1t dt = ln b − ln 1 = ln b
3
(b) Explain how the logarithm of a number greater than 1 may be
interpreted as the area of a region under a curve.(What is the curve?)
Rb
Solution We know ,from (a), ln b = 1 1t dt So we may interpret it
as a area under the curve y = 1t from t = 1 to t = b.
• 6.3 #50 For each number x > 2, let A(x) be the area of the region
under the curve y = x3 from 2 to x. Find A0 (6).
A(x) =
R x 3Solution We can express A(x) as definite integral, that is,
3 . Taking
t
dt
=
F
(x)
−
F
(2)
where
F
(x)
is
an
antiderivative
of
t
2
the derivative, we get
A0 (x) = F 0 (x) = x3
So, A0 (6) = 63 = 216.
• 6.4 #2 Write down a definite integral that gives the area of the shaded
region of Figure 11.
Solution The area of the shaded portion is given by
Z 3
A=
[f (x) − g(x)]dx.
2
Note that this rule doesn’t depend on whether f (x) and/or g(x) are
positive or negative. It is valid so long as the graph of f (x) lies above
the graph of g(x) for all x from x = 2 to x = 3.
• 6.4 #8 Find the area of the region.
Solution Note that 2 ≥Rx(2 − x) for 0 ≤ x ≤
R 22. So the area is
2
given by a definite integral 0 2 − x(2 − x)dx = 0 2 − 2x + x2 dx =
[2x − x2 + x3 /3]20 = 8/3
• 6.4 #10 Find the area of the region.
Solution Note that y = e2x is an increasing function, and its value
at x = 0 is 1. On the other hand, y = 1 − x is a decreasing function,
and its value at x = 0 is also 1. So both the functions have the same
value at x = 0. But after that the first function increases continuously
to e2 at x = 1, while the second function goes to 0 at x = 1. Hence
for all x between x = 0 to x = 1, e2x ≥ 1 − x. So the area between the
4
curves from x = 0 to x = 1 is
Z 1
A =
[e2x − (1 − x)]dx
0
Z 1
[e2x + x − 1]dx
=
0
h e2x
i¯1
x2
¯
−x ¯
2
2
0
e2 1
1
= ( + − 1) − ( )
2
2
2
e2
=
−1
2
=
+
• 6.4 #18 Find the area of the region bounded by two graphs; y = 4/x
and y = 5 − x
Solution We first find the points of the intersection. By setting
= 5 − x and solving for x, we get x = 1, 4. Noting that 5 − x
is greaterR than 4/x for x between 1 and 4, the area is given by this
4
integral; 1 [5 − x] − x4 . Therefore the desired area is
4
x
Z
4
[5 − x] −
1
4
= [5x − x2 /2 − 4 ln x]41 = 15/2 − 8 ln 2
x
• 6.4 #20
Solution (a) Note that y = x2 is an increasing function, and its
value at x = 1 is 1. On the other hand, y = 1/x2 is a decreasing
function, and its value at x = 1 is also 1. So both the functions have
the same value at x = 1. But after that the first function increases
1
continuously to 16 at x = 4, while the second function goes to 16
at
2
2
x = 4. Hence for all x between x = 1 to x = 4, x ≥ 1/x . So the area
between the curves from x = 1 to x = 4 is
Z 4
A =
[x2 − 1/x2 ]dx
1
Z 4
=
[x2 − x−2 ]dx
1
=
h x3
i¯4
¯
+ x−1 ¯
3
1
64 1
1
81
= ( + ) − ( + 1) =
3
4
3
4
5
(b) The moment we go to the left of 1, things become little tricky.
To the left of 1, 1/x2 is greater than x2 , since 1/x2 is a decreasing
function, going to ∞ as x → 0, while x2 is an increasing function,
going to 0 as x → 0 (Draw the graphs of the two functions and you
will see this for yourself). So the area of the region between the two
curves from x = 21 to x = 4 is given by
Z 1
Z 4
1
1
2
A=
[ − x ]dx +
[x2 − 2 ]dx = A1 + A2 , say.
1 x2
x
1
2
81
. So let’s
We have already found the value A2 in part (a) to be
4
calculate the value of A1 .
Z 1
A1 =
[−x2 + 1/x2 ]dx
1
2
1
Z
=
[−x2 + x−2 ]dx
1
2
=
h −x3
i¯1
¯
− x−1 ¯ 1
3
2
1
1
17
= (− − 1) − (− − 2) =
3
24
24
Hence, A = A1 + A2 =
17 81
503
+
=
.
24
4
24
• 6.4 #22
Solution Note that y = 1/x is a decreasing function, while y = 4x
and y = x/2 are linear increasing functions. The area enclosed by the
three curves looks similar to Figure 14 on page 355. To find the area of
the enclosed region, we first need to find out the intersection points of
the three curves. Clearly, y = 4x and y = x/2 intersect at the origin.
Solving 4x = 1/x, we get, x2 = 1/4, i.e., x = 1/2. Hence the curves
y = 4x and y = 1/x intersect at x = 1/2.
√
Solving x/2 = 1/x, we get, x2 = 2, i.e.,
√ x = 2. Hence the curves
y = x/2 and y = 1/x intersect at x = 2.
Now to find the total area, divide the region into two parts
√ - one from
x = 0 to x = 1/2, and the other from x = 1/2 to x = 2. The first
region is bounded by the curves y = 4x, y = x/2 and x = 1/2, with
4x ≥ x/2, while the second region is bounded by the curves y = 1/x,
6
y = x/2 and x = 1/2, with 1/x ≥ x/2. Hence the total area of the
region is given by
Z √2
x
1 x
[4x − ]dx +
[ − ]dx
1
2
x 2
0
2
√
h
x2 i¯¯ 2
x2 i¯¯ 12 h
2
2x −
¯ + ln |x| −
¯
4 0
4 21
h1
i h³ √
1
1
1
1 ´i
−
− 0 + ln 2 − ) − (ln −
2 16
2
2 16
√
1
1
3
ln 2 − ln = ln 2 + ln 2 = ln 2
2
2
2
Z
A =
=
=
=
1
2
• 6.4 #24
Solution It is similar to Problem 22. So first lets find the intersection points. The curves y = x and y = 4x cleary intersect at the
origin.
Letting 4x = 12 − x2 , we get x2 + 4x − 12 = 0, i.e., (x + 6)(x − 2) = 0,
i.e., x = −6 or x = 2. Since we are dealing with positive x-values only,
we have that the curves y = 4x and y = 12 − x2 intersect at x = 2.
Letting x = 12 − x2 , we get x2 + x − 12 = 0, i.e., (x + 4)(x − 3) = 0,
i.e., x = −4 or x = 3. Since we are dealing with positive x-values only,
we have that the curves y = x and y = 12 − x2 intersect at x = 3.
Now to find the total area, divide the region into two parts - one from
x = 0 to x = 2, and the other from x = 2 to x = 3. The first region is
bounded by the curves y = 4x, y = x and x = 2, with 4x ≥ x, while
the second region is bounded by the curves y = 12 − x2 , y = x and
x = 2, with 12 − x2 ≥ x. Hence the total area of the region is given by
Z 2
Z 3
[4x − x]dx +
[12 − x2 − x]dx
A =
0
2
Z 2
Z 3
=
3xdx +
[12 − x2 − x]dx
0
2
h 3x2 i¯2 h
x3 x2 i¯¯3
¯
=
−
¯ + 12x −
¯
2 0
3
2 2
h
i h³
´i
9
8
= 6 − 0 + 36 − 9 − ) − (24 − − 2
2
3
9 8
55
= 11 − + =
2 3
6
7