HW 3 – Due Sep 12, Wed
1. Find the area enclosed by the graph of f (x) and x-axis on the interval [a, b].
(a) f (x) = −x2 + 4;
[−1, 2] Soln:
• Find all the roots: −x2 + 4 = 0, x2 = 4, x = ±2
• -2 and 2 divide [−1, 2] into 1 region.
Z 2
2 !
3
x
(−x2 + 4)dx = abs − + 4x
• Area= =9
3
−1
x
−1
2
Total
−1
x3
− + 4x abs of difference
3
−3.67
5.33
9
9
(b) f (x) = −3x2 + 12;
[0, 4]
2
• −3x + 12 = 0, x2 = 4, x = ±2
• x = ±2 divide [0, 4] into 2 regions [0, 2] and [2, 4]
Z 2
Z 4
2
2
• Area=
−3x + 12dx + −3x + 12dx = 16 + 32 = 48
2
Z0
F (x) = −3x2 + 12dx = −x3 + 12x.
−x3 + 12x abs of difference
x
0
0
2
16
16
4
-16
32
Total
48
(c) f (x) = 6(x − 3)(x + 2); [−3, 3]
• 6(x − 3)(x + 2) = 0, x = 3, −2
• x = 3, −2 divide [−3, 3] into [−3, −2], [−2, 3]
Z −2
Z 3
• Area= 6(x − 3)(x + 2)dx + 6(x − 3)(x + 2)dx
−3
−2
Z
Z
2
•
6(x − 3)(x + 2)dx = 6 x − x − 6dx = 2x3 − 3x2 − 36x
• Area= 141
x
2x3 − 3x2 − 36x abs difference
-3
27
-2
44
17
3
-81
125
Total
141
1
(d) f (x) = x(x2 − 1)2 ;
[−1, 2]
2
• x(x − 1) = 0, x = −1, 0, 1
• [−1, 2] → [−1, 0], [0, 1], [1, 2]
Z 2
Z 1
Z 0
f (x)dx
f (x)dx + f (x)dx + • Area= 1
0
−1
Z
Z
1
(x2 − 1)3
u3
2
2
2
•
x(x − 1) dx, u = x − 1, du = 2xdx,
=
u2 dx =
2
6
6
29
• Area=
6
(x2 − 1)3
x
abs difference
6
-1
0
0
-0.167
0.167
1
0
0.167
2
4.5
4.5
Total
4.667
2. sketch the graph and find the area of the region completely enclosed by the graphs
of the given function f and g
(a) f (x) = −x2 + 4x and g(x) = 2x − 3
• Solve f (x) = g(x), −x2 + 4x = 2x − 3, (x − 3)(x + 1) = 0,
Z 3
Z 3
2
• Area=
f (x) − g(x)dx = −x + 2x + 3dx
−1
Z−1
x3
• F (x) = −x2 + 2x + 3dx = − + x2 + 3x.
3
• Area=10.67
x = −1, 3
(b) f (x) = x3 − 6x2 + 9x and g(x) = x2 − 3x
• Solve f (x) = g(x), x(x2 − 7x + 12) = x(x − 4)(x − 3) = 0,
Z 3
Z 4
• Area=
f (x) − g(x)dx + f (x) − g(x)dx
3
Z0
x4 7x3
3
2
−
+ 6x2
• F (x) = x − 7x + 12xdx =
4
3
• Area=11.83
√
(c) f (x) = x and g(x) = x2
• f (x) = g(x), x = x4 , x(x3 − 1) = 0,
Z 1
• Area= f (x) − g(x)dx
0
2
x = 0, 3, 4
x(x − 1)(x2 + x + 1),
x = 0, 1
Z
• F (x) =
√
2
x3
x − x2 dx = x3/2 −
3
3
1
3
√
(d) f (x) = 2x and g(x) = x x + 1.
√
√
• √
f (x) = g(x), 2x = x x + 1, x(2 − x + 1) = 0, x = 0, or 2 =
x + 1 ⇒ x = 3
Z
3
f (x) − g(x)dx
• Area= Z0
Z
√
√
2(x + 1)5/2
2
• F (x) =
2x − x x + 1dx = x − (u − 1) udu = x2 −
+
5
{z
}
|
• Area=
u=x+1
3/2
2(x + 1)
3
• Area= 1.26
3. (Consumer’s and Product’s Surplus) The management of the Titan Tire Company
has determined that the quantity demanded x of their Super Titan tires/week is
related to the unit price p by the relation
p = 144 − x2
where p is measured in dollars and x is measured in units of a thousand. Titan will
make x units of the tires available in the market if the unit price is
1
p = 48 + x2
2
dollars. Determine the consumer’s surplus and the producer’s surplus when the
market unit price is set at the equilibrium price.
Soln:
1
• D(x) = 144 − x2 , S(x) = x2 + 48
2
• x̄ : equilibrium point.
3 2
1
x = 96, x = ±8, x̄ = 8
144 − x2 = x2 + 48,
2
2
• p̄ = D(x̄) = D(8) = 80 dollars
8
Z x̄
x3 • Customer’s surplus=
D(x)dx − p̄x̄ = 144x − − 640 = 981.33 − 640
3 0
0
3
8
Z x̄
x
• Producer’s surplus= p̄x̄ −
S(x)dx = 640 −
+ 48x = 640 − 469.33 =
6
0
0
170.67
3
4. (Present Value of an Investment) Suppose you put $800 monthly into a bank account for 12 yr and the account earns interest at the rate of 10% year compounded
continuously. Find the present value.
• t: in month.
0.1
monthly
12
• T = 12 years = 12 × 12 month = 144 month
Z T
Z T
r(T −t)
rT
rT
800e−rt dt
800e
dt = e
• Pe =
• r = 0.1 annually =
Z
0
144
• P =
0
800e−0.1/12t dt = −800 × 120e−0.1/12t |144
= −28914.64 − (−96000) =
0
0
67085.36
4