CALCULUS 2 – Worksheet #34
1997 Free Response (3)
Let R be the region enclosed by the graphs of y = ln (x2 + 1) and y = cos x.
(a)
Find the area of R.
(b)
Write an expression involving one or more integrals that gives the length of the boundary of the
region R. Do not evaluate.
(c)
The base of a solid is the region R. Each cross section of the solid perpendicular to the x-axis is
an equilateral triangle. Write an expression involving one or more integrals that gives the volume of
the solid. Do not evaluate.
Answer:
(a)
(b)
(c)
intersection: cos x = ln(x2 + 1) ––> cos x – ln(x2 + 1) = 0 ––>
a
Area = ⌠
⌡(cos x – ln(x2 + 1)) dx = 1.168
–a
a
⌠
dy
2x
dy
⎛ 2x ⎞2
1+⎜ 2
⎟ dx +
dx = x2 + 1 and dx = – sin x L = ⎮
⎝x + 1⎠
⌡
–a
Area of equilateral triangle = 43 s 2
The volume is the integral of the area of the cross-section.
a
Therefore: V =
"
!a
(
)
2
3
cos x ! ln(x 2 + 1) dx
4
x ≈ .91585766 = a
a
⌠
⌡ 1 + (– sin x)2 dx
–a
CALCULUS 2 – Worksheet #34
AND ALSO DO THE FOLLOWING:
1.
2.
Suppose that the growth of a population y = y(t) is given by the logistic equation
1000
y=
1 + 999e–0.9 t
(a) What is the population at time t = 0?
1
(b) What is the carrying capacity, A ?
1000
(c) What is the constant, k ?
.0009 or 9/10,000
(d) When does the population reach 75% of the carrying capacity? t = 8.895
⌠2x + 1 dx=
⎮
⌡4 + x2
A) ln(x2 + 4) + C
x
B) ln(x2 + 4) + tan–1 2 + C
1
x
•D) ln(x2 + 4) + 2 tan–1 2 + C
3.
1
x
C) 2 tan–1 2 + C
E) none of these
⌠
⌡x tan–1x dx =
x2
x
1
A) 2 tan–1 x – 2 + C B) 2 [x2tan–1x + ln(1 + x2)] + C
1
1
D) 2x (tan–1x) + C •E) 2 [(x2 + 1)tan–1x – x] + C
⎛x2 + 1⎞
C) ⎜⎝ 2 ⎟⎠ tan–1x + C
4.
A particle moves along a line with acceleration 2 + 6t at time t. When t = 0, its velocity equals 3 and it
is at position s = 2. When t = 1, it is at position s =
A) 2 B) 5 C) 6 •D) 7 E) 8
5.
The population of a country increases at a rate proportional to the existing population. If the population
doubles in 20 years, then the factor (constant) of proportionality is
1
1
A) ln 2 B) ln 20 •C) 20 ln 2 D) 2 ln 20 E) 2e–20
6.
dy
y
If dx =
and y = 1 when x = 4, then
2 x
(A) y2 = 4 x – 7 (B) ln y = 4 x – 8 (C) ln y = x – 2 (D) y = e
x
•(E) y = e
x–2
7.
The curve that passes through the point (1,1) and whose slope at any point (x,y) is equal
3y
to x has the equation
(A) 3x – 2 = y (B) y3 = x •(C) y = |x3| (D) 3y2 = x2 + 2 (E) 3y2 – 2x = 1
8.
If a substance decomposes at a rate proportional to the amount of the substance present, and if the
amount decreases from 40 gm to 10 gm in 2 hr, then the constant of proprtionality is
1
1
1
1
•(A) –ln 2 (B) – 2 (C) – 4 (D) ln 4 (E) ln 8