Math 1B Final Exam Problems – A sampling of fair game
1 − x2
.
1 + x2
⎛π x ⎞
Suppose the area in the first quadrant bound by y = − x + sin ⎜
⎟ is revolved around the y-axis.
⎝ 2 ⎠
Assume all variables represent length in meters.
Sketch a diagram showing the volume and sketch in a representative cylindrical shell.
Set an integral to compute the volume generated.
Evaluate the integral to find the volume generated. .
1. Find the area above the x-axis and below the curve y =
2.
a.
b.
c.
3. Find the length of the curve 5y 3 = x 2 that lies inside the circle of radius 6 centered at the origin.
4. If a spherical loaf of bread is cut into slices of equal
width, each slice will have the same amount of crust.
To verify this, suppose that a semicircle y = r 2 − x 2
is revolved about the x-axis to generate a sphere. Let
AB be the arc of a semicircle that stands above an
interval of length h on the x-axis. Show that the
surface area swept out by AB does not depend on the
location of the interval. That is, the length of AB in
the diagram is independent of a.
5. A spherical tank or radius r is filled with water. Find the minimum work required to empty the tank
through a hole at its top. Be clear to show how your tank is oriented in a coordinate system.
6. Show that the area of the region between the curve y =
1
and the entire x-axis is the same as the
1 + x2
area of the unit disk, x 2 + y 2 ≤ 1 .
7. Find the first three non-zero terms in the Maclaurin series for f ( x ) = cos3 x
a. By cubing the Maclaurin series for y = cos x
1
3
b. By using the identity cos3 x = cos 3x + cos x and adding terms.
4
4
8. Approximate the integral
∫
0.1
−0.1
e − x dx using
2
a. Simpson’s rule with n = 4.
b. The Taylor polynomial of degree 6.
9. Use a Maclaurin series to approximate
nearest 5 × 10−13 ?
3
. How many terms are necessary to approximate this to the
e
10. Use integration by parts to find the area below y = e − x cos π x between x = 0 and x = ½ .
Hint: To find an antiderivative, you’ll need to do parts twice to see the integral recur.
11. Consider the area between the curves y = cos x and y = x 2 cos x .
a. Find the exact value of the area.
b. Use a trapezoidal approximation with n = 2 to approximate the area.
c. Use a midpoint approximation with n = 2 to approximate the area.
d. Find the Simpson approximation with n = 4.
12. Set up integrals to compute the volume of revolution generated by revolving the region above the x2
axis and below y = e − x around the y-axis.
a.
Use the shell method.
b. Use the disc method.
c. Evaluate one of these (note that both are improper integrals.)
13. A right circular cone has height h and base radius r and is buried
so that its base is horizontal and its tip is at ground level. If the
cone is filled to half its height with water, how much work is
required to pump all the water out of the cone?
14. Find the x coordinate of the centroid of the portion of the unit
circle in the first quadrant.
15. Solve the initial value problem:
dy
π
y ( 0 ) = and
= sin 3 x cos 2 y
4
dx
Hint sin 3 x = sin x sin 2 x = sin x (1 − cos 2 x )
∞
16. Use the integral test to determine whether the series converges or diverges:
17. Find a Maclaurin series for ln (1 + x 2 ) . Hint: integrate its derivative.
∑
n =1
1
n3
What is the radius of convergence?
18. Find the first two non-zero terms in the Taylor series for g ( x ) = ∫ arctan ( t 2 ) dt ,
x2
1
expanding about x = 1.
19. Use a binomial series to approximate 3 9 . How many terms are required to approximate to the
⎛ 1⎞
nearest thousandth? Hint: 9 = 8 ⎜1 + ⎟
⎝ 8⎠
Math 1B Final Exam Problems Solutions –
1 − x2
.
1 + x2
ANS: The curve lies above the axis on (-1,1) so the area is
1
2
2
1 1− x
11− x
1
2
=
2
=
2
−
1
+
=
−
2
+
4
arctan
dx
dx
dx
x
x
( ) =π −2
∫−1 1 + x 2
∫0 1 + x 2
∫0 1 + x 2
0
Note that the usual partial fractions method requires the division as above first. If you feel
compelled to use a trig-substitution, x = tan θ yields 1 + x 2 = sec 2 θ and dx = sec 2 θ dθ so
1. Find the area above the x-axis and below the curve y =
2
π / 4 ⎛ 1 − tan θ
1 − x2
dx
2
=
∫0 ⎜⎝ sec2 θ
0 1 + x2
2∫
1
π /4
⎞ 2
2
⎟ sec θ dθ = 2 ∫0 1 − tan θ dθ
⎠
= 2∫
π /4
0
2 − sec 2 θ dθ = 4θ − 2 tan θ
π /4
0
=π −2
⎛π x ⎞
2. Suppose the area in the first quadrant bound by y = − x + sin ⎜
⎟ is revolved around the y-axis.
⎝ 2 ⎠
Assume all variables represent length in meters.
a. Sketch a diagram showing the volume and sketch in a representative cylindrical shell.
b. Set up an integral to compute the volume generated.
1 ⎛
⎛ π x ⎞⎞
2π ∫ x ⎜ − x + sin ⎜
⎟ ⎟ dx
0
⎝ 2 ⎠⎠
⎝
c. Evaluate the integral to find the volume generated.
1
1 ⎛
1
2π 3
⎛ π x ⎞⎞
⎛π x ⎞
dx
x
=
−
+ 2π ∫ x sin ⎜
2π ∫ x ⎜ − x + sin ⎜
⎟
⎟ dx
⎟
0
0
3
⎝ 2 ⎠⎠
⎝ 2 ⎠
⎝
0
2 (π 2 − 12 )
2π
8 2π
⎛ 2x
⎛π x ⎞ 4
⎛ π x ⎞⎞
=−
+ 2π ⎜ − cos ⎜
=−
⎟ + 2 sin ⎜
⎟⎟ = −
3
3
3π
⎝ 2 ⎠ π
⎝ 2 ⎠⎠ 0 π
⎝ π
1
3. Find the length of the curve 5y 3 = x 2 that lies inside the
circle of radius 6 centered at the origin.
The points of intersection are where 5y 3 = x 2 and
x 2 + y 2 = 36 , that is where 5 y 3 + y 2 = 36 .
As the graph below shows, the point of intersection is
approximately ( 5.70228,1.86656 ) . Thus The length is
approximately
2∫
1.86656
0
1 + ( x ' ) dy = 2 ∫
2
2
1.86656
⎛ 15 y 2 ⎞
225 y 4
1+ ⎜
2
1
+
dy
⎟ dy = ∫0
4 (5 y 3 )
⎝ 2x ⎠
1.86656
0
= 2∫
1.86656
45 y + 4
2
3/ 2
dy = ∫
45 y + 4dy =
( 45 y + 4 )
0
4
135
0
1.86656
1.86656
0
≈ 12.01103
4. If a spherical loaf of bread is cut into slices of equal
width, each slice will have the same amount of crust.
To verify this, suppose that a semicircle y = r 2 − x 2
is revolved about the x-axis to generate a sphere. Let
AB be the arc of a semicircle that stands above an
interval of length h on the x-axis. Show that the
surface area swept out by AB does not depend on the
location of the interval. That is, the length of AB in
the diagram is independent of a.
ANS:
2π ∫
a +h
a
2
r −x
2
2
a +h
a +h
⎛ −x ⎞
r2
2
2
=
−
dx
r
x
dx = 2π r ∫ dx = 2π r ( a + h − a ) = 2π rh
1+ ⎜
2
π
2
2
∫
2
2 ⎟
a
a
r −x
⎝ r −x ⎠
5. A spherical tank or radius r is filled with water. Find the minimum work required to empty the tank
through a hole at its top. Be clear to show how your tank is oriented in a coordinate system.
W = ∫ dW = 9810 ∫
r
−r
( r − y ) π ( r 2 − y 2 ) dy
r
⎛
y3 ⎞
= 9810rπ ∫ ( r − y ) dy = 9810rπ ⎜ r 2 y − ⎟
−r
3 ⎠− r
⎝
⎛ 4r 3 ⎞
4
= 9810rπ ⎜
⎟ = 13080π r Joules
⎝ 3 ⎠
r
2
2
6. Show that the area of the region between the curve
1
and the entire x-axis is the same as the area of
y=
1 + x2
the unit disk, x 2 + y 2 ≤ 1 .
∞
∞ dx
dx
=
2
arctan ( x ) = π
∫−∞ 1 + x 2 ∫0 1 + x 2 = 2 lim
b→∞
7. Find the first three non-zero terms in the Maclaurin series for f ( x ) = cos3 x
a. By cubing the Maclaurin series for y = cos x
3
2
⎛
⎞ ⎛
⎞ ⎛
⎞
x2 x4
x6
x2 x4
x6
x2 x4
x6
1
1
1
−
+
−
+
"
=
−
+
−
+
"
−
+
−
+ "⎟
⎜
⎟
⎜
⎟
⎜
2 24 720
2 24 720
2 24 720
⎝
⎠ ⎝
⎠ ⎝
⎠
⎛
⎞⎛
⎞
x4 2 x6
x2 x4
x6
3x 2 7 x 4
= ⎜1 − x 2 + −
+ "⎟ ⎜1 − +
−
+ "⎟ ≈ 1 −
+
6
45
2 24 720
2
8
⎝
⎠⎝
⎠
1
3
cos 3x + cos x and adding terms.
4
4
2
1
3
1 ⎛ 9 x 81x 4 ⎞ 3 ⎛
x2 x4 ⎞
12 x 2 84 x 4
cos3 x = cos 3x + cos x ≈ ⎜ 1 −
+
+
1
−
+
=
1
−
+
⎟
⎜
⎟
4
4
4⎝
2
24 ⎠ 4 ⎝
2 24 ⎠
8
96
b. By using the identity cos3 x =
8. Approximate the integral
∫
0.1
−0.1
e − x dx using
2
a. Simpson’s rule with n = 4.
0.1
Δx
− x2
∫−0.1 e dx = 3 ( y0 + 4 y1 + 2 y2 + 4 y3 + y4 )
1
= ( e −0.01 + 4e −0.0025 + 2e0 + 4e −0.0025 + e −0.01 ) ≈ 0.19933541078
60
Or, using symmetry,
0.1
0.1
1
− x2
− x2
−0.000625
+ 2e −0.0025 + 4e −0.005625 + e −0.01 ) ≈ 0.199335333707
∫−0.1 e dx = 2∫0 e dx ≈ 60 (1 + 4e
b. The Taylor polynomial of degree 6.
0.1
⎛
⎛
x2 x4 x6 ⎞
x3 x5 x7 ⎞
∫−0.1 e dx ≈ ∫−0.1 ⎜⎝1 − 2 + 6 − 24 ⎟⎠ dx = 2 ⎜⎝ x − 6 + 30 − 168 ⎟⎠ ≈ 0.199667332
0
3
9. Use a Maclaurin series to approximate . How many terms are necessary to approximate this to the
e
−13
nearest 5 ×10 ?
n
3 ( −1)
3
3
3 3 3
3
3
−1
ANS: = 3e = 3 − + −
is alternating so that
≈ 1.4 × 10−13
+
−
+"+
2 6 24 120 720
16!
e
n!
n
15
−1)
(
suggests the approximation 3∑
≈ 1.10363832351 is accurate to the nearest 5 × 10−13 .
n!
n =0
10. Use integration by parts to find the area below y = e − x cos π x between x = 0 and x = ½ .
0.1
0.1
− x2
u = e− x
dv = cos π xdx
−x
du = −e dx
u = e− x
−x
du = −e dx
v=
sin π x
π
I = ∫ e cos π xdx =
−x
dv = sin π xdx
v=
− cos π x
=
e − x sin π x
π
π
+
e − x sin π x
π
+
1
e
π∫
1 ⎡ −e − x cos π x 1
−
π ⎢⎣
π
π
−x
sin π xdx
⎤
I⎥
⎦
e− x ⎛
1 ⎞
cos π x ⎞
π e− x ⎛
cos π x ⎞
⎛
x
I
⇔ ⎜1 + 2 ⎟ I =
−
⇔
=
sin
π
⎜
⎟
⎜ sin π x −
2
π ⎝
π ⎠
π +1⎝
π ⎟⎠
⎝ π ⎠
1/ 2
Thus the area is
∫
0
π
1/ 2
cos π x ⎞
π ⎛ 1 1⎞
⎛
e cos π xdx = 2
e ⎜ sin π x −
+
⎟ =
2
π +1 ⎝
π ⎠0
(π + 1) ⎜⎝ e π ⎟⎠
−x
−x
11. Consider the area between the curves y = cos x , y = x 2 cos x , x = –1 and x = 1 .
a. Find the exact value of the area.
∫
1
−1
cos x − x cos xdx = ∫ (1 − x ) cos xdx
1
2
−1
SOLN: = (1 − x 2 ) sin x
1
−1
u = 1 − x2
dv = cos xdx
du = −2 xdx
v = sin x
2
1
u=x
dv = sin xdx
−1
du = dx
v = − cos x
+ 2 ∫ x sin xdx
= 0 − 0 − 2 x cos x −1 + 2 ∫ cos xdx = 4 ( sin (1) − cos (1) ) ≈ 1.20467
1
1
−1
b. Use a trapezoidal approximation with n = 2 to approximate the area.
The trapezoidal area, as depicted above, is exactly 1 square unit: the sum of two triangular areas,
each with base 1 (along the y axis) and height 1 (parallel to x-axis.)
c. Use a midpoint approximation with n = 2 to approximate the area.
3
⎛1⎞
SOLN: At the midpoints the height is cos(0.5) (1 − 0.52 ) = cos ⎜ ⎟ ≈ 0.65819 so that the
4
⎝2⎠
midpoint area is then Δx ( h1 + h2 ) = 1( 0.65819 + 0.65819 ) ≈ 1.32
d. Find the Simpson approximation with n = 4.
SOLN: The Simpson approximation is a weighted average of the midpoint and trapezoid values:
2 (1.32 ) + 1
≈ 1.21
3
12. Set up integrals to compute the volume of revolution generated by revolving the region above the x2
axis and below y = e − x around the y-axis.
a. Use the shell method.
∞
∞
−x
−u
−u
−b
∫ 2π rhdx = 2π ∫ xe dx = π ∫ e dx = −π lim e = −π lim e − 1 = π
0
2
b
b →∞
0
0
x →∞
b. Use the disc method.
1
∫ π r dy = π ∫ − ln ydy = −π lim y ln y − y
2
0
b → 0+
1
b
= π + π lim+
b →0
ln y
1/ y
= π + π lim+
=π
0
→
b
1/ y
−1/ y 2
c. Evaluate one of these (note that both are improper integrals.)|
SOLN: Both are shown above.
13. A right circular cone has height h and base radius r and is buried so that its
base is horizontal and its tip is at ground level. If the cone is filled to half
its height with water, how much work is required to pump all the water out
of the cone?
SOLN: Place the origin of the x axis at the tip of the cone and increasing
down along the axis of the cone. By similar triangles the radius R of the
cross-section at x satisfies R/x = r/h so the volume is
∫ Force density × depth × element of volume =
9800π r
⎛ rx ⎞
9800π ∫ x ⎜ ⎟ dx =
h/2
h2
⎝h⎠
h
2
9800π r 2 x 4
=
4
h2
h
=
h/2
2
∫
h
h/2
x3 dx
18375π h 2 r 2
8
14. Find the x coordinate of the centroid of the portion of the unit circle in the first quadrant.
SOLN:
∫
x=
1
0
x 1 − x 2 dx
π /4
=
4
π
∫
π /2
0
sin t cos tdt = −
2
4
π
1
∫
0
1
4u 3
4
u du =
=
3π 0 3π
2
15. Solve the initial value problem:
π
dy
y ( 0 ) = and
= sin 3 x cos 2 y
dx
4
dy
= sin 3 x cos 2 y ⇔ ∫ sec 2 ydy = ∫ sin 3 xdx ⇔ tan y = ∫ sin x − sin x cos 2 xdx
dx
SOLN:
⎛ 7 − cos3 x
⎞
1
⇔ tan y = − cos x − cos3 x + c ⇔ y = arctan ⎜
− cos x ⎟
3
3
⎝
⎠
∞
16. Use the integral test to determine whether the series converges or diverges:
∑
n =1
SOLN The terms are positive and decreasing and
∫
∞
1
1
n3
b
x −3/ 2 dx = −2 lim x −1/ 2 = 2 is finite, so the
b →∞
1
series is convergent.
17. Find a Maclaurin series for ln (1 + x 2 ) . Hint: integrate its derivative.
What is the radius of convergence?
∞
∞
d
2x
n
2 n
dx
=
x
−
x
dx
=
ln (1 + x 2 ) = ∫ ln (1 + x 2 ) dx = ∫
2
2
( −1) x 2 n+1dx
(
)
∑
∑
2
∫
∫
dx
1+ x
n=0
n=0
SOLN:
2n+2
2n+2
∞
∞
∞
n
n x
n x
= 2∑ ∫ ( −1) x 2 n +1dx = 2∑ ( −1)
= ∑ ( −1)
n +1
2n + 2 n = 0
n=0
n=0
The radius of convergence is 1.
18. Find the first two non-zero terms in the Taylor series for g ( x ) = ∫ arctan ( t 2 ) dt ,
x2
1
expanding about x = 1.
SOLN: c0 = g (1) = ∫ arctan ( t 2 ) dt = 0 , c1 = 2 x arctan ( x 2 )
1
1
c2 = 2 arctan ( x 2 ) +
4x2
1 + x2
=
x =1
π
2
x =1
=
π
2
+ 2 so g ( x ) = ∫ arctan ( t 2 ) dt ≈
19. Use a binomial series to approximate
x2
1
3
1/ 3
and
π
2
( x − 1) + ⎛⎜
π
2
⎞
+ 1⎟ ( x − 1)
⎝4 ⎠
9 . How many terms are required to approximate to the
∞ 1/ 3
⎛
⎞⎛ 1 ⎞
2
22
225
⎡ ⎛ 1 ⎞⎤
⎛ 1⎞
9 = ⎢8 ⎜1 + ⎟ ⎥ = 2 ⎜ 1 + ⎟ = 2∑ ⎜
2
≈
+
−
+
⎟⎜ ⎟
3 ⋅ 8 2 ⋅ 3282 3!⋅ 3383 .
⎝ 8⎠
n =0 ⎝ n ⎠ ⎝ 8 ⎠
⎣ ⎝ 8 ⎠⎦
nearest thousandth?
1
1
5
= 2+ −
+
12 288 20736
Evidently, this series is alternating so, since the 4th term is less than half of a thousandth, only 3
1
1
599
terms are needed to approximate to the nearest thousandth: 3 9 ≈ 2 + −
=
≈ 2.080
12 288 288
3
1/ 3
n