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001 10.0 points Evaluate the integral
I =
∫ (^) π/ 2
0
(2 sin(θ) + sin^3 (θ)) dθ.
1. I = 3
2. I =
correct
3. I =
4. I = 1
5. I =
6. I =
Explanation: Since
sin^2 (θ) = 1 − cos^2 (θ)
we see that
2 sin(θ) + sin^3 (θ) = sin(θ)(2 + sin^2 (θ))
= sin(θ)(2 + 1 − cos^2 (θ))
= sin(θ)(3 − cos^2 (θ)).
Thus
I =
∫ (^) π/ 2
0
sin(θ)(3 − cos^2 (θ)) dθ
As the integral is now of the form
sin(θ) f (cos(θ)) , f (x) = 3 − x^2 ,
set x = cos(θ). Then
dx = − sin(θ) dθ ,
while θ = 0 =⇒ x = 1 ,
θ =
π 2
=⇒ x = 0.
In this case
I = −
1
(3 − x^2 ) dx =
0
(3 − x^2 ) dx.
Consequently,
I =
[
3 x −
x^3
] 1
0
002 10.0 points Determine the integral
I =
(3 sin(θ) − 2 sin^3 (θ)) dθ.
- I = − cos(θ) +
sin^3 (θ) + C
- I = cos(θ) +
sin^3 (θ) + C
- I = cos(θ) −
cos^3 (θ) + C
- I = cos(θ) −
sin^3 (θ) + C
- I = − cos(θ) −
cos^3 (θ) + C correct
- I = − cos(θ) +
cos^3 (θ) + C
Explanation: After simplification, we see that
3 sin(θ) − 2 sin^3 (θ) = sin(θ)(3 − 2 sin^2 (θ)).
On the other hand,
sin^2 (θ) = 1 − cos^2 (θ).
Thus the integrand can be rewritten as
sin(θ)(3 − 2(1 − cos^2 (θ)))
= sin(θ)(1 + 2 cos^2 (θ)).
As this is now of the form sin(θ) f (cos(θ)), the subsitution x = cos(θ) is suggested. For then
dx = − sin(θ) dθ ,
in which case
I = −
(1 + 2x^2 ) dx = −
x +
x^3
+ C.
Consequently,
I = − cos(θ) −
cos^3 (θ) + C
with C an arbitrary constant.
keywords: trig identity, trig function, integral
003 10.0 points
Evaluate the integral
I =
∫ (^) π/ 2
0
sin^3 x dx.
1. I =
2. I =
3. I =
correct
4. I = 1
5. I =
Explanation: Since sin^2 x = 1 − cos^2 x ,
we see that
I =
∫ (^) π/ 2
0
(1 − cos^2 x) sin x dx.
This suggests using the substitution u = cos x. For then du = − sin x dx, while
x = 0 =⇒ u = 1 ,
x =
π 2
=⇒ u = 0.
In this case,
I = −
1
(1 − u^2 ) du =
0
(1 − u^2 ) du.
Thus
I =
[
u −
u^3
] 1
0
004 10.0 points
Determine the integral
I =
∫ (^) π/ 2
0
(3 cos(θ) + cos^3 (θ)) dθ.
1. I = 4
2. I =
3. I = 3
4. I =
correct
5. I =
Explanation: Since
cos^2 (θ) = 1 − sin^2 (θ) ,
the integrand can be rewritten as
3 cos(θ) + cos^3 (θ) = cos(θ)(3 + cos^2 (θ))
= cos(θ)(3 + (1 − sin^2 (θ))) = cos(θ)(4 − sin^2 (θ))
Thus
I =
∫ (^) π/ 2
0
cos(θ)(4 − sin^2 (θ)) dθ.
As the integrand is now of the form
cos(θ) f (sin(θ)) , f (x) = 4 − x^2 ,
Since
sin^2 (x) cos^3 (x) = (sin^2 (x) cos^2 (x)) cos(x)
= sin^2 (x)(1 − sin^2 (x))cos(x)
= (sin^2 (x) − sin^4 (x))cos(x) ,
the integrand is of the form cos(x)f (sin(x)), suggesting use of the substitution u = sin(x). For then du = cos(x) dx ,
while x = 0 =⇒ u = 0
x =
π 2
=⇒ u = 1.
In this case
I =
0
3(u^2 − u^4 ) du.
Consequently,
I =
[
u^3 −
u^5
] 1
0
keywords: Stewart5e, indefinite integral, powers of sin, powers of cos, trig substitu- tion,
007 10.0 points Find the value of the integral
I =
2
9 + (x − 2)^2
dx.
1. I =
π correct
- I = 3π
3. I =
π
4. I =
5. I = 3
6. I =
π
Explanation: Set 3 tan u = x − 2. Then
9 + (x − 2)^2 = 9 + (3 tan u)^2
= 9(1 + tan^2 u) = 9 sec^2 u ,
while
3 sec^2 u du = dx.
Also x = 2 =⇒ u = 0 , and x = 5 =⇒ u =
π 4
In this case
I =
∫ (^) π/ 4
0
3 sec^2 u 9 sec^2 u
du =
∫ (^) π/ 4
0
du.
Consequently,
I =
[
u
]π/ 4 0
π.
008 10.0 points
To which one of the following does the inte- gral
I =
x^3 √ x^2 − 1
dx
reduce after an appropriate trig substitution?
1. I =
sin^4 (θ) dθ
2. I =
tan^3 (θ) sec^4 (θ) dθ
3. I =
sin^3 (θ) sec^4 (θ) dθ
4. I =
sin^4 (θ) sec^3 (θ) dθ
5. I =
tan^4 (θ) dθ
6. I =
sec^4 (θ) dθ correct
Explanation: Set x = sec(θ). Then
dx = sec(θ) tan(θ) dθ,
sec^2 (θ) − 1 = tan(θ).
In this case,
I =
sec^3 (θ) tan(θ)
sec(θ) tan(θ) dθ.
Consequently,
I =
sec^4 (θ) dθ.
009 10.0 points
To which one of the following does the inte- gral
I =
x^2 √ x^2 + 1
dx
reduce after an appropriate trig substitution?
1. I =
sin^3 (θ) dθ
2. I =
sin^2 (θ) sec^3 (θ) dθ correct
3. I =
tan^3 (θ) dθ
4. I =
sec^3 (θ) dθ
5. I =
tan^2 (θ) sec^3 (θ) dθ
6. I =
sin^3 (θ) sec^2 (θ) dθ
Explanation: Set x = tan(θ). Then
dx = sec^2 (θ) dθ,
tan^2 (θ) + 1 = sec(θ).
In this case,
I =
tan^2 (θ) sec(θ)
sec^2 (θ) dθ
tan^2 (θ) sec(θ) dθ.
Consequently,
I =
sin^2 (θ) sec^3 (θ) dθ.
010 10.0 points Find the area bounded by the graphs of f (x) = e^4 x, g(x) = e−^8 x and the line y = 6.
- Area =
(6 ln 6 − 5) sq. units correct
- Area =
(6 ln 6 − 5) sq. units
- Area =
(ln 6 − 1) sq. units
- Area =
(ln 6 − 1) sq. units
- Area =
(ln 6 + 1) sq. units
- Area =
(6 ln 6 + 5) sq. units
Explanation: The graph of f is an exponentially increas- ing function, while the graph of g is an expo- nentially decreasing function. The y-intercept of both graphs is at y = 1. Thus the required area is the shaded area in the figure
g : f :
- revenue = $236. 44
Explanation: The revenue function is given by
R(x) =
4 x + 3 −
3 x + 8
dx, R(0) = 0.
Thus
R(x) = 2x^2 + 3x − ln(3x + 8) + C
where the arbitrary constant C is determined by the condition R(0) = 0. Consequently,
R(x) = 2x^2 + 3x − ln
( (^3) x + 8 8
and so the weekly revenue from the sale of 10 printer cartridges is given by
R(10) = $
230 − ln
013 (part 1 of 2) 10.0 points The weekly Marginal Revenue (in dollars) from the sale of x pairs of running shoes is given by
R′(x) = 8x − 3 +
2 x + 3
, R(0) = 0
where R(x) is the revenue in dollars.
(i) Find the revenue function.
- R(x) = 4x^2 − 3 x − ln
( 2 x + 3 3
- R(x) = 4x^2 − 3 x + ln
( (^2) x + 3 3
correct
- R(x) = 4x^2 − 3 x + 2 ln
( (^2) x + 3
3
- R(x) = 4x^2 − 3 x + ln
( 2 x 3
- R(x) = 4x^2 + 3x + 2 ln
( (^2) x + 3
3
Explanation:
The revenue function R is the indefinite integral
R(x) =
8 x − 3 +
2 x + 3
dx
= 4x^2 − 3 x + ln(2x + 3) + C
of the Marginal Revenue R′, the arbitrary constant C being determined by the condition R(0) = 0. Thus
R(x) = 4x^2 − 3 x + ln
( 2 x + 3
3
014 (part 2 of 2) 10.0 points (ii) Find the revenue from the sale of 10 pairs of shoes.
- $ 372. 03 correct
Explanation: To find the revenue from the sale of 10 pairs of shoes we substitute in
R(x) = 4x^2 − 3 x + ln
( (^2) x + 3
3
for x = 10. Thus
revenue = $372. 03.
015 10.0 points
Determine the indefinite integral
I =
e^3 x √ 1 − e^6 x^
dx, (x < 0).
1. I =
arctan e^3 x^ + C
2. I =
ln
1 − e^3 x 1 + e^3 x
∣ +^ C
- I = e^6 x
1 − e^6 x^ + C
4. I =
1 − e^6 x^ + C
5. I =
arcsin e^3 x^ + C correct
Explanation: Set u = e^3 x. Then
du = 3 e^3 x^ dx ,
while
e^6 x^ = e^3 xe^3 x^ = u^2.
In this case
I =
1 − u^2
du.
Consequently,
I =
arcsin e^3 x^ + C.
016 10.0 points
Evaluate the integral
I =
1 / 3
x (1 + x)
dx.
1. I = 2
- I = 2π correct
3. I = 6
- I = 6π
5. I = 1
- I = 1π
Explanation: Set u^2 = x. Then 2u du = dx, while
x =
=⇒ u =
x = 3 =⇒ u =
In this case,
I = 12
1 /
√ 3
1 + u^2
du
[
tan−^1 u
]√ 3
1 /
√ 3
Thus
I = 12
(π 3
π 6
= 2π.
017 10.0 points Determine the integral
I =
3 − x √ 1 − x^2
dx.
1. I =
sin−^1 x −
1 − x^2 + C
2. I =
tan−^1 x −
1 − x^2 + C
3. I =
tan−^1 x +
1 − x^2 + C
- I = 3 sin−^1 x +
1 − x^2 + C
- I = 3 sin−^1 x +
1 − x^2 + C correct
- I = 3 tan−^1 x −
1 − x^2 + C
Explanation: We deal with the two integrals
I 1 =
1 − x^2
dx, I 2 =
x √ 1 − x^2
dx
separately.
- Area =
(ln 2) sq. units
- Area =
(ln 2)^2 sq. units correct
- Area =
(ln 2)^2 sq. units
- Area = 5(ln 2) sq. units
- Area = 5(ln 2)^2 sq. units
- Area =
(ln 2) sq. units
Explanation: The region bounded by the graphs of
f (x) =
5 ln x x
, y = 0, x = 2
is similar to the shaded region in
(not drawn to scale). Now the lower limit of integration is the x-intercept of the graph of f , i.e., the solution of f (x) = 0. But
f (x) =
5 ln x x
= 0 =⇒ x = 1.
Thus
Area =
1
5 ln x x
dx.
The integral can be evaluated using the sub- stitution u = ln x. For then
du =
x
dx ,
while
x = 1 =⇒ u = 0,
x = 2 =⇒ u = ln 2.
Consequently,
Area =
∫ (^) ln 2
0
5 u du =
(ln 2)^2.
020 10.0 points
Find the volume, V , of the solid of revolu- tion generated by rotating the graph of
y = 2 ln x
about the y-axis between y = 0 and y = 8.
- V = e^4 π cu.units
- V = e^8 π cu.units
- V =
e^4 − 1
π cu.units
- V =
e^8 − 1
π cu.units correct
- V = 2
e^4 − 1
π cu.units
- V = 2
e^8 − 1
π cu.units
Explanation: The volume, V , of the solid of revolution generated by rotating the graph of x = f (y) about the y-axis between y = a and y = b is given by
V = π
∫ (^) b
a
f (y)^2 dy.
To apply this we have first to express x as a function of y since initially y is given in terms of x by y = 2 ln x. But after exponentiation, x = e
1 2 y. Thus
V = π
0
ey^ dy = π
[
ey^
] 8
0
Consequently,
V =
e^8 − 1
π cu.units.
021 10.0 points Find the volume of the solid of revolution generated by rotating the graph of y = 6 ln x about the y-axis between y = 0 and y = 24.
- V = 8963. 87 π cu.units
- V = 8945. 87 π cu.units
- V = 8957. 87 π cu.units
- V = 8951. 87 π cu.units
- V = 8939. 87 π cu.units correct
Explanation: The volume, V , of the solid of revolution generated by rotating the graph of x = f (y) about the y-axis between y = a and y = b is given by
V = π
∫ (^) b
a
f (y)^2 dy.
To apply this we have first to express x as a function of y since initially y is given in terms of x by y = 6 ln x. But after exponentiation, x = e
(^16) y
. Thus
V = π
0
e
(^13) y dy = 3π
[
e
(^13) y ]^24 0
Conequently,
V = 8939. 87 π cu.units.
022 (part 1 of 2) 10.0 points
Use integration by parts to evaluate the integral
I =
∫ (^) f (a)
f (0)
f −^1 (x) dx ,
assuming that f has an inverse, f −^1 , on [0, a] and that f ′^ exists. (Hint: set x = f (u).)
1. I =
∫ (^) a
0
f (x) dx
- I = af (a)
- I = af (a) −
∫ (^) a
0
f (x) dx correct
- I = af (a) +
∫ (^) a
0
f (x) dx
- I = af (a) +
∫ (^) a
0
xf (x) dx
- I = af (a) −
∫ (^) a
0
xf (x) dx
Explanation: Set x = f (u), so that u = f −^1 (x). Then
dx = f ′(u) du,
while
x = f (0) =⇒ u = 0,
x = f (a) =⇒ u = a.
Then I =
∫ (^) a
0
uf ′(u) du.
But integration by parts now shows that
I =
[
uf (u)
]a
0
∫ (^) a
0
f (u) du.
Consequently,
∫ (^) f (a)
f (0)
f −^1 (x) dx = af (a) −
∫ (^) a
0
f (x) dx.
023 (part 2 of 2) 10.0 points
Use your answer to part 1 to compute the value of
I =
0
sin−^1 x dx.
