**Unformatted text preview: **young (toy68) – HW05 – villafuerte altu – (53615)
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001 002 1 10.0 points The shaded region in 10.0 points y Evaluate the integral
I = π/2 Z x cos3 x dx . 0 π
2 1. I = 2
correct
3 2. I = 5
6 3. I = 1
6 π 3π
2 is bounded by the graph of
f (x) = 2 sin3 (x) 4. I = 1
5. I = on [0, 3π/2] and the the x-axis. Find the area
of this region. 1
3 1. area = 4π Explanation:
Since
cos2 x = 1 − sin2 x ,
we see that
Z I = π/2
0 (1 − sin2 x) cos x dx . This suggests using the substitution u = sin x.
For then du = cos x dx, while
x = 0
x = =⇒ π
2 =⇒ u = 0, 2. area = 6π
3. area = 3
4. area = 6
5. area = 4 correct
6. area = 3π
Explanation:
The area of the shaded region is given by u = 1.
I = In this case,
I = Z 0 1
2 (1 − u ) du . Thus
I = 1
u − u3
3 i1
0 = 2
.
3 3π/2
0 |2 sin3 (x)| dx which as the graph shows can in turn be written as
Z π
Z 3π/2
3
I =
2 sin (x) dx −
2 sin3 (x) dx .
0 h Z Since π sin2 (x) = 1 − cos2 (x) , young (toy68) – HW05 – villafuerte altu – (53615)
we thus see that
n Z π Z 3π/2 o
I =
−
2 sin(x)(1−cos2 (x)) dx .
π 0 To evaluate these integrals, set u = cos(x).
For then
du = − sin(x) dx , 4. I = −1
5. I = − 3
2 Explanation:
After division
cos x − 2 sin x
cos3 x in which case
Z π
2 sin(x)(1 − cos2 (x)) dx = sec2 x − 2 tan x sec2 x 0 = −2 −1 Z 1 (1 − u2 ) du = 2 1
= 2 u − u3
3
while
Z 3π/2
π h 2 i1 Z −1 = −1 = (1 − 2 tan x) sec2 x . 1 (1 − u2 ) du Thus 8
,
3 I = Z π/4
0 (1 − 2 tan x) sec2 x dx . Let u = tan x ; then
2 sin(x)(1 − cos2 (x)) dx
= −2 Z 0 −1 (1 − u2 ) du du = sec2 x dx
so
I = Z 1 0 4
1 3 i0
= − .
= −2 u − u
3
3
−1
h Consequently,
1
(1 − 2 u) du = u − u2 0 . Consequently, the shaded region has I = 0 . area = 4 . keywords:
003 10.0 points Evaluate the definite integral
I = Z 0 1. I = − π/4 cos x − 2 sin x
dx .
cos3 x 1
2 2. I = −2
3. I = 0 correct 004 10.0 points Evaluate the indefinite integral
Z
I =
2 cos4 2t dt .
1. I =
2. I =
3. I =
4. I =
5. I =
1
1
3t+sin 4t+ sin 8t +C correct
4
8
1
1
3t + cos 4t + cos 8t + C
4
8
1
1
3t − sin 4t + sin 8t + C
4
8
1
1
3t + sin 4t − sin 8t + C
4
8
1
1
3t + cos 4t − cos 8t + C
4
8 young (toy68) – HW05 – villafuerte altu – (53615)
1
1
3t − cos 4t + cos 8t + C
4
8 6. I = Explanation:
Since
1
1
+
cos
2θ
,
cos θ =
2 3 √
3. I = 4 2
4. I = 4
√
5. I = 4 6 2 the integrand can be rewritten as 2
1
2 cos 2t =
1 + cos 4t
2
1
1 + 2 cos 4t + cos2 4t .
=
2 6. I = 2
Explanation:
By a double angle formula for cos(2x), 4 But in turn, this last expression can be rewritten as
o
1
1n
1 + 2 cos 4t +
1 + cos 8t .
2
2 Thus 2 cos4 2t =
and so
1
I =
2
13
1
+ 2 cos 4t + cos 8t ,
2 2
2 Z
3
1
+ 2 cos 4t + cos 8t dt .
2
2 Consequently,
I =
1
1
3t + sin 4t + sin 8t + C
4
8 with C an arbitrary constant.
005 10.0 points Evaluate the integral
I = 2 Z 0 2π/3 p 1 + cos(θ) dθ . Hint: use a double angle formula to express
1 + cos(θ) in terms of cos2 (θ/2).
√
1. I = 2 6 correct
√
2. I = 2 2 cos(2x) = 2 cos2 (x) − 1 ,
so, taking x = θ/2, we see that
s
p
√
θ
θ
= 2 cos
1 + cos(θ) = 2 cos2
2
2 when cos(θ/2) ≥ 0, hence for 0 ≤ θ ≤ π.
Thus
i
2π/3
√ Z 2π/3
√ h
θ
θ
I = 2 2
dθ = 4 2 sin
cos
2
2 0
0 Consequently, √
I = 2 6 . keywords: definite integral, double angle formula
006 10.0 points Find the value of
Z π/4
I=
2 tan4 x dx .
0 1. I = 1
(3π − 2)
4 2. I = 1
(3π − 2)
6 3. I = 1
(3π − 8)
4 4. I = 1
(3π − 4)
6 5. I = 1
(3π − 8) correct
6 young (toy68) – HW05 – villafuerte altu – (53615)
1
(3π − 4)
4 6. I = 1
5
5
3. I = x2 + x sin(2x) + cos(2x) + C
2
2
4
correct Explanation:
Since
tan2 x = sec2 x − 1 ,
we see that
tan4 x = tan2 x(sec2 x − 1)
2 2 4. I = 1
5 2 1
x + x sin(2x) + cos(2x) + C
2
2
4 5. I = 1
5 2 1
x − x cos(2x) − sin(2x) + C
2
2
4 Explanation:
Since 2 = tan x sec x − tan x
= tan2 x sec2 x − (sec2 x − 1) .
Thus
2 2 tan x = (tan x − 1) sec x + 1 .
In this case,
I=2 cos2 (x) = 1
(1 + cos(2x))
2 sin2 (x) = 1
(1 − cos(2x)) ,
2 and
4 Z 0 π/4 4 we see that (tan2 x − 1) sec2 x dx + Z π/4 2 dx . Z
1
x (6 (1 + cos(2x))
I=
2
−4 (1 − cos(2x))) dx
Z
Z
= x dx + 5
x cos(2x) dx 0 To evaluate the first of these integrals, set
u = tan x. Then
Z 1
i1
h1
4
2
(u2 − 1) du = 2 u3 − u = − .
3
3
0
0
On the other hand
Z π/4
h iπ/4
1
= π.
2 dx = 2x
2
0
0
Consequently,
I =
007 1
(3π − 8) .
6
10.0 points Determine the indefinite integral
Z
I = x (6 cos2 (x) − 4 sin2 (x)) dx .
1. I = 5
1 2 5
x − x cos(2x) − sin(2x) + C
2
2
4 2. I = 5 2 1
5
x − x sin(2x) + cos(2x) + C
2
2
4 1
= x2 + 5
2 Z x cos(2x) dx . But after integration by parts,
Z
1
x cos(2x) dx = x sin(2x)
2
Z
1
sin(2x) dx
−
2
1
1
= x sin(2x) + cos(2x) + C .
2
4
Consequently,
I = 1 2 5
5
x + x sin(2x) + cos(2x) + C .
2
2
4
008 10.0 points The shaded region in young (toy68) – HW05 – villafuerte altu – (53615)
y 5 On the other hand,
sin2 (x) = 1
(1 − cos(2x)) .
2 Thus π
2 x
π π/2
1 1
− cos(2x) + 2 sin(x) dx
V = π
2 2
0
h1
iπ/2
1
= π x − sin(2x) − 2 cos(x)
.
2
4
0
Z Consequently,
(not drawn to scale) is the one below the
graph of
f (x) = sin(x)
on the interval [0, π/2]. Find the volume of
the 3D solid obtained by rotating this region
about the dashed line y = −1.
1. Volume = 1
1
π − cu.units
2
2 2. Volume = 1 2
π + 2π cu.units
2 1
3. Volume = π 2 + 2π cu.units correct
4
1
1
4. Volume = π − cu.units
4
2 Volume = keywords:
009 I = 1. I =
2. I = 6. Volume = 1 2 1
π − π cu.units
2
2 3. I = = π Z π/2
f (x)2 + 2f (x) dx 0 =π Z π/2
0
sin2 (x) + 2 sin(x) dx . Z 0 1
π + 2 cu.units
4 0 10.0 points Evaluate the integral 5. Volume = Explanation:
The 3D solid obtained by rotating the
shaded region about the line y = −1 is given
by the integral
Z π/2
I = π
(f (x) + 1)2 − (1)2 dx 1 2
π + 2π cu.units .
4 4. I =
5. I =
6. I = π/3 sec(x) tan(x)
dx .
3 + sec(x)
5
correct
ln
4
5
− ln
6
5
− ln
4
4
ln
3
5
ln
6
4
− ln
3 Explanation:
Since
d
(sec(x)) = sec(x) tan(x) ,
dx young (toy68) – HW05 – villafuerte altu – (53615)
use of the substitution
u = 3 + sec(x) and deal with I1 , I2 separately. For I1 , set
u = cos(x). Then
du = − sin(x) dx is suggested. For then
du = sec(x) tan(x)dx ,
while x = 0 =⇒ π
3 x = =⇒ I = 5 4 Consequently, while u = 4, x = 0
π
x =
2 u = 5. Thus
Z h
i5
1
du = ln(u) .
u
4 10.0 points Evaluate the integral
Z π/2
I =
sin(x) (f ′ (cos(x)) − 2 sin(x)) dx =⇒
=⇒ u = 1,
u = 0. Thus
′ f (u) du = 1 h = 0 Z I1 = −
5
I = ln
.
4
010 6 Z 1 f ′ (u) du 0 i1 f (u) 0 = (f (1) − f (0)) = 3. On the other hand,
sin2 (x) =
so
1
1 − cos(2x) ,
2 0 when f (0) = 2 and f (1) = 5. I2 = Z π/2 0 1. I = 3 − π = 2. I = 5 + π
1 − cos(2x) dx
h x− iπ/2
1
1
= π.
sin(2x)
0
2
2 Consequently,
3. I = 4 + π
1
I = 3− π .
2 1
4. I = 5 − π
2
1
5. I = 4 + π
2 011 Find the value of the integral 1
6. I = 3 − π correct
2
Explanation:
To evaluate the integral we write I as the
sum
Z π/2
I = I1 + I2 =
sin(x) f ′ (cos(x)) dx
0 −2 Z 0 10.0 points π/2 sin2 (x) dx I = Z 2 1. I = 3
2. I = 1
π
3 3. I = 3π 5 1
dx .
9 + (x − 2)2 young (toy68) – HW05 – villafuerte altu – (53615)
4. I = 1
6 1
π correct
5. I =
12
6. I = 1
π
6 7 4. I = Z sin5 (θ) sec6 (θ) dθ 5. I = Z sin5 (θ) dθ correct Explanation:
Set x = sin(θ). Then Explanation:
Set 3 tan u = x − 2. Then
9 + (x − 2)2 = 9 + (3 tan u)2
2 In this case 3 sec2 u du = dx . Consequently, Also
x = 2 =⇒ u = 0, x = 5 =⇒ u = and I = π
.
4 In this case
Z
Z π/4
1 π/4
3 sec2 u
I =
du =
du.
9 sec2 u
3 0
0
1 h iπ/4
1
u
π .
=
3
12
0
10.0 points To which of the following does the integral
Z
x5
√
dx
I =
1 − x2
reduce after an appropriate trig substitution?
Z
1. I =
sec5 (θ) sin6 (θ) dθ
2. I = tan(θ) sec5 (θ) dθ 3. I = Z sin5 (θ) sec5 (θ) dθ Z 013 sin5 (θ) dθ . 10.0 points Evaluate the integral
I = Consequently, Z sin5 (θ)
cos(θ) dθ .
cos(θ) Z I = while 012 1 − sin2 (θ) = cos(θ) . 2 = 9(1 + tan u) = 9 sec u , I = q dx = cos(θ) dθ, Z 2
0 √ 1. I = 1
2 2. I = 1
π correct
2 3. I = 3
4 4. I = 3
π
4 3
dx .
16 − x2 5. I = π
6. I = 1
Explanation:
Set x = 4 sin u; then
dx = 4 cos u du young (toy68) – HW05 – villafuerte altu – (53615) 8 and
16 − x2 = 16(1 − sin2 u) = 8 cos2 u ,
while x = 0
x = 2 In this case
Z
I = 3 π/6 0 =⇒ u = 0, =⇒ π
u = .
6 cos u
du = 3
cos u Z 015 10.0 points Evaluate the integral
Z I = 2 (4 +
0 p 4 − x2 ) dx . 1. I = 4 + π π/6 du . 0 2. I = 8 + 2π Consequently
I = 1
π .
2 3. I = 4 − 2π
4. I = 8 + π correct 014 10.0 points Determine the integral
Z
3t
√
dt .
I =
1 − t4
1. I = 3 tan−1 t2 + C
3
tan−1 t2 + C
2
3p
3. I =
1 − t4 + C
2 2. I = 4. I = 5. I = 8 − 2π
6. I = 4 − π
Explanation:
Since
Z 2
Z
I=
4 dx +
0 3
+C
1 − t4
6. I = 3 sin−1 t2 + C Explanation:
Set t2 = sin(u). Then p 0 4 − x2 dx = I1 + I2 , we evaluate the two integrals separately. Now 3 −1 2
sin
t + C correct
2 5. I = √ 2 I1 = 2 Z 4 dx = 8 . 0 On the other hand, to evaluate the second
integral we set x = 2 sin(u). For then
dx = 2 cos(u) du ,
while 4 − x2 = 4 cos2 (u) , x = 0 =⇒ u = 0, x = 2 =⇒ u = 2t dt = cos(u) du ,
in which case
1
I =
2 Z 3 cos(u)
3
du = u + C
cos(u)
2 with C an arbitrary constant. Consequently,
3
I = sin−1 (t2 ) + C .
2 π
.
2 In this case
I2 = 4 Z π/2 cos2 (u) du 0 = 2 Z 0 π/2 (1 + cos(2u)) du . young (toy68) – HW05 – villafuerte altu – (53615)
Thus
h
iπ/2
1
I2 = 2 u + sin(2u)
= π.
0
2 Consequently, and so
Z
I= π/4 0 9 h 2 (sec θ − 1) dθ = tan θ − θ I = 1− . π
4 . 10.0 points Evaluate the integral
Z 1
x2
I =
dx .
2 3/2
0 (2 − x )
√
π
1. I = 2 3 +
3
π
2. I = 1 +
4
√
π
3. I = 2 3 −
3
√
π
4. I = 2 2 +
3
√
π
5. I = 2 −
4
π
6. I = 1 − correct
4
Explanation:
√
Let x = 2 sin θ. Then
√
dx = 2 cos θ dθ , 2 − x2 = 2 cos2 θ ,
while 0 Consequently, I = I1 + I2 = 8 + π .
016 iπ/4 x = 0 =⇒ θ = 0, x = 1 =⇒ θ = π
.
4 In this case,
Z π/4 √
2 2 sin2 θ cos θ
√
I =
dθ
2 2 cos3 θ
0
Z π/4
Z π/4
sin2 θ
dθ =
tan2 θ dθ .
=
2θ
cos
0
0
Now
tan2 θ = sec2 θ − 1 , d
tan θ = sec2 θ ,
dθ 017 10.0 points Evaluate the integral
I = 1 Z √ 0 1. I = √ 1
x2 +1 dx . 2−1 √
2. I = ln( 2 − 1 )
3. I = √ 2 (1 + 4. I = ln(1 +
5. I = √ 6. I = √ √ √ 2) 2 ) correct √
2( 2−1)
2 ln(1 + √ 2) Explanation:
Set x = tan(u) , then
dx = sec2 (u) du , x2 + 1 = sec2 (u) , while
x = 0 =⇒ u = 0, x = 1 =⇒ u = π
.
4 In this case
Z π/4
Z π/4
sec2 (u)
I =
du =
sec(u) du .
sec(u)
0
0
On the other hand,
Z
sec(u) du = ln (| sec(u) + tan(u)|) + C . young (toy68) – HW05 – villafuerte altu – (53615)
Thus 10 On the other hand, by Pythagoras,
I = ln (| sec(u) + tan(u)|) π/4 . 0 Consequently, 018 √ 10.0 points x
.
x2 + 9 with C an arbitrary constant. Explanation:
Set x = 3 tan θ. Then
p
dx = 3 sec2 θ dθ,
x2 + 9 = 3 sec θ . In this case
Z
2
sec2 θ
I =
dθ
3
tan θ sec θ
Z sec θ
2
dθ =
tan θ
3 019 10.0 points Evaluate the integral
I = Z 1 (x2 0 2
dx .
+ 3)3/2 1. I = 2
2. I = 1
correct
3 3. I = 1
4. I = 1
2 5. I = 2
3 Explanation:
√
Set x = 3 tan(u). Then
dx = √ 3 sec2 (u) du , while Z csc θ dθ . But
Z sin θ = √ =⇒ √
2 x2 + 9 − 3 I = ln +C
3
x 2) . Determine the integral
Z
2
√
dx .
I =
x x2 + 9
√
2 x2 + 9 + 3 1. I = ln +C
3
x
√
3 x2 + 9 − 3 2. I = ln +C
2
x
√
2 x2 + 9 + x 3. I = ln +C
3
3
√
3 x2 + 9 + x 4. I = ln +C
2
3
√
3 x2 + 9 − x 5. I = ln +C
2
3
√
2 x2 + 9 − 3 6. I = ln + C correct
3
x 2
3 x
3 Consequently,
I = ln(1 + = tan θ = csc θ dθ = ln | csc θ − cot θ| + C . x = 0 =⇒ u = 0, x = 1 =⇒ u = π
.
6 On the other hand,
(x2 + 3)3/2 = 3(tan2 (u) + 1)
√
= 3 3 sec3 (u) . 3/2 young (toy68) – HW05 – villafuerte altu – (53615)
Thus 11 while I = 2 Z √
3 sec2 (u)
√
du
3 3 sec3 (u) π/6 0 2
=
3 Z π/6 cos(u) du =
0 iπ/6
2h
.
sin(u)
3
0 Consequently
I = 1
.
3 x = 0 =⇒ x = 1 =⇒ In this case
Z π/4
Z π/4
3 sec2 u
I =1+
du = 1 +
3 du .
sec2 u
0
0
Consequently,
3
I = 1+ π .
4
021 keywords:
020 10.0 points Evaluate the integral
Z 1 2
x +4
I =
dx .
2
0 1+x
π
1. I = 4 1 −
4
3
2. I = 1 + π correct
4
3
3. I = 1 − π
4 u = 0,
π
u = .
4 10.0 points Which one of the following functions is an
antiderivative of f when
1
f (x) = 2
?
x − 4x+ 5 x − 5 1. F (x) = ln x + 2 2. F (x) = ln(x2 − 4 x + 5)
3. F (x) = − 5 (x − 2)
− 4 x + 5)2 (x2 4. F (x) = sin−1 (x − 2)
5. F (x) = tan−1 (x − 2) correct
Explanation:
The indefinite integral
Z
Z
I = f (x) dx = 4. I = 4(1 − π) 1
dx
− 4x+ 5
consists of all antiderivatives of f . But by
completing the square we see that
1
1
=
x2 − 4 x + 5
(x2 − 4 x + 4) + 1
π
5. I = 4 1 +
4
6. I = 3 + 4π
Explanation:
Since
x2 + 1 + 3
3
x2 + 4
=
=
1
+
,
1 + x2
1 + x2
1 + x2
the integral can be written as
Z 1
Z 1
3
3
dx
=
1+
dx .
I =
1+
2
1 + x2
0 1+x
0
In the last integral, set x = tan u. Then
2 dx = sec u du, =
ThusZ
I= x2 1
.
(x − 2)2 + 1 1
dx = tan−1 (x − 2) + C .
(x − 2)2 + 1
Consequently, only
F (x) = tan−1 (x − 2) is an antiderivative of f , as can be checked by
differentiating F . ...

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- Calculus, Derivative, Fundamental Theorem Of Calculus, Cos