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Name _________________________ Period _____
Calculus 2 Chapter 7 Test Practice
1) Identify the integration method you would use for each of the following problems. Do NOT integrate!
1
3x + 4
a.
b.
c.
dx
dx
64 − x 2 dx
2
2
x + 8x + 25
x + 2x −15
∫
________________
€
d.
∫x
2
cos(x)dx
________________
________________
€
e.
∫ cos (x)sin (x)dx
x→0
€
sin(3x)
4x
4
3
________________
€
________________
€
€
2) Find the limits using L’Hospital’s Rule, when necessary.
a. lim
∫
∫
f.
∫ 5x
25 − x 2 dx
________________
€
________________
sin −1 (2x)
b. lim
x→0
x
________________
€
c. lim( x x )
x→0
________________
€
% 2 ( 3x
d. lim '1 − *
x → +∞ &
x)
________________
€
(
e. lim x e
x→∞
€
1
x
)
−1
________________
3) Integrate each indefinite integral:
a. x ln(x)dx
∫
________________
€
b.
∫ x e dx
2 x
________________
€
c.
∫x
2
∫x
3
∫x
5
4 − x 3 dx
________________
€
d.
4 − x 2 dx
________________
€
e.
4 x 2 −16 dx
________________
€
f.
∫e
2x
cos(3x)dx
________________
€
g.
∫ 5cos (x)sin (x)dx
8
3
________________
€
h.
∫ 6sec (2x)tan (2x)dx
4
3
________________
€
i.
4
∫ (x − 3)(x +1)
2
dx
________________
€
Name _________________________ Period _____
Calculus 2 Chapter 7 Test Practice
1) Identify the integration method you would use for each of the following problems. Do NOT integrate!
1
3x + 4
a.
b.
c.
dx
dx
64 − x 2 dx
2
2
x + 8x + 25
x + 2x −15
∫
tan-1(x)
€
d.
∫x
∫
∫
2
partial fractions
€
cos(x)dx
e.
integration by parts
€
∫ cos (x)sin (x)dx
4
3
€
f.
trigonometric integration
2
∫ 5x
25 − x 2 dx
u-substitution___
€
€
2) Find the limits using L’Hospital’s Rule, when necessary.
sin(3x)
3cos(3x) 3
a. lim
= lim
=
x→0
x→0
4x
4
4
sin −1 (2x)
b. lim
= lim
x→0
x→0
x
trig substitution__
(x = 8sin(θ))
€
sin(3x)
3
=
x→0
4x
4
lim
2
2
1 − 4 x 2 = lim
= =2
2
x
→
0
1
1
1− 4x
sin −1 (2x)
=2
lim €
x→0
x
€
c. lim( x x )
x→0
$ 1 '
$ 1 '
$ x 2 '€
$ ln x '
x
x
&
)
&
)
= lim
= lim& − ) = lim( −x ) = 0
lny= lim( x ln x ) = lim& −1 ) = lim
x→0
x → 0% x
( x → 0&% −x −2 )( x → 0&% − 1 2 )( x → 0% x ( x → 0
x
0
lim( x x ) = 1
lny=0 → y = e = 1
€
€
x→0
€
€
€
€
% 2 ( 3x
d. lim '1 − *
x → +∞ &
x)
% x − 2(
% x (
ln'
*
'
*⋅ (2x −2 )
% 2(
& x )
& x − 2)
lny = lim 3x ln'1 − * = 3 lim
= 3 lim
=
1
−1 2
x → +∞
x → +∞
x → +∞
& x)
x
x
3
−2x
−2x
= 3 lim 2
= 3 lim
= 3⋅ −2 = −6
x → +∞ x ( x − 2)
x → +∞ ( x − 2)
-6
€ lny= -6 → y = e€
≈ 0.00248
€
€
€
e −1)
e ⋅ −x )
e )
(
(
(
e. lim x (e −1) = lim
= lim
= lim
= lim
€e
1
−x
1
x
1
1
1
−2
x
x
x
x→∞
x→∞
1
x
−2
x→∞
x→∞
% 2 ( 3x -6
lim '1 − * = e ≈ 0.00248
x → +∞ &
x)
x→∞
1
x
= e0 = 1
(
lim x e
x→∞
€
€
€
€
€
1
x
)
−1 = 1
3) Integrate each indefinite integral:
x 2 ln x
x
x 2 ln x x 2
a. x ln(x)dx =
−
dx =
− +C
2
2
2
4
2
x
u = ln x
v=
2
1
€
du = dx dv = xdx
x
∫
€ b.
∫
2 x
2 x
x
€
x 2 4 − x 3 dx = −
1
3
∫
u = 4 − x 3,
du = −3x 2 dx
€
1
− du = x 2 dx
3
d.
€
∫
∫x
3
2 x
v = ex
dv = e x dx
€
€
€
∫ x e dx = x e − ∫ 2xe dx = x e − [2xe − ∫ 2e dx] = x e
u = x2
du = 2xdx
c.
x 2 ln x x 2
− +C
2
4
x
x
u = 2x
du = 2dx
2 x
v = ex
dv = e x dx
€
3&
2# 3&
2#
udu = − %u 2 ( + C = − %(4 − x 3 ) 2 ( + C
9$ '
9$
'
€
€
€
4 − x 2 dx
x = 2sin(θ ),dx = 2cos(θ )dθ
∫ 8sin (θ ) 4 − 4 sin (θ )2cos(θ )dθ =16 ∫ sin (θ )cos(θ ) 4 cos (θ )dθ
= 32 ∫ sin (θ )cos (θ )dθ = 32 ∫ [1 − cos (θ )]cos (θ )sin(θ )dθ
€
$ cos (θ ) cos (θ ) '
+
= 32 ∫ [cos (θ ) − cos (θ )]sin(θ )dθ = 32 & −
)+C
€
3
5 (
%
=
3
2
3
€
3
2
2
2
2
3
2
€
#
%
= 32 %
%$
€
€
€e.
∫x
5
(
5
4 − x 2 € 32
)
€
€
€
4 − x2
€
)
3
3
&
8(
( + C=
('
(
4 − x2
5
)
5
4
−
(
4 − x2
3
)
3
+C
€
x = 2sec(θ ),dx = 2tan(θ )sec(θ )dθ
∫ 32sec (θ ) 4(4 sec (θ )) −16⋅ 2sec(θ )tan(θ )dθ = ∫ 64 sec (θ ) 16(sec (θ ) −1)⋅ tan(θ )dθ
= 256 ∫ sec (θ ) tan (θ )⋅ tan(θ )dθ = 256 ∫ sec (θ )tan (θ )dθ
€
= 256 ∫ (1+ tan (θ )) tan (θ )sec (θ )dθ = 256 ∫ (1+ 2tan (θ ) + tan (θ )) tan (θ )sec (θ )dθ
€
" u 2u
u %
+ '
= 256 ∫ (tan (θ ) + 2tan (θ ) +€tan (θ )) sec (θ )dθ = 256 ∫ ( u + 2u + u ) du = 256$ +
5
7&
#3
5
2
6
2
2
€
(
−
5
6
€
5
4
4 x 2 −16 dx
=
€
− 2xe x + 2e x + C = x 2e x − 2xe x + 2e x + C
6
2
2
2
2
2
2
4
2
2
3
2
4
6
2
2
4
6
## 2
€
&3
# x 2 − 4 &5 # x 2 − 4 &7 &
x
−
4
%
(
2
%
%
2 ('
2 (' %$
2 (' (
# tan 3 θ 2tan 5 θ tan 7 θ &
$
$
%
+
+
+
+
= 256%
( = 256% €
(
5
7 '
3
5 €
7
$ 3
%
(
$
'
5
7
#
%
= 256%
%
$
(
x2 − 4
24
3
) +(
x2 − 4
80
5
) +(
7&
3
5
7&
#
x 2 − 4 ( % 32( x 2 − 4 ) 2 16( x 2 − 4 ) 2 2( x 2 − 4 ) 2 (
+
+
=
(+C
896 (( %%
3
5
7
(
'
' $
)
e 2x
3 2x
e 2x 3 $
€
sin(3x)
−
=
+
−
e
sin(3x)dx
cos(3x)
&
∫ 2
2
2 2%
e 2x
e 2x
u = cos(3x)
v=
u = sin(3x)
v=
2
2
2x
2x
du = −3sin(3x)dx
dv
=
e
dx
du
=
3cos(3x)dx
dv
=
e
€
€
€
€ € dx
2x
2x
e
3
e
9
So, ∫ e 2x cos(3x)dx = cos(3x)
+ sin(3x)
− ∫ e 2x cos(3x)dx ,
2 4
2 4
2x
2x
13
€ e + 3 sin(3x) e ,
e 2x cos(3x)dx = cos(3x)
∫
4
2 4
2
2
3
2x
2x
€ =
cos(3x)e
€
+
sin(3x)e 2x + C,
∫ e cos(3x)dx
13
26
€
€
€
€ €
e 2x
–
2
g.
∫ 5cos (x)sin (x)dx = 5 ∫ cos (x)[1 − cos (x)] sin(x)dx = 5 ∫ (cos (x) − cos (x)) sin(x)dx
€
−cos (x)
−cos (x)
€
€
= 5 ∫ cos (x)sin(x)dx − 5 ∫ cos (x)sin(x)dx = 5
−5
+C
9
11
8
€
€
3
8
2
8
=
€
€
h.
€
€
∫ 6sec (2x)tan (2x)dx = 6 ∫ sec (2x)tan (2x)sec(2x)tan(2x)dx
= 6 ∫ sec (2x)(sec (2x) −1) sec(2x)tan(2x)dx = 6 ∫ (sec (2x) − sec (2x)) sec(2x)tan(2x)dx
#u
#u
# sec (2x) 3sec (2x) &
u &
3u &
1
−
= 6 ∫ ( u − u€) du = 3% − ( + C = % −
(+C = %
(+C
4'
4 '
2
4
2
$6
$2
$
'
4
3
3
3
OR
€
€
2
2
5
4
6
3
4
6
4
3
∫ 6sec (2x)tan (2x)dx = 6 ∫
4
3
€
sec (2x)tan 3 (2x)⋅ sec 2 (2x)dx
2
∫ tan (2x)€ (tan (2x) +1)⋅ sec€ (2x)dx = 6 ∫ (tan€ (2x) + tan (2x))⋅ sec (2x)dx
"u
"u
" tan (2x) 3tan (2x) %
u %
3u %
1
+
= 6 ∫ (u + u ) €
du = 3$ + ' + C = $ +
'+C = $
'+C
4&
4 &
2
4
2
#6
#2
#
&
=6
3
2
2
6
5
€
€
€
11
€
6
i.
€
2x
10
9
5cos11 (x)
€ − 5cos (x) + C
11
9
5
€
∫ 2e
10
9
8
€
€
'
cos(3x)dx )
(
∫ e 2x cos(3x)dx = cos(3x)
€
€
3
€
f.
∫
4
5
6
4
3
2
6
3
€
4
dx
(x − 3)(x +1) 2
€
€
4
A
B€
C
=
+
+
(x − 3)(x +1) 2 (x − 3) (x +1) (x +1) 2
4 = A(x +1) 2 + B(x − 3)(x +1) + C(x − 3)
A = 1/4, B = -1/4, C = -1
# 1
ln x − 3 ln x +1
1
1
1 &
−
+
+C
−
−
%
2 ( dx =
4
4
(x +1)
$ 4(x − 3) 4(x +1) (x +1) '
∫
€
4