calculus 2 chapter 7 test practice - saint paul public ... · calculus 2 chapter 7 test practice 1)...
TRANSCRIPT
Calculus 2 Chapter 7 Test Practice 1) Identify the integration method you would use for each of the following problems. Do NOT integrate!
a.
€
1x 2 + 8x + 25
dx∫ b.
€
3x + 4x 2 + 2x −15
dx∫ c.
€
64 − x 2dx∫
________________ ________________ ________________ d.
€
x 2 cos(x)dx∫ e.
€
cos4 (x)sin3(x)dx∫ f.
€
5x 25 − x 2dx∫ ________________ ________________ ________________
2) Find the limits using L’Hospital’s Rule, when necessary.
a.
€
limx→ 0
sin(3x)4x
________________
b.
€
limx→ 0
sin−1(2x)x
________________
c.
€
limx→ 0
xx( ) ________________
d.
€
x→+∞lim 1− 2
x%
& '
(
) * 3x
________________
e.
€
x→∞lim x e
1x −1( )
________________
Name _________________________ Period _____
3) Integrate each indefinite integral: a.
€
x ln(x)dx∫ ________________
b.
€
x 2exdx∫ ________________
c.
€
x 2 4 − x 3dx∫ ________________
d.
€
x 3 4 − x 2dx∫ ________________
e.
€
x 5 4x 2 −16∫ dx ________________
f.
€
e2x cos(3x)dx∫ ________________
g.
€
5cos8(x)sin3(x)dx∫ ________________
h.
€
6sec4 (2x)tan3(2x)dx∫ ________________
i.
€
4(x − 3)(x +1)2
dx∫
________________
Calculus 2 Chapter 7 Test Practice 1) Identify the integration method you would use for each of the following problems. Do NOT integrate!
a.
€
1x 2 + 8x + 25
dx∫ b.
€
3x + 4x 2 + 2x −15
dx∫ c.
€
64 − x 2dx∫
tan-1(x) partial fractions trig substitution__
(x = 8sin(θ)) d.
€
x 2 cos(x)dx∫ e.
€
cos4 (x)sin3(x)dx∫ f.
€
5x 25 − x 2dx∫ integration by parts trigonometric integration u-substitution___
2) Find the limits using L’Hospital’s Rule, when necessary.
a.
€
limx→ 0
sin(3x)4x
= limx→ 0
3cos(3x)4
=34
€
limx→ 0
sin(3x)4x
=
€
34
b.
€
limx→ 0
sin−1(2x)x
= limx→ 0
21− 4x 21
= limx→ 0
21− 4x 2
=21
= 2
€
limx→ 0
sin−1(2x)x
= 2
c.
€
limx→ 0
xx( )
lny=
€
limx→ 0
x ln x( )=
€
limx→ 0
ln xx −1
$
% &
'
( ) =
€
limx→ 0
1x
−x −2$
%
& &
'
(
) ) = limx→ 0
1x
− 1x2
$
%
& &
'
(
) ) = limx→ 0
−x 2
x$
% &
'
( ) = lim
x→ 0−x( ) = 0
lny=0 → y = e0 = 1
€
limx→ 0
xx( ) = 1
d.
€
x→+∞lim 1− 2
x%
& '
(
) * 3x
lny =
€
x→+∞lim 3x ln 1− 2
x%
& '
(
) * =
€
3x→+∞lim
ln x − 2x
%
& '
(
) *
1x
=
€
3x→+∞lim
xx − 2%
& '
(
) * ⋅ 2x −2( )
−1x 2
=
€
= 3x→+∞lim
−2x 3
x 2 x − 2( )= 3
x→+∞lim
−2xx − 2( )
= 3⋅ −2 = −6
lny= -6 → y = e-6 ≈ 0.00248
€
x→+∞lim 1− 2
x%
& '
(
) * 3x
= e-6 ≈ 0.00248
e.
€
x→∞lim x e
1x −1( ) =
€
x→∞lim
e1x −1( )1x
=
€
x→∞lim
e1x ⋅ −x −2( )−x −2
=
€
x→∞lim
e1x( )1
=x→∞lim e
1x = e0 =1
€
x→∞lim x e
1x −1( ) = 1
Name _________________________ Period _____
3) Integrate each indefinite integral:
a.
€
x ln(x)dx∫ =x 2 ln x2
−x2dx∫ =
x 2 ln x2
−x 2
4+C
€
x 2 ln x2
−x 2
4+C
€
u = ln x v =x 2
2du =
1xdx dv = xdx
b.
€
x 2exdx∫ = x 2ex − 2xexdx∫ = x 2ex − 2xex − 2exdx∫[ ] = x 2ex − 2xex + 2ex +C=
€
x 2ex − 2xex + 2ex +C
€
u = x 2 v = ex
du = 2xdx dv = exdx
€
u = 2x v = ex
du = 2dx dv = exdx
c.
€
x 2 4 − x 3dx∫ =
€
−13
udu∫ =
€
−29u32
#
$ %
&
' ( +C =
€
−29(4 − x 3)
32
#
$ %
&
' ( +C
€
u = 4 − x 3,du = −3x 2dx
−13du = x 2dx
d.
€
x 3 4 − x 2dx∫
€
x = 2sin(θ),dx = 2cos(θ )dθ
=
€
8sin3(θ ) 4 − 4sin2(θ )2cos(θ )dθ∫ =
€
16 sin3(θ)cos(θ) 4cos2(θ )dθ∫
=
€
32 sin3(θ)cos2(θ)dθ∫ =
€
32 [1− cos2(θ )]cos2(θ)sin(θ)dθ∫
=
€
32 [cos2(θ ) − cos4 (θ )]sin(θ)dθ∫ =
€
32 −cos3(θ )3
+cos5(θ )5
$
% &
'
( ) +C
=
€
324 − x 2( )
532
5−
4 − x 2( )38
3
#
$
% % %
&
'
( ( (
+C=
€
4 − x 2( )5
5−4 4 − x 2( )
3
3+C
e.
€
x 5 4x 2 −16∫ dx
€
x = 2sec(θ ),dx = 2tan(θ )sec(θ )dθ
=
€
32sec5(θ ) 4(4 sec2(θ )) −16 ⋅ 2sec(θ )tan(θ )dθ∫ =
€
64sec6(θ ) 16(sec2(θ ) −1) ⋅ tan(θ)dθ∫
=
€
256 sec6(θ) tan2(θ) ⋅ tan(θ)dθ∫ =
€
256 sec6(θ)tan2(θ )dθ∫
=
€
256 1+ tan2(θ)( )2 tan2(θ)sec2(θ)dθ∫ =
€
256 1+ 2tan2(θ) + tan4 (θ )( ) tan2(θ )sec2(θ )dθ∫
=
€
256 tan2(θ) + 2tan4 (θ ) + tan6(θ)( )sec2(θ )dθ∫ =
€
256 u2 + 2u4 + u6( )du∫ =
€
256 u3
3+2u5
5+u7
7"
# $
%
& '
=
€
256 tan3θ3
+2tan5θ5
+tan7θ7
#
$ %
&
' ( =
€
256
x 2 − 42
#
$ %
&
' ( 3
3+2 x 2 − 4
2#
$ %
&
' ( 5
5+
x 2 − 42
#
$ %
&
' ( 7
7
#
$
% % % %
&
'
( ( ( (
=
€
256x 2 − 4( )
3
24+
x 2 − 4( )5
80+
x 2 − 4( )7
896
#
$
% % %
&
'
( ( ( =
€
32 x 2 − 4( )32
3+16 x 2 − 4( )
52
5+2 x 2 − 4( )
72
7
#
$
% % %
&
'
( ( (
+C
f.
€
e2x cos(3x)dx∫ =
€
cos(3x)e2x
2 –
€
−32e2x sin(3x)dx∫ =
€
cos(3x)e2x
2+
€
32
€
sin(3x)e2x
2−
32e2x cos(3x)dx∫
$
% &
'
( )
€
u = cos(3x) v =e2x
2du = −3sin(3x)dx dv = e2xdx
€
u = sin(3x) v =e2x
2du = 3cos(3x)dx dv = e2xdx
So,
€
e2x cos(3x)dx∫ =
€
cos(3x)e2x
2+
€
34sin(3x)
e2x
2−94
e2x cos(3x)dx∫ ,
€
134
€
e2x cos(3x)dx∫ =
€
cos(3x)e2x
2+
€
34sin(3x) e
2x
2,
€
e2x cos(3x)dx∫ =
€
213
€
cos(3x)e2x +
€
326sin(3x)e2x + C,
g.
€
5cos8(x)sin3(x)dx∫ =
€
5 cos8(x) 1− cos2(x)[ ]sin(x)dx∫ =
€
5 cos8(x) − cos10(x)( )sin(x)dx∫
=
€
5 cos8(x)sin(x)dx∫ − 5 cos10(x)sin(x)dx∫ =
€
5−cos9(x)
9− 5
−cos11(x)11
+C
=
€
5cos11(x)11
−5cos9(x)
9+C
h.
€
6sec4 (2x)tan3(2x)dx∫ =
€
6 sec3(2x)tan2(2x)sec(2x)tan(2x)dx∫
=
€
6 sec3(2x) sec2(2x) −1( )sec(2x)tan(2x)dx∫ =
€
6 sec5(2x) − sec3(2x)( )sec(2x)tan(2x)dx∫
=
€
6 u5 − u3( ) 12 du∫ =
€
3 u6
6−u4
4#
$ %
&
' ( +C =
€
u6
2−3u4
4#
$ %
&
' ( +C =
€
sec6(2x)2
−3sec4 (2x)
4#
$ %
&
' ( +C
OR
€
6sec4 (2x)tan3(2x)dx∫ =
€
6 sec2(2x)tan3(2x)⋅ sec2(2x)dx∫
=
€
6 tan3(2x) tan2(2x) +1( )⋅ sec2(2x)dx∫ =
€
6 tan5(2x) + tan3(2x)( )⋅ sec2(2x)dx∫
=
€
6 u5 + u3( ) 12 du∫ =
€
3 u6
6+u4
4"
# $
%
& ' +C =
€
u6
2+3u4
4"
# $
%
& ' +C =
€
tan6(2x)2
+3tan4 (2x)
4"
# $
%
& ' +C
i.
€
4(x − 3)(x +1)2
dx∫
€
4(x − 3)(x +1)2
=A
(x − 3)+
B(x +1)
+C
(x +1)2
€
4 = A(x +1)2 + B(x − 3)(x +1) +C(x − 3) A = 1/4, B = -1/4, C = -1
€
14(x − 3)
−1
4(x +1)−
1(x +1)2
#
$ %
&
' ( dx∫ =
€
ln x − 34
−ln x +14
+1
(x +1)+C