calculus diff chapter 7 test review answers
TRANSCRIPT
Calculus Diff Chapter 7 Test Review Answers
1. Let
€
u = 2xdu = 2 dx
giving
€
12
11+ u2∫ du = 1
2arctan 2x( )+C
2. Let
€
u =1+ cos xdu = −sin x dx
giving
€
− u2∫ du
€
= − 13 1+ cos x( )3 +C
3. Let
€
u = ln xdu = 1
x dx giving
€
1u2∫ du
€
=−1ln x
+C
4. Let
€
u = −x2
du = −2x du giving
€
−12
eu−4
−16∫ du =
−12eu
−16−4
=−12e−16 − −1
2e−4 =
−12e16
+12e4
5. Let
€
u = x − 3du = dx
giving
€
1udu
0
3∫ =
u12
12
3
0= 2 3 − 2 0 = 2 3
6. Let
€
u = 2xdu = 2 dx
giving
€
12
cosu0
π3∫ du =
12
sinu
π3
0=
12
sin π3( ) − 1
2sin 0( ) =
12
32
%
& '
(
) * =
34
7. Use
€
u = 9x w = 15 sin 5x( )
du = 9 dx dw = cos 5x( ) dx to get 9x( ) 1
5 sin 5x( )( )− 15 sin 5x( ) ⋅9 dx∫
€
= 9x5 sin 5x( ) + 9
25cos 5x( ) +C
8. Use
€
u = sin−1 x w = xdu = 1
1−x2 dx dw = dx to get
€
sin−1 x( ) dx∫
€
= sin−1 x( ) x( ) − x1−x2
dx∫ , use
€
u =1− x2
du = −2x dx to get
€
sin−1 x( ) x( ) − −12( ) 1
u du∫
€
= x sin−1 x + 12 ⋅u12
12
+C
€
= x sin−1 x + 1− x2 +C
9. Use partial fractions,
€
9x + 2x + 6( ) x −1( )
=A
x + 6+
Bx −1
giving
€
9x + 2 = A x −1( ) + B x + 6( ), the two
equations are
€
9 = A + B2 = −A + 6B
, solving gives
€
A =527
and
€
B =117
. The integral is now
€
527
x + 6+
117
x −1∫ dx
€
=527ln x + 6 +
117ln x −1 +C
10. Use partial fractions,
€
x2 − x + 4x x − 5( ) x + 4( )
=Ax
+B
x − 5+
Cx + 4
giving
€
x2 − x + 4 = A x − 5( ) x + 4( ) + Bx x + 4( ) +Cx x − 5( ) or
x2 − x + 4 = Ax2 − Ax − 20A+Bx2 + 4Bx +Cx2 − 5Cx , the three equations are 1= A+B+C
−1= −A+ 4B− 5C4 = −20A
,
solving last equation gives
€
A = −15 , then have
1= − 15 +B+C
−1= − − 15( )+ 4B− 5Cwhich gives
€
B = 815 and
€
C = 23 . The integral is now
€
−15x
+8
15x − 5
+23
x + 4∫ dx
€
=−15ln x +
815ln x − 5 +
23ln x + 4 +C
11. Use partial fractions with unfactorable quadratic,
€
x2 + 2x + 3x x2 +1( )
=Ax
+Bx +Cx2 +1
giving
€
x2 + 2x + 3 = A x2 +1( ) + Bx +C( )x , the three equations are
€
1 = A + B2 = C3 = A
giving
€
B = −2 . The integral
is now
€
3x
+−2x + 2x2 +1
dx∫
€
=3x
+−2xx2 +1
+2
x2 +1 dx∫ , in the second integral use
€
u = x2 +1du = 2x dx
giving 3 1x
dx∫ −1u
du∫ + 2 1x2 +1
dx∫
€
= 3ln x − lnu + 2tan−1 x +C
€
= 3ln x − ln x2 +1 + 2tan−1 x +C
12. Use partial fractions with a repeated factor,
€
x2 + x − 7x2 x − 3( )
=Ax
+Bx2
+Cx − 3
giving
€
x2 + x − 7 = Ax x − 3( ) + B x − 3( ) +Cx2, the three equations are
€
1 = A +C1 = −3A + B−7 = −3B
, solving gives
€
A = 49 ,
€
B = 73 and
€
C = 59 . The integral is now
€
49x
+73x2 +
59
x − 3∫ dx
€
=49ln x − 7
3x−1 +
59ln x − 3 +C
13. Complete the square to get 3
x2 −10x + −5( )2( )+ 29− −5( )2 dx∫
€
=3
x − 5( )2 + 4 dx∫
€
=34
1x−5( )2
4 +1 dx∫ , let
€
u = x−52
du = 12 dx
giving
€
34⋅ 2 1
u2 +1 du∫
€
= 32 tan
−1 u +C
€
= 32 tan
−1 x−52( ) +C
14. Complete the square to get 5x +3x2 + 6x +32( )+10−32
dx∫
€
=5x + 3x + 3( )2 +1
dx∫ , write as two
integrals 5xx +3( )2 +1
dx∫ +3
x +3( )2 +1 dx∫ =
5 x +3( )x +3( )2 +1
dx∫ +3− 5( ) 3( )x +3( )2 +1
dx∫ , on the first
use
€
u = x + 3( )2 +1du = 2 x + 3( ) dx
and on the second use
€
w = x + 3dw = dx
giving
€
52
1u
du∫ −12 1w2 +1
dw∫
€
=52lnu −12tan−1w +C
€
=52ln x + 3( )2 +1 −12tan−1 x + 3( ) +C
15. Let
€
x = 3sinθ and
€
dx = 3cosθ dθ , also
€
9 − x2 = 3cosθ . Then
€
1
x2 9 − x2dx∫
€
=1
3sinθ( )2 3cosθ( )3cosθ dθ∫
€
=19
1sin2θ
dθ∫
€
=19
csc2θ dθ∫
€
=−19cotθ + C
€
=− 9 − x2
9x+ C
16. Let
€
x = tanθ and
€
dx = sec2θ dθ , also
€
1+ x2 = secθ . Then 1
1+ x2( )32dx∫ = 1
secθ( )3sec2θ dθ∫ = 1
secθ dθ∫ = cosθ∫ dθ
= sinθ +C = x
1+ x2+C
17. x
x2 − 5 3
∞∫ dx = lim
b→∞ x
x2 − 5 3
b∫ dx , let
€
u = x2 − 5du = 2x dx
giving limb→∞
12
1u 3
b∫ du
= limb→∞
12⋅u
12
12 x=3
x=b
= limb→∞
x2 − 53
b = ∞2 − 5 − 32 − 5 =∞− 2 =∞ , diverges.
18.
€
2x2 +1
dx1
∞∫ = lim
b→∞ 2 1
x2 +1 dx
1
b∫ = lim
b→∞ 2 tan−1 x
1
b
€
= 2 tan−1 ∞( ) − tan−1 1( )( )
€
= 2 π2−π4
$
% &
'
( ) = 2
π4$
% & '
( ) =
π2
, converges.
19. 1x 2− x( )
dx3
4∫ = lim
b→4 1
x 2− x( ) dx
3
b∫ , let
u = 2− x
du = − 12 x
−12 dx
giving
limb→4
− 2 1u
du 3
b∫ = lim
b→4 − 2 ln u
x=3x=b = lim
b→4 − 2 ln 2− x
3
b = −2 ln 2− 4 + 2 ln 2− 3
= −2 ln 0( )+ 2 ln 2− 3( ) = −2 −∞( )+ 2 ln 2− 3( ) =∞+ 2 ln 2− 3( ) =∞ , diverges
20.
€
sin x
cos2 x 0
π2∫ dx = limb→π
2
sin xcos2 x
dx 0
b∫ , let
€
u = cos xdu = −sin x dx
giving limb→π
2
− 1u2 du
0
b∫
= limb→π
2
−u−1
−1x=0
x=b
= limb→π
2
1u x=0
x=b = lim
b→π2
1cos x 0
b
€
=1
cos π2( )−
1cos 0( )
€
=10−11
€
= ∞ −1 = ∞ , diverges
21.
€
11+ x( ) x
dx1
∞
∫ =1
x + x3
2 dx
1
∞
∫ . Since
€
x12 ≥ 0 then
€
x12 + x
32 ≥ x
32 giving
€
1
x12 + x
32≤1
x32
.
€
1
x3
2 dx
1
∞
∫ is a p function with
€
p = 32 which is greater than 1, therefore
€
1
x3
2 dx
1
∞
∫ converges.
Since
€
1
x3
2 dx
1
∞
∫ converges and
€
1
x12 + x
32≤1
x32
, then by the direct comparison test
€
11+ x( ) x
dx1
∞
∫ also converges.
22. Since cos x ≤1 then −cos x ≥ −1 then
€
2 − cos x ≥1 then
€
2 − cos x ≥1 and
€
2 − cos xx
≥1x
.
1x
dx0
1∫ is a p function with p = 1 and therefore diverges. Since
€
2 − cos xx
≥1x
and 1x
dx0
1∫
diverges, then by the direct comparison test
€
2 − cos xx
dx0
1∫ also diverges.