Calculus 6-1 - 6-4 review worksheet answers 6-1 - 6-4...Sections 6/1 – 6/4 Review worksheet...

Calculus BC

Sections 6/1 – 6/4

Review worksheet answers

Find the area of the region bounded by :

1. y = x2 y = x – 3

x = -1 x = 2

A = ∫ ⋅−b

adxxgxf )]()([

= ∫− ⋅−−2

1

2 )]3([ dxxx

= ∫− ⋅+−2

1

2 )3( dxxx

=

2

1

23

323

−

+− x

xx

=

−+−

−−

+− 3

2

1

3

1)62

3

8

= 5.106

63

6

23

3

20==

−− sq. units

2. y = x2 - 5 y = 3 - x

2

Find the points of intersection.

x2 – 5 = 3 – x

2

2x2 = 8

x2 = 4

x = ±2

A = ∫ ⋅−b

adxxgxf )]()([

= ∫− ⋅−−−2

2

22 )]5()3[( dxxx

= ∫− ⋅−2

2

2 )28( dxx

=

2

2

3

3

28

−

−

xx

=

+−−

−

3

1616

3

1616

=

−−

3

32

3

32 =

3

64 square units

3. y = x3 – x y = x

2 – x

Find the points of intersection.

x3 – x = x

2 – x

x3 – x

2 – 2x = 0

x(x2 – x – 2) = 0

x(x – 2)(x + 1) = 0

x = -1, 0 , 2

A = [ ]∫− +−−0

1

23 )()( dxxxxx

+ [ ]∫ +−−2

0

23 )()( dxxxxx

= ∫− −−0

1

23 )2( dxxxx + ∫ −−2

0

23 )2( dxxxx

=

0

1

234

34−

−− x

xx +

2

0

234

34

−− x

xx

Calculus BC



= ( )

−+− 1

3

1

4

10 + )0(4

3

84 −

−−

= 12

32

12

5

3

8

12

5

3

8

12

5+=+=

−+

= 12

37 square units

Find the volume generated by rotating the

given region around the given line.

4. Region bounded by :

y = x2

+ 1, x = -1, x = 2, x-axis

Axis of rotation :

x-axis

V = dxRb

a⋅∫

2π (disc method)

R = y = x2

+ 1

R2 = (x

2 + 1)

2

= x4 + 2x

2 + 1

a = -1 b = 2

V = dxxx ⋅++∫− )12( 22

1

4π

= dxxx ⋅++∫− )12( 22

1

4π

=

2

1

35

3

2

5−

++ x

xxπ

=

−−

−−

++ 1

3

2

5

12

3

16

5

32π

=

+++++ 1

3

2

5

12

3

16

5

32π

=

++ 3

3

18

5

33π

= π(6.5 + 6 + 3)

= 15.5π or 2

31π cubic units


y = 6 – 2x, x = 0, y = 0

Axis of rotation :

y-axis

V = dyRd

c⋅∫

2π (disc method)

R = x

Since y = 6 – 2x, x = 3 – ½ y

So, R = 3 – ½ y

R2 = (3 – ½ y)

2

= 9 – 3y + ¼ y2

c = 0 d = 6

V = dyy

y ⋅

+−∫

6

0

2

439π

= dyy

y ⋅

+−∫

6

0

2

439π

=

6

0

32

122

39

+−

yyyπ

= ( )[ ])000(185454 +−−+−π

= 18π cubic units

Calculus BC




y = 4 – x, y = x, x = 0

Axis of rotation :

x = 0

V = ( ) ( )[ ] dyRRd

c⋅−∫

2

2

2

1π (washer method)

R1 = x1

Since y = 4 – x on that line, x = 4 – y

R1 = 4 – y

R2 = x2

Since y = x on that line, x = y

R2 = y

(R1)2 = (4 - y)

2 = 16 – 8y + y

2

(R2)2 = y

2

(R1)2 - (R2)

2

= (16 – 8y + y2) - (y

2)

= 16 – 8y

Finding c and d

4 – x = x

4 = 2x

2 = x

Since x = 2, y = 4 – 2 = 2

c = 0 d = 2

V = dyy ⋅−∫2

0)816(π

= 2

0

2 ]416[ yy −π

= π[(32 – 16) – (0 – 0)]

= π(32 – 0)

= π(32)

= 32π cubic units


y = 6 – x2, y = x

Axis of rotation :

y = -4

V = ( ) ( )[ ] dxRRb

a⋅−∫

2

2

2

1π (washer method)

R1 = 4 + y1 = 4 + (6 - x2) = 10 – x

2

R2 = 4 + y2 = 4 + x

(R1)2 = (10 – x

2)2 = 100 – 20x

2 + x

4

(R2)2 = (4 + x)

2 = 16 – 8x + x

2

(R1)2 - (R2)

2

= (100 – 20x2 + x

4) - (16 – 8x + x

2)

= x4 – 21x

2 + 8x + 84

Finding a and b

6 – x2 = x

0 = x2 + x – 6

0 = (x+3)(x-2)

x = -3 x = 2

a = -3 b = 2

V = ( ) ( )[ ] dxRRb

a⋅−∫

2

2

2

1π

= ( ) dxxxx ⋅++−∫−2

3

24 84821π

=

2

3

235

84475

−

++− xxx

xπ

= ( ) ( )[ ]252361896.4816816564.6 −++−−++−π

= π[(134.4) – (-75.6)]

= π(210) = 210π cubic units

Calculus BC




y = 4x – x2, x = 0

Axis of rotation :

y-axis

V = ∫ ⋅⋅⋅b

adxhpπ2 (shell method)

p = x

h = y = 4x – x2

p·h = x(4x – x2) = 4x

2 – x

3

Finding a and b

4x – x2 = 0

x(4 – x) = 0

x = 0 x = 4

V = ∫ ⋅−⋅4

0

32 )4(2 dxxxπ

=

4

0

43

43

42

−

xxπ

=

−−

− )00(64

3

2562π

=

3

642π

= 3

128π cubic units


y = x , y = 2

x = 6, y = 0

Axis of rotation :

x-axis

V = ∫ ⋅⋅⋅d

cdyhpπ2 (shell method)

p = y

h = 6 - x

Since y = x , y2 = x

h = 6 - y2

p·h = y(6 - y2) = 6y – y

3

Finding a and b

x = 2

x = 4

If x = 4, y = 4 = 2

c = 0 d = 2

V = ∫ ⋅−⋅2

0

3 )6(2 dxyyπ

=

2

0

42

432

−

yyπ

= 2π[(12 - 4) – (0 – 0)]

= 2π(8)

= 16π cubic units

Calculus BC




y = 2

2 xe

−+ , y = x

Axis of rotation :

y-axis

V = ∫ ⋅⋅⋅b

adxhpπ2 (shell method)

p = x

h = y1 – y2 = 2

2 xe

−+ – x

p·h = x(2

2 xe

−+ - x) = 2x + x

2x

e−

⋅ - x2

Finding a and b 2

2 xe

−+ = x

2

2 xe

−+ – x = 0

To approximate x, graph y = 2

2 xe

−+ – x

on a graphing calculator and see where it

crosses the x-axis (i.e. where y = 0 using

TRACE).

Or use the calculator’s table function.

Either way, x is about 2.017

So, a = 0 and b = 2.017.

V = ( )∫ ⋅−⋅+⋅−

017.2

0

22

22 dxxexxxπ

=

017.2

0

32

322

2

−−

−xe

xx

π

= 2π[(1.324489379) – (-0.5)]

= 2π(1.824489379)

= 5.73 cubic units

Find the length of the arc specified.

11. y = ( )2

3

93

2+x

-1 < x < 6

L = dxdx

dyb

a⋅

+∫

2

1

dx

dy = ( )2

1

92

3

3

2+⋅ x

= ( )2

1

9+x 2

dx

dy= x + 9

1 +

2

dx

dy = 1 + x + 9 = x + 10

L = dxx ⋅+∫−6

110

= ( )6

1

2

3

103

2

−

+x

Calculus BC



= ( ) ( )

− 2

3

2

3

9163

2

= [ ]27643

2−

= [ ]373

2 =

3

74 units

12. y = ¼ x2 – ½ ln x

4 < x < 16

L = dxdx

dyb

a⋅

+∫

2

1

dx

dy =

x

x

xx

x

x

x

2

1

2

1

22

1

2

22−

=−=−

2

dx

dy=

2

24

4

12

x

xx +−

1 +

2

dx

dy =

2

24

4

121

x

xx +−+

= 2

24

2

2

4

12

4

4

x

xx

x

x +−+

= 2

24

4

12

x

xx ++

= ( )

2

22

)2(

1

x

x +

=

22

2

1

+

x

x

L = dxx

x⋅

+∫

16

4

22

2

1

= dxx

x⋅

+∫

16

4

2

2

1

= dxxx

x⋅

+∫

16

4

2

2

1

2

= dxx

x ⋅

⋅+∫

16

4

1

2

1

2

1

= dxx

x ⋅

+∫

16

4

1

2

1

=

16

4

2

ln22

1

+ x

x

= ( ) ( )[ ]4ln816ln1282

1+−+

= [ ]4ln816ln1282

1−−+

=

+

4

16ln120

2

1

= [ ]4ln1202

1+

= 60 + ½ ln 4

= 60 + ln 2 (about 60.7 units)

Find the surface area swept out when the

function is rotated around the given axis.

13. y = x3

0 < x < 2

around x-axis

S = dxdx

dyR

b

a⋅

+∫

2

12π

R = y = x3

dx

dy = 3x

2

Calculus BC



2

dx

dy = 9x

4

1 +

2

dx

dy = 1 + 9x

4

S = dxxx ⋅+⋅∫2

0

43 912π

= dxxx ⋅⋅+∫2

0

32

1

4 )91(2π

u = 1 + 9x4

du = 36x3·dx

36

1du = x

3·dx

If x = 2, u = 1 +9·24 = 145

If x = 0, u = 1 +9·04 = 1

S =

⋅∫ duu

36

12 2

1145

1π

= duu ⋅∫ 2

1145

118

π

=

145

1

2

3

3

2

18

u

π

=

145

13

2

18

uu

π

= [ ]145

13

2

18uu⋅

π

= ( )1114514527

−π

= ( )114514527

−π

(about 203 sq. units)

14. y = 21 x− 0 < x < 1 around y-axis

S = dydy

dxR

d

c⋅

+∫

2

12π

R = x

Since y = 21 x− , x = 21 y−

R = 21 y−

x = 21 y−

dy

dx =

21 y

y

−

−

2

22

1 y

y

dy

dx

−=

1+2

22

11

y

y

dy

dx

−+=

1+2

2

2

22

11

1

y

y

y

y

dy

dx

−+

−

−=

1+2

2

1

1

ydy

dx

−=

S = dyy

y ⋅−

⋅−⋅∫1

0 2

2

1

112π

= dyy

y ⋅−

⋅−⋅∫1

0 2

2

1

112π

= dy⋅∫1

02π

= [ ]10

2 yπ

= 2π·1 - 2π·0 = 2π square units

Calculus 6-1 - 6-4 review worksheet answers 6-1 - 6-4...Sections 6/1 – 6/4 Review worksheet...

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Transcript of Calculus 6-1 - 6-4 review worksheet answers 6-1 - 6-4...Sections 6/1 – 6/4 Review worksheet...