Review Problems Integration 1. Find the instantaneous rate of change of the function at x = -2 _ 1

Review Problems Integration

1. Find the instantaneous rate of change of the function

at x = -2

_

2

1( )g x

x

2

3

3

( )

' ( ) 2

1'(2) 2( 2)

4

g x x

g x x

g

1


2. One of these curves is the graph of a function f, another is the graph of f’, and the third is the graph of f”. Which is which?

_

A is f

B is f’ when C crosses the x-axis B has an extrema, so B is f’

C is f” cannot be f because it has a minimum where no other curve has a zero. Same for f’

2

A

B

C


3. One of these three curves represents the position of a particle moving in a straight line, another represents the particle’s velocity, and the third represents its acceleration. Which curve is which and why?

Curve A is acceleration

Curve B is velocity

Curve C is the position

Neither B nor C crosses t at the points where A has extrema, therefore, A is not acceleration, nor position, so A is s”(t)

It crosses the t-axis at the point where B has an extrema and C doesn’t, so B is s’(t), therefore, C is the position function

3

AB

C


4. The graphs of f and g are shown. If h is defined by h(x) = f(x) g(x), find h’(1)

4

3

2

1

-2 -1 1 2 3 4 5 6

h’(x) = f’(x) g(x) + f(x) g’(x)

h’(1) = f’(1) g(1) + f(1) g’(1)

= 2 * 1 + 2 * -1

h’(1) = 0

4


x 1 2 3

f(x) 3 1 7

g(x) 2 8 2

f’(x) 4 5 7

g’(x) 6 7 9

h(x) = f(g(x))

h’(x) = f’(g(x)) (g’(x))

h’(1) = f’(g(1)) (g’(1))

= [f’(2)] (6)

= 5 * 6

= 30

5

5. The functions f and g are differentiable and defined for all real numbers. The function h is given by h(x) = f(g(x)). Using the values of f, g, f’ and g’ in the table, find h’(1)


6. The table shows a few values of the function f and its derivative f’. If h is a function given by

What is h’(-1)?

6

2( ) ( )h x f x

x -1 0 1

f(x) 3 1 7

f’(x) 4 2 1

-2

2

2

( ) ( )

' ( ) ' ( )(2 )

' ( 1) ' (( 1) )(2( 1))

' (1)( 2)

(1)( 2) 2

h x f x

h x f x x

h f

f


7. Find the derivative of the function

-

7

2( ) 3f x x x

122

122

( ) ( 3 )

1'( ) ( 3 ) (2 3)

2

f x x x

f x x x x


8. Find the derivative of the function f(x)=sin(cos x)

f(x) = sin(cos x)

= cos(cos x) * -sin x

= -sin x * cos (cos x)

8



_

9

sin( ) xf x esin

sin

( )

' ( ) cos

x

x

f x e

f x e x


Find the derivative of the function

_

10

2( ) sin (2 )f x x2( ) [sin(2 )]

' ( ) 2[sin(2 )][cos(2 )](2)

' ( ) 4[sin(2 )][cos(2 )]

f x x

f x x x

f x x x



_

11

2( ) ln(1 )f x x

2

2

2

( ) ln(1 )

1'( ) 2

12

1

f x x

f x xx

x

x


12. Based on the data in the chart below, estimate

by using five subintervals of equal length

A. By left-hand Riemann sums

Intervals:

8 + 28 + 48 + 44 + 24 = 152

15

14

13 (8,12)

12 (12,11)

11

10

9

8

7 (4,7)

6 (16,6)

5

4

3 (2,0)

2

1

4 8 12 16 20

12

20

0( )v t dt

t O 2 4 6 8 10 12 14 16 18 20

v() 2 4 7 9 12 15 11 9 6 5 3

20 04

5

48 44

24

8

28




B. By Right-hand Riemann Sums

Intervals:

28 + 48 + 44 + 24 + 12 = 156

15

14

13

12 (8,12)

11 (12, 11)

10

9

8

7 (4,7)

6 (16,6)

5

4

3 (20,3)

2

1

4 8 12 16 20

13

20

0( )v t dt

t O 2 4 6 8 10 12 14 16 18 20

v() 2 4 7 9 12 15 11 9 6 5 3

20 04

5

44

24

1228 48




C. By Midpoint Rule

Intervals:

16 + 36 + 60 + 36 + 20 = 168

(10, 15)

15

14

13

12 (14, 9)

11 (6, 9)

10

9

8

7 (18, 5)

6

5

4 (2, 4)

3

2

1

4 8 12 16 20

14

20

0( )v t dt

t O 2 4 6 8 10 12 14 16 18 20

v() 2 4 7 9 12 15 11 9 6 5 3

20 04

5

60

36

2016

36


Based on the midpoint rule, find an estimate of the average velocity over the time interval 0 to 20 inclusive

Average Velocity =

15

20

0

1( )

201

[168] 8.420

v t dt


13. A particle moves along a number line such that its position s at any time t, t>0, is given by

A. Find the average velocity over the time interval

Average velocity

7 moving to the left16

3 2( ) 2 15 24 1s t t t t

1 2t

2 1

2 13 2

3 2

( ) ( )

(2) 2(2) 15(2) 24(2) 1

2 8 15 4 24 2 1

16 60 48 1

5

(1) 2(1) 15(1) 24(1) 1

2 1 15 1 24 1 1

2 15 24 1

12

(2) (1) 5 127

2 1 1

s t s t

t t

s

s

s s



B. Find the instantaneous velocity at t = 2

Instantaneous velocity

Moving to the left 12

17

3 2( ) 2 15 24 1s t t t t

3 2

2

2

( ) 2 15 24 1

( ) ' ( ) 6 30 24

(2) 6(2) 30(2) 24

6 4 30 2 24

24 60 24

12

s t t t t

v t s t t t

v



C. When is the particle at rest?

At rest when v(t) = 0

18

3 2( ) 2 15 24 1s t t t t

2

26( 5 4) 0

6( 4)( 1) 0

4 0 or

6

1 0

4 or

30 2 0

1

4

t t

t t

t t

t t

t t



D. What is the total distance traveled by the particle over the time interval

Use the endpoints 0,5 and when particle stops 1,4

Total distance:

t=0 to t=1 12-1=11

t=1 to t=4 12-(-15) = 27

t=4 to t=5 -4 – (-15) = 11

Distance = 49

t=5

s=-4

t=4

s=-15

t=0 t=1

s=1 s=12

-15 -4 0 1 12

19

3 2( ) 2 15 24 1s t t t t

0 5t

3 2

3 2

3 2

3 2

at t=0 2 0 15 0 24 0 1 1

at t=1 2 1 15 1 24 1 1 12

at t=4 2 4 15 4 24 4 1 15

at t=5 2 5 15 5 24 5 1 4


14. Consider the differential equation

and let y = f(x) be the solution

A. On the axis provided, sketch a slope field on the 14 points indicated

-

1

-2 -1 0 1 2

-1

20

1' 0

2yy x

1' 0

2' 2 0

' 2

2

yy x

yy x

yy x

dy xdx y

x,y -2 -1 0 1 2

1 -4 -2 0 2 4

0 Inf Inf Inf Inf Inf

-1 4 2 0 -2 -4




B. For the particular solution with the initial condition f(2)= -1, write the equation of the tangent line to the graph of f at x = 2

21

1' 0

2yy x

at point (2,-1)

equation: Y + 1 = -4(x – 2)

4( )dy

slopedx




C. Write the particular solution to the given differential equation with the initial condition f(1) = 1

-

22

1' 0

2yy x 2 2

22

22

2 2

2

2

2

22 2

2at x= 1 f (1) = 1

11

21

21

2 22 1

2 1

dy xdx y

ydy xdx

y xC

yx C

C

C

yx

y x

y x

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Review Problems Integration 1. Find the instantaneous rate of change of the function at x = -2 _ 1