The Definite Integral

Day 6

Motion Problems

Strategies for Finding Total Area

ARRIVAL---HW Questions Working in PODS

Additional Practice—Packet p. 13 and 14

•Make good use of your time!

•Practice makes perfect!

•Ask ME questions, ask your

CLASSMATES questions!

Area Problems

Review Problem ……

1. Set up an integral that represents the shaded region.

2. Evaluate the integral using the Fundamental Theorem of

Calculus THEN confirm your answer with “ fnInt”

f x( ) = x2

x2

0

3

ò dx = 9

Find the area of the shaded region f x( ) = x2

27 - 9 = 18

Discuss with your partner what we need to do

to find the area of the shaded region.

Important to Remember:

Integrals find area between the curve

and the x-axis.

Remember ….. • When evaluating integrals, “areas” beneath the x-axis are negative.

• When evaluating total area, all areas are positive.

a) Evaluate the integral

a) Find the total area of the graph

12 4

f(x)

12

0

( )f x dx

12 4 12

0 0 4

( ) ( ) ( ) 3 7 4f x dx f x dx f x dx

12 4 12

0 0 4

( ) ( ) ( ) 3 7 10f x dx f x dx f x dx

Practice—Find the area of the shaded region.

NO CALCULATOR

YOU TRY!!

Check your answers…

16

316

19

3

8

3

Today’s Learning Outcomes…

Apply antiderivatives to motion problems

Recognize the relationship between displacement and the total distance traveled by an object

Motion Problems

Remember …….

Given a position function:

Then Velocity: '

And Acceleration: ' ''( )

s t

v t s t

a t v t s t

Given the velocity function,

what could we determine …….

Given velocity v t

Position ( ) + C= s t v t dt

BECAUSE s(t) is the ANTIDERIVATIVE of v(t)

Remember our earlier example in this unit ……

( ) 55 from 2 : 00 to 5:00

Distance traveled = (55)(3) miles

v t mph

55

2 2

OR

55 55 55(5 2) 55(3) milesdt t

Graphically: What does this look like?

Position of object at time t.

Terminology …..

Displacement—how far away from the starting point the

object is at the end of a given time interval

Distance Traveled—amount of movement by the object in

the positive and negative direction

The graph below shows the velocity of a particle over time.

The area between the curve and the x-axis represents

distance traveled.

Area above the x-axis

Distance traveled in the positive direction

Movement away from the starting location

Area below the x-axis

Distance traveled in the negative direction

Movement back towards the starting location

The graph shows the velocity of a particle over time.

a) What is the displacement of the particle from 0 to 20

seconds?

1 1a) 13 6 + 7 -4 = 39 -14 = 25 feet displacement

2 2

The graph shows the velocity of a particle over time.

b) What is the total distance traveled from 0 to 20

seconds?

1 1b) 13 6 + 7 -4 = 39+14

2 2

= 53 feet total distance traveled

Another Example …… The velocity of a particle, in ft/sec, is given by 2 .

Find the displacement and total distance traveled from 2 to 4.

v t t

t

4 4

2 2 2v t dt t

42

2t

2 24 2 16 4 12

Displacement is 12 feet.

Velocity is positive in interval,

therefore total distance also 12 feet.

Before you start: Draw a sketch of the velocity

curve to understand visually what is involved.

The velocity of a particle, in ft/sec, is given by 2 6.

Find the displacement and total distance traveled from 2 to 7.

v t t

t

7 7

2 2 2 6v t dt t

72

26t t

2 27 6 7 2 6 2 7 8 15

Displacement is 15 feet.

But what about total distance?

Before you start: Draw a sketch of the velocity

curve to understand visually what is involved.

The velocity of a particle, in ft/sec, is given by 2 6.

Find the displacement and total distance traveled from 2 to 7.

v t t

t

7

2 v t dt

3 72 2

2 36 6t t t t

2 2 2 23 6 3 2 6 2 7 6 7 3 6 3

1 16 17

Total distance traveled is 17 feet.

3 7

2 3 v t dt v t dt

How confident are you in your

ability to… Apply antiderivatives to motion problems

Recognize the relationship between displacement and the total distance traveled by an object