Unit 8 Day 1

Integral as Net Change

Free Response Practice Packet p.1

Terminology

Displacement of an object is the distance from an arbitrary starting point at the end of some time interval.

Actual Distance Traveled is how far an object travels regardless of direction.

Review: Position and Velocity

If s(t) is the position of an object then the velocity of the object will be s’(t).

The area between the velocity curve and the x-axis represents the total displacement or change in position.

Displacement '( ) b b

a as t dt v t dt

Example Sketch:

Review: Position and Velocity

To find the actual distance traveled integrate the absolute value of the velocity function.

Actual Distance b

av t dt

Example Sketch:

Velocity measured in feet per second:

3 2( ) 10 24 [0,6]v t t t t on t

Use a sign line to determine when the object is moving with

positive displacement and negative displacement.

When will the object be stopped?

Example:

Use integration to determine the net or total

displacement.

3 2( ) 10 24 [0,6]v t t t t on t

Example:

Use integration to find the total distance traveled.

Draw the path of the particle on a number line. Then

determine the total distance traveled.

3 2( ) 10 24 [0,6]v t t t t on t

Generalization

2

1

'

2 1( ) ( ) ( )

t

t

s t dt s t s t

Integral of the

rate of change

Net, or total, change over [t1,t2]

Accumulation Example

Rate of Change

Function

Real Life Meaning

Interpret

N’(t)

Industry Rate at which pollutants enter

a lake, measured in pounds per month

Total number of pounds of pollutants that

enter the lake over a period of “x” months

rate of change function

b

a

dt

0

N'(t)

x

dt

The rate at which pollutants enter a lake from a factory is

where N is the total number of pounds of

pollutants in the lake at time t. How many pounds of pollutants

enter the lake in 16 months? What is the average rate pollutants

enter the lake over the 16 month time period? Include units.

3/2'( ) 280N t t

Example Problem

Starting at 7:00 water leaks from a tank at a rate of (2+0.25t) gal/hr. How much water leaks out between 9:00 and 11:00?

Example Problem #2—Using data table and

trapezoidal approximation.

The number of cars passing an observation point is called the rate of traffic flow in cars/hour. The table below represents the traffic flow for a road. Use the data to estimate the number of cars using the road from 7:00-9:00.

Time 7:00 7:15 7:30 7:45 8:00 8:15 8:30 8:45 9:00

Rate r(t) 1,044 1,297 1,478 1,844 1,451 1,378 1,155 802 542

Example Problem #2—Using data table and

trapezoidal approximation.

Time 7:00 7:15 7:30 7:45 8:00 8:15 8:30 8:45 9:00

Rate r(t) 1,044 1,297 1,478 1,844 1,451 1,378 1,155 802 542

Solution:

Practice (Recommend Doing at Some Point)

Packet p. 2

Moving right=positive displacement

Moving left=negative displacement

BE SURE to TRY this PAGE!!!

Answers: Packet p. 2

1a. Stopped at t = 0, π/3 seconds

1b. Moving right (0, π/3)

1c. Moving left (π/3, π/2]

2. Distance moved right

3. Distance move left

4. Displacement from origin = 2 m

5. Total distance = 6 m

Practice – packet p.3

1a. Stopped at t = π/2, 3π/2 seconds

1b. Displacement = 0 m

1c. Total distance = 20 m

2a. Stopped at t = 0, π/2, 3π/2, 2π seconds

2b. Displacement = 0 m

2c. Total distance = 20/3 m

3a. Stopped at t = 0, 2, π seconds

3b. Displacement = -1.449 m

3c. Total distance = 1.913 m

AREA BETWEEN CURVES

Applications of Integrals

Recall:

S f xa

b

dx

Integration is used to find the area between

the curve and the x-axis.

Today. . . finding the area between two curves

dx

=

b

top bottoma

b

a

A y y

f x g x dx

A

Given for then the area of region A is: ( ) ( ) 0f x g x [ , ]x a b

NOT really any different . . . .

Area between curve and x-axis can be

thought of as area between curve f(x) and

curve g(x)=0.

=

= ( ) 0 ( )

b

top bottoma

b

a

b b

a a

A y y dx

f x g x dx

f x dx f x dx

22topy x

bottomy x

Find the area between these two equations. First

consider a very thin vertical strip.

The length of the strip is:

top bottomy y or 22 x x

Since the width of the strip is a very

small change in x, we could call it dx.

22topy x

bottomy x

top bottomy y

dx

Since the strip is a long thin rectangle, the

area of the strip is:

2length width 2 x x dxRepeat with more thin strips then add them together:

22

12 x x dx

9

2Note: The boundaries of integration are the x-coordinates of the

intersection points of the equations.

Example Problem #1: Calculate the area of the region between and over the interval [1,3]

Step #1—Determine if f(x)>g(x) or g(x)>f(x).

For this problem we will use the calculator!!

2( ) 4 10f x x x 2( ) 4g x x x

Example Problem #2: Calculate the area of the region between and

Step #1—Determine whether f(x)>g(x) or g(x)>f(x).

For this problem we will draw the sketch by hand!!!

( ) 4g x2( )f x x

Hint: You will need to find

where the two functions

intersect to find the

integration boundaries!!

Ex: Calculate the area between and on [-2,5]

NO CALCULATOR! Trick, tricky!

A. Graph f(x)—find x and y-intercepts, minimum value (there is one because

the function is an upwards facing parabola!)

B. Graph g(x)—find the x and y-intercepts

C. Find the intersection of the two curves by setting them equal to each other.

2( ) 5 7f x x x

( ) 12g x x

2( ) 5 7f x x xEx: Calculate the area between and

on [-2,5] A. Graph f(x)—x-intercepts: y-intercept: (0,-7)

minimum: (2.5, -13.25)

B. Graph g(x)—x-intercept: (12,0) y-intercept: (0,-12)

C. Find intersection points : (5,-7) and (1, -11)

SKETCH:

( ) 12g x x

(-1.14,0) and (6.14, 0),

dx

x value

x valuefunction in x dx

a b

a

b

dy

y value

y valuefunction in y dy

Why should x always have all the fun?

Let’s integrate with respect to y

Find the area bounded by , the x-axis, and by the line

y x

2y x

y x

2y x

y x

2y x

If we try vertical strips, we have to

integrate in two parts:

dx

dx2 4

0 2 2x dx x x dx

lefty x

2righty x

y x

2y x

dx

dx

We can find the same area using a

horizontal strip.

dy

lefty x

2righty x

dySince the width of the strip is dy, we

find the length of the strip by solving

for x in terms of y.

y x

2y x

2y x

2y x

22

02 y y dy

length of strip

width of strip

10

3 NOTE: The boundaries of integration are

the y-coordinates of the intersection

points of the equations.

( )

ending y

starting y

right left dy

3y = x

y = x + 2

3

2

x = y

x = y - 2

Sometimes dy Integration Works Best

Both equations need to be in terms of y. This means solve

both equations for x.

1.7930037

1

23 ( 2)y y dy

4.215

Using the Calculator as a Tool

Watch the demo and try on your calculator.

Summary

( )

ending x

starting x

top bottom dx ( )

ending y

starting y

right left dy

Use the way that gives you the easiest integral to evaluate

Both equations need to be in terms of y. This

means solve both equations for x.

Both equations need to be in terms of x. This

means solve both equations for y.

General Strategy for Area Between Curves:

1

Decide on vertical or horizontal strips. (Pick whichever is easier to write

formulas for the length of the strip, and/or whichever will let you

integrate fewer times.)

Sketch the curves.

2

3 Write an expression for the area of the strip.

• If the width is dx, the length must be in terms of x.

• If the width is dy, the length must be in terms of y.

4 Find the limits of integration. (If using dx, the limits are x values; if

using dy, the limits are y values.)

5 Integrate to find area.