Integral as Accumulation

What information is given to us by the area underneath a curve?

Velocity and Distance

𝑣 (𝑡 )=3𝑚/ 𝑠

𝑠 (1 )=¿

𝑠 (2 )=¿

𝑠 (3 )=¿

𝑡

You are given a velocity function where velocity is constant . Find the distance traveled at any time, .

3

3

So we can clearly see that the distance traveled is equal to the area under the graph of this constant velocity function.

Since we know: Distance = Velocity Time

Velocity and Distance

What if velocity is not constant?

Will the area underneath the graph still represent the distance?

To answer this, again think about the area as a Riemann Sum.

To think about this, we can approximate the area over any interval by using a Riemann Sum and initially allowing

+

The velocity of a function is given as . What is the distance traveled from to ?

𝒕

𝒗

To make this more exact, we can increase the number of rectangles and divide the time into smaller and smaller pieces.

∆ 𝑡=0.01∆ 𝑡=0.1∆ 𝑡=0.25∆ 𝑡=0.5

𝑑=𝑣 ∙ ∆𝑡

∆ 𝑡

𝑣

The area of each rectangle would represent the distance traveled over that time interval

(We will use a midpoint riemann sum where the height will represent an approximation of the average velocity over this interval).

𝑑

if we traveled at a constant velocity represented by the height.

𝑡

𝑣

Distance as Integral of VelocityWhen given the graph of velocity and asked to find the distance, we can divide up the area into rectangles where the area of each will represent the distance traveled. If we allow , we will get closer and closer to the actual distance.

We already know the area under a graph can be found by integrating the function and therefore….

¿∫𝑎

𝑏

𝑣 (𝑡 ) 𝑑𝑡𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒[𝑎 ,𝑏 ]

Average Velocity and DistanceUse the Velocity function, to find the average velocity from .

13∫0

3

(3 𝑡2−2 )𝑑𝑡¿ [𝑡3−2 𝑡 ]03

3

¿27−63

=7

-We know the average velocity from is equal to 7.

-Therefore if we travel for 3 hours.

-We know the distance traveled will be 21 (the same as the area under the graph).

Distance vs. PositionI used distance in order to simplify things. We can really only use distance in the previous slides when the area is always above the

Consider the distance traveled in this function.

𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒=∫0

4

𝑣 (𝑡 )𝑑𝑡

Then the distance traveled (assuming ) would be 0.

𝑣

Since velocity is not always 0, this makes no sense.

If…𝐴

𝐵

The correct terminology is…

h𝐶 𝑎𝑛𝑔𝑒 𝑖𝑛𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛=∫0

4

𝑣 (𝑡 )𝑑𝑡

So the change in Position (assuming is 0.

𝑣

The distance traveled ¿∫0

2

𝑣 (𝑡 ) 𝑑𝑡+|∫2

4

𝑣 (𝑡 )𝑑𝑡|

The particle moved to the right units and then moved to the left B units. Since , the particle is where it started and the change in position is 0.

𝐴𝐵

𝐴+¿𝐵∨¿

h𝐶 𝑎𝑛𝑔𝑒 𝑖𝑛𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 [𝑎 ,𝑏 ]=∫𝑎

𝑏

𝑣 (𝑡 )𝑑𝑡 *Position is also sometimes referred to as displacement.

∫2

4

|𝑣 (𝑡 )|𝑑𝑡

Integration of any rate of change can be used to find the Net Change.

• If you are given a graph or function that represents the rate of flow of water (liters/hour), the integral will give you the total amount or net change of water.

• If you are given a rate of growth/decline of a population (people/year), then the integral will give you the net change in population.

The area of the rectangle…..

Velocity and Position

Velocity

Time

Net Change in Position

(Displacement)

Water Flow

Time Time

Population Growth

WaterFlow Net water flow

Population Growth

Net Change in Population

Water Flow Water flows out of a cylindrical pipe at a rate given by liters/second, After 20 seconds, how much water has left the pipe?

𝑇𝑜𝑡𝑎𝑙𝑊𝑎𝑡𝑒𝑟 (𝑙𝑖𝑡𝑒𝑟𝑠)=∫0

20

√𝑡+1𝑑𝑡

Homework

Net Change/Accumulation Worksheet

(1, 3, 12-16, 19, 21, 27)

Initial Position

When finding the current position, you must know where something started.

Imagine a particle moved to the left 5 units. It’s final position will depend upon its initial position.

Give that g, find the following:

Acceleration and Velocity

𝑎 (𝑡 )=3𝑚/𝑠2

𝑡

3

The area under the graph would represent the Net Change in velocity.

If a graph of acceleration is given, what would the integral represent?

If the initial velocity, ,Calculate the following velocities…

Think about the Rectangle