Area Project 2015

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CHAPTER 5: PROJECT Name: Part I FUNDAMENTAL THEOREM OF CALCULUS Let’s start with a simple example that you can calculate easily. Suppose you ride a scooter at a constant speed of 5 miles per hour. How far do you move from time t = 1 to t = 3 hours? Now, let’s determine how to find the answer using Calculus so we can apply our method to more complicated problems. Recall that velocity is the distance traveled per time elapsed. In other words, v = s’(t). If you want to find the distance function then calculate the anti- derivative of the velocity: s ( t ) = s' ( t)dt = vdt = If we assume that the starting position is zero we have s(0) = 0 and can find C in the distance function. s(t) = Now, use this function to find the distance you traveled from the start to time t = 1 and from the start to t = 3. s(1) = s(3) = Subtract these values to find the distance traveled over the two hour interval.

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Investigate how definite integrals are related to area under a curve

Transcript of Area Project 2015

Page 1: Area Project 2015

CHAPTER 5: PROJECT Name:Part I

FUNDAMENTAL THEOREM OF CALCULUS

Let’s start with a simple example that you can calculate easily. Suppose you ride a

scooter at a constant speed of 5 miles per hour. How far do you move from time t = 1 to t = 3 hours?

Now, let’s determine how to find the answer using Calculus so we can apply our method to more complicated problems.

Recall that velocity is the distance traveled per time elapsed. In other words, v = s’(t). If you want to find the distance function then calculate the anti-derivative of the velocity:

s ( t )=∫ s '(t)dt=∫ v dt=∫❑

If we assume that the starting position is zero we have s(0) = 0 and can find C in the distance function.

s(t) =

Now, use this function to find the distance you traveled from the start to time t = 1 and from the start to t = 3.

s(1) = s(3) =

Subtract these values to find the distance traveled over the two hour interval.

Note that this difference is equivalent to calculating the following:

∫1

3

s ' ( t )dt=s (3 )−s(1)

We call this a definite integral and generalize it to apply to any anti-derivative with known endpoints.

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∫a

b

f ' ( x )dx= f (b )− f (a)

Use this Fundamental Theorem of Calculus to evaluate the following definite integrals:

1. ∫2

5

3dt

2. ∫0

6

3dx

3. ∫0

3

9dx

4. ∫0

2

3 x+1dx

5. ∫1

3

4−2x dx

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6. ∫0

3

x2dx

7. ∫0

2

4−x2dx

8. ∫0

4

√x dx

9. ∫2

3

1+4 t+3t 2dt

10. ∫−1

1

x4+3 x3+1dx

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Check your answers using the graphing calculator. To evaluate a definite integral you use the command: fnInt (Math menu)

Now, draw graphs to correspond to each derivative function in problems 1-5.Shade the area underneath the function ONLY over the interval x = a to x = b.

1. 2.

3. 4.

5.

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What is the shaded area on each graph? Write your answer beneath your sketch.

Why does finding the anti-derivative over specific intervals relate at all to area?!