State Standard – 16.0a Students use definite integrals in problems involving area. Objective – To be able to use the 2 nd derivative test to find concavity

State Standard

– 16.0a Students use definite integrals in problems involving area.

Objective – To be able to use the 2nd derivative test to find concavity and points of inflection.

For Instance:

We can easily calculate the exact area of regions with straight sides.

l

w

A = lw

b

h

A = ½bhHowever, it isn’t so easy to find the area of a region with curved sides.

Inscribed rectangles are all below the curve:Lower Sum

Circumscribed rectangles are all above the curve:Upper sum

Lower Sum Upper Sum

Let’s find the area under the curve by using rectangles using right endpoints.

–3 –2 –1 1 2 43

1

2

3

4

5

6

7

10

9

8

y = x2 from [0,4]

What is the Area of each rectangle?

A = (1)(12) +(1)(22) +(1)(32) +(1)(42)

A = 1 + 4 + 9 + 16

A = 30 sq units

n = 4b a

xn

4 0

4

1

Let’s find the area under the curve by using rectangles using left endpoints.

–3 –2 –1 1 2 43

1

2

3

4

5

6

7

10

9

8

y = x2 from [0,4]


A = (1)(02) +(1)(12) +(1)(22) +(1)(32)

A = 0 + 1 + 4 + 9

A = 14 sq units

n = 4b a

xn

4 0

4

1

–3 –2 –1 1 2 43

1

2

3

4

5

6

7

10

9

8

A = 30 sq units

–3 –2 –1 1 2 43

1

2

3

4

5

6

7

10

9

8

A = 14 sq units A = (30 + 14) / 2

A = 22 sq units


–3 –2 –1 1 2 43

1

2

3

4

5

6

7

10

9

8

y = x2 from [0,4]


A = + + +

= 17.5 sq units

n = 8b a

xn

4 0

8

1

2

210

2

21 1

2 2

211

2

21 3

2 2

+ + + + 212

2

21 5

2 2

213

2

21 7

2 2

1 1 9 25 9 490 2

8 2 8 8 2 8A

Approximate the area from (a) the left side (b) the right side

1)

2)

3)

2 2 [0,5] 5y x n

3 [0,3] 3y x n

[0,1] 4y x n

State Standard

– 13.0a Students know the definition of the definite integral using by using Riemann sums.

Objective – To be able to use definite integrals to find the area under a curve.

Let’s find the area under the curve by using rectangles using right endpoints.

–3 –2 –1 1 2 43

1

2

3

4

5

6

7

10

9

8

y = x2 from [0,4]


A = (1)(42) +(1)(32) +(1)(22) +(1)(12)

A = 16 + 9 + 4 + 1

A = 30 sq units

n = 4b a

xn

4 0

4

1

We could also use a Right-hand Rectangular Approximation Method (RRAM).


–3 –2 –1 1 2 43

1

2

3

4

5

6

7

10

9

8

y = x2 from [0,4]


A = (1)(02) +(1)(12) +(1)(22) +(1)(32)

A = 0 + 1 + 4 + 9

A = 14 sq units

n = 4b a

xn

4 0

4

1

This is called the Left-hand Rectangular Approximation Method (LRAM).

–3 –2 –1 1 2 43

1

2

3

4

5

6

7

10

9

8

y = x2 from [0,4]


A =(1)(0.52) +(1)(1.52) +(1)(2.52) +(1)(3.52)

A = 0.25 + 2.25 + 6.25 + 12.25

A = 21 sq units

n = 4b a

xn

4 0

4

1

Another approach would be to use rectangles that touch at the midpoint. This is the Midpoint Rectangular Approximation Method (MRAM).

Mid-Point Rule:

2 31 2

1

...2 2

2n n

x xx xf f

M hx x

f

( h = width of subinterval )

Another approach would be to use rectangles that touch at the midpoint. This is the Midpoint Rectangular Approximation Method (MRAM).

1.031251.28125

1.78125

Approximate area:

6.625

2.53125t v

1.031250.5

1.5 1.28125

2.5 1.78125

3.5 2.53125

In this example there are four subintervals.As the number of subintervals increases, so does the accuracy.

211

8V t

1w

' .5,1.5,2.5,3.5 for midptuse t s

' 0,1, 2,3,4t s

211

8V t

Approximate area:6.65624

t v

1.007810.25

0.75 1.07031

1.25 1.19531

1.382811.75

2.25

2.75

3.25

3.75

1.63281

1.94531

2.32031

2.75781

13.31248 0.5 6.65624width of subinterval

With 8 subintervals:

.5w

.25,

.75,

1.25,

1.75,

2.25,

t

etc

' 0,.5,1, ,1.5,2,2.5,t s etc

Approximate the area from the midpoint MRAM

1)

2)

3)

2 2 [0,5] 5y x n

3 [0,3] 3y x n

[0,1] 4y x n

State Standard

– 13.0a Students know the definition of the definite integral using by using Riemann sums.

Objective – To be able to use definite integrals to find the area under a curve.

–1 1 2 43

1

2

3

4

5

6

7

10

9

8

–3 –2 –1 1 2 43

1

2

3

4

5

6

7

9

8

–3 –2 –1 1 2 43

1

2

3

4

5

6

7

10

9

8

RRAMLRAM

MRAM

Right Rectangular Approximation Method

IntegrationSymbol

lower limit of integration

upper limit of integration

integrandvariable of integration

(dummy variable)

It is called a dummy variable because the answer does not depend on the variable chosen.

Definite Integrals

b

af x dx F b F a

Definition Area Under a Curve (as a Definite Integral)

If y = f(x) is nonnegative and can be integrated over a closed interval [a,b], then the area under the curve y = f(x) from a to b is the integral of f from a to b,

b

aA f x dx

If the velocity varies:1

12

v t

Distance:21

4s t t

After 4 seconds: 116 4

4s

8s

1Area 1 3 4 8

2

The distance is still equal to the area under the curve!

Notice that the area is a trapezoid.

4

0

11

2t dt

1. 0a

af x dx

If the upper and lower limits are equal, then the integral is zero.

2. b a

a bf x dx f x dx

Reversing the limits changes the sign.

b b

a ak f x dx k f x dx 3. Constant multiples can be

moved outside.

4.b b b

a a af x g x dx f x dx g x dx

Integrals can be added and subtracted.

Properties of Definite Integrals

a b c

y f x

( )c b c

a a bf x dx f x dx f x dx 5.

Integrals can be separated

3

0

14f x dx 6

3

6f x dx Given

6

0

f x dx

14 6 20

3 6

0 3

f x dx f x dx

p. 391

# 21 – 25 all

5.3 The Fundamental Theorem of Calculus

Morro Rock, California

The First Fundamental Theorem of Calculus

If f is continuous at every point of , and if

F is any antiderivative of f on , then

,a b

b

af x dx F b F a

,a b

(Also called the Integral Evaluation Theorem)

Integrals such as are called indefinite integrals

because we can not find a definite value for the answer.

2x dx

2x dx31

3x C

When finding indefinite integrals, we always include the “plus C”.

Integrals such as are called definite integrals

because we can find a definite value for the answer.

4 2

1x dx

4 2

1x dx

43

1

1

3x C

3 31 14 1

3 3C C

64 1

3 3C C

63

3 21

The constant always cancels when finding a definite integral, so we leave it out!

b

af x dx F b F a

Area from x=0to x=1

2y x

Find the area under the curve from x=1 to x=2.

2 2

1x dx

23

1

1

3x

31 12 1

3 3

8 1

3 3

7

3

Area from x=0to x=2

Area under the curve from x=1 to x=2.

211

8y x

43

0

1

24A x x

4 2

0

11

8A x dx

0 4x

Actual area under curve:

20

3A 6.6

31(4) 4

24A

1(0) 0

24

Find the area between the

x-axis and the curve

from to .

cosy x

0x 3

2x

2

3

2

3

2

0cos x dx

3 / 2

0sin x

3sin sin 0

2

1 0 1

pos.

neg.

1

2

5.3 Homework

Pg 402

19 – 30 and 35

5 2

22 3x x dx

53 2

2

13

3x x x

3 2 3 21 15 5 3 5 2 2 3 2

3 3

125 825 15 4 6

3 3

125 810 2

3 3

11712

3

117 36

3 3

15351

3

5.3b The 2nd Fundamental Theorem of Calculus

Morro Rock, California

The 2nd Fundamental Theorem of Calculus

If f is continuous on , then the function ,a b

x

aF x f t dt

has a derivative at every point in , and ,a b

x

a

dF df t dt f x

dx dx

x

a

df t dt f x

dx

Second Fundamental Theorem:

1. Derivative of an integral.

a

xdf t dt

xf x

d

2. Derivative matches upper limit of integration.



a

xdf t dt f x

dx



3. Lower limit of integration is a constant.


x

a

df t dt f x

dx




New variable.


cos xd

t dtdx cos x 1. Derivative of an integral.



sinxd

tdx

sin sind

xdx

0

sind

xdx

cos x

The long way:Second Fundamental Theorem:

20

1

1+t

xddt

dx

2

1

1 x




2

0cos

xdt dt

dx

2 2cosd

x xdx

2cos 2x x

22 cosx x

The upper limit of integration does not match the derivative, but we could use the chain rule.

53 sin

x

dt t dt

dx The lower limit of integration is not a constant, but the upper limit is.

53 sin

xdt t dt

dx

3 sinx x

We can change the sign of the integral and reverse the limits.

5.3b Homework

Pg. 402

# 7 – 17

5.5 Integration by Substitution

The chain rule allows us to differentiate a wide variety of functions, but we are able to find antiderivatives for only a limited range of functions. We can sometimes use substitution to rewrite functions in a form that we can integrate.

Example 1:

52x dx Let 2u x

du dx5u du61

6u C

62

6

xC

The variable of integration must match the variable in the expression.

Don’t forget to substitute the value for u back into the problem!

Example 2:

21 2 x x dx One of the clues that we look for is if we can find a function and its derivative in the integrand.

The derivative of is .21 x 2 x dx

1

2 u du3

22

3u C

3

2 22

13

x C

2Let 1u x

2 du x dx

Note that this only worked because of the 2x in the original.Many integrals can not be done by substitution.

Example 3:

4 1 x dxLet 4 1u x

4 du dx1

4du dx

Solve for dx.1

21

4

u du3

22 1

3 4u C

3

21

6u C

3

21

4 16

x C

Example 4:

cos 7 5 x dx7 du dx

1

7du dx

1cos

7u du

1sin

7u C

1sin 7 5

7x C

Let 7 5u x

Example 6:

2 3sin x x dx3Let u x

23 du x dx21

3du x dx

We solve for because we can find it in the integrand.

1sin

3u du

1cos

3u C

31cos

3x C

2 x dx

5.5 Homework

Pg. 420 – 4211 – 5, 7 – 14, 19, 21, 23,

25, and 31

Documents

State Standard – 16.0a Students use definite integrals in problems involving area. Objective – To be able to use the 2 nd derivative test to find concavity