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5.1a Areas and Distances. 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 and points of inflection. Area. We can easily calculate the exact area of regions with straight sides. - PowerPoint PPT Presentation
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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]
What is the Area of each rectangle?
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
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]
What is the Area of each rectangle?
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]
What is the Area of each rectangle?
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).
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]
What is the Area of each rectangle?
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]
What is the Area of each rectangle?
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.
Second Fundamental Theorem:
1. Derivative of an integral.
a
xdf t dt f x
dx
1. Derivative of an integral.
2. Derivative matches upper limit of integration.
3. Lower limit of integration is a constant.
Second Fundamental Theorem:
x
a
df t dt f x
dx
1. Derivative of an integral.
2. Derivative matches upper limit of integration.
3. Lower limit of integration is a constant.
New variable.
Second Fundamental Theorem:
cos xd
t dtdx cos x 1. Derivative of an integral.
2. Derivative matches upper limit of integration.
3. Lower limit of integration is a constant.
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
1. Derivative of an integral.
2. Derivative matches upper limit of integration.
3. Lower limit of integration is a constant.
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