math 5a: area under a curve. the problem: find the area of the region below the curve f(x) = x 2 +1...
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![Page 1: Math 5A: Area Under a Curve. The Problem: Find the area of the region below the curve f(x) = x 2 +1 over the interval [0, 2]](https://reader034.vdocument.in/reader034/viewer/2022051621/56649dea5503460f94ae5db1/html5/thumbnails/1.jpg)
Math 5A: Area Under a Curve
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The Problem: Find the area of the
region below the curve f(x) = x2+1 over the interval [0, 2]. QuickTime™ and a
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IDEAS?
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IDEA
Approximate Area by:
Area Triangle + Area Rectangle = (2)(1) + (1/2)(2)(4)
= 6
Actual Area _______ 6
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IDEA: Vertical strips Cut the region into
vertical strips. Cut the top
horizontally to make rectangles.
Approximate the area under the curve using the rectangles.
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4 rectangles
Each rectangle Width= = =1/2 Height determined by
functional value. Area of rectangles=
Actual Area ____3.75
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€
=1 12( ) + 5
412( ) + 2 1
2( ) + 134
12( ) = 3.75
€
f (0)Δx + f 12( )Δx + f 1( )Δx + f 3
2( )Δx
€
B − A
4
€
Δx
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4 rectangles-right endpoint
Each rectangle Width= = =1/2 Height determined by
functional value at rt endpt. Area of rectangles=
Actual Area ____5.75€
f 12( )Δx + f 1( )Δx + f 3
2( )Δx + f (2)Δx€
Δx
€
B − A
4
€
= 54
12( ) + 2 1
2( ) + 134
12( ) + 5 1
2( ) = 5.75
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Caution Left endpoint does not always yield an
underestimate, nor right an overestimate. Consider f(x) = sinx on [0, ]
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4 rectangles-midpoint
Each rectangle Width= = =1/2 Height determined by functional
value at midpoint. Area of rectangles=
Actual Area ____4.625€
f 14( )Δx + f 3
4( )Δx + f 54( )Δx + f 7
4( )Δx€
Δx
€
B − A
4
€
=1716
12( ) + 25
1612( ) + 41
1612( ) + 65
1612( ) = 4.625
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Improving Area Estimate
Left Endpoint - 8 Rectangles Width = = =1/4 Area of Rectangles
€
Δx
€
B − A
8
€
f 0( )Δx + f 14( )Δx + f 1
2( )Δx + f ( 34 )Δx +
f 1( )Δx + f 54( )Δx + f 3
2( )Δx + f ( 74 )Δx =
4.1875
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More Rectangles-Left Endpoint
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# rect=16 32 64
Area= 4.421875 4.542969 4.604492
![Page 12: Math 5A: Area Under a Curve. The Problem: Find the area of the region below the curve f(x) = x 2 +1 over the interval [0, 2]](https://reader034.vdocument.in/reader034/viewer/2022051621/56649dea5503460f94ae5db1/html5/thumbnails/12.jpg)
More Rectangles-Right Endpoint
# rect= 16 32 64
Area=4.921875 4.792969 4.729492
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Estimate Appears to Approach 4.6666…Regardless of Point Chosen
Left Endpoint Right Endpoint Midpoint
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Choose Random Point
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Best Estimate As the number of rectangles, n,
increases, the area in the rectangles appears to approach the area under the curve, regardless of point chosen.
DefineArea Under the Curve =
€
limn→∞
(AreaRectangles)
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Generalizing… f(x) conts. and f(x) 0 on [a,b] Partition [a,b]
a random point in
€
≥
€
a = x0 < x1 < ...< xn = b
€
x i*
€
[x i−1,x i ]
€
Δx =b− a
n
€
AREA = limn→∞
f (x i*)Δx
i=1
n
∑
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Formalize Example
If choose right endpoint
€
Δx =b− a
n=
€
i 2n( )
€
=limn→∞
i 2n( )( )
2+1[ ]
i=1
n
∑ 2
n€
f (x i*) = f i 2
n( )( ) = i 2n( )( )
2+1
€
AREA = limn→∞
f (x i*)Δx
i=1
n
∑
€
2
n
€
x i* = x i = a+ iΔx =
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Example - continued
€
L =limn→∞
i 2n( )( )
2+1[ ]
i=1
n
∑ 2
n=
€
=limn→∞
8
n3i2 +
2
ni=1
n
∑i=1
n
∑ ⎡
⎣ ⎢
⎤
⎦ ⎥=
€
limn→∞
8
n3i2 +
2
n
⎡ ⎣ ⎢
⎤ ⎦ ⎥
i=1
n
∑
€
limn→∞
8
n3
n(n +1)(2n +1)
6
⎛
⎝ ⎜
⎞
⎠ ⎟ +
2
nn
⎡
⎣ ⎢
⎤
⎦ ⎥
BIG STEP HERE !!What happened?
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Example continued
€
L =limn→∞
8
n3
n(n +1)(2n +1)
6
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥+ limn→∞
2
nn
⎡ ⎣ ⎢
⎤ ⎦ ⎥
So…AREA =
€
4 23
€
=16
6+ 2 =
14
3
Corresponds to earlier approximation of 4.6666….
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Things to Think About What if f(x) <0 on all or part of [a,b]?
Can we always find a closed form expression for ?
What do we do when we can’t?
€
f (x i*)Δx
i=1
n
∑
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Remember the big step?