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Given = a 2D figure R and we would like to find its area.  
Take a ruler x and measure R from one end to another, say from x =3D a to x =3D b.
a=3Dx
b=3Dx
R<= /b> <= /span>
More= on Area
Take a ruler x and measure R from one end to another, say from x =3D a to x =3D b.
Let L(x) =3D cross–sectional length at x.
------=_NextPart_01C97FE2.59D46D00 Content-Location: file:///C:/8E4AA093/2FundamentalTheoremofCalaculus-MoreAreas_files/slide0158.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Given = a 2D figure R and we would like to find its area.  
Take a ruler x and measure R from one end to another, say from x =3D a to x =3D b.
a=3Dx
b=3Dx
R<= /b> <= /span>
More= on Area
L(x= )
Take a ruler x and measure R from one end to another, say from x =3D a to x =3D b.
Let L(x) =3D cross–sectional length at x.
------=_NextPart_01C97FE2.59D46D00 Content-Location: file:///C:/8E4AA093/2FundamentalTheoremofCalaculus-MoreAreas_files/slide0156.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Given = a 2D figure R and we would like to find its area.  
Take a ruler x and measure R from one end to another, say from x =3D a to x =3D b.
a=3Dx
b=3Dx
R<= /b> <= /span>
More= on Area
L(x= )
Take a ruler x and measure R from one end to another, say from x =3D a to x =3D b.
Let L(x) =3D cross–sectional length at x.
Let the sequence {a=3Dx0, x<= /span>1<= /span>, x2, ..= xn<= /span>=3Db} be a regular partition of [a, b]
------=_NextPart_01C97FE2.59D46D00 Content-Location: file:///C:/8E4AA093/2FundamentalTheoremofCalaculus-MoreAreas_files/slide0160.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Given = a 2D figure R and we would like to find its area.  
Take a ruler x and measure R from one end to another, say from x =3D a to x =3D b.
Let L(x) =3D cross–sectional length at x.
a=3Dx0
b=3Dxn
Let the sequence {a=3Dx0, x<= /span>1<= /span>, x2, ..= xn<= /span>=3Db} be a regular partition of [a, b]
x1
x2
xi–1
xi
R<= /b> <= /span>
More= on Area
L(x= )
------=_NextPart_01C97FE2.59D46D00 Content-Location: file:///C:/8E4AA093/2FundamentalTheoremofCalaculus-MoreAreas_files/slide0159.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Given = a 2D figure R and we would like to find its area.  
Take a ruler x and measure R from one end to another, say from x =3D a to x =3D b.
Let L(x) =3D cross–sectional length at x.
a=3Dx0
b=3Dxn
Let the sequence {a=3Dx0, x<= /span>1<= /span>, x2, ..= xn<= /span>=3Db} be a regular partition of [a, b] and select an arbitrary point xi<= /span> in = each sub-interval
[x i–1, xi].
*
x1
x2
xi–1
xi
R<= /b> <= /span>
More= on Area
L(x= )
------=_NextPart_01C97FE2.59D46D00 Content-Location: file:///C:/8E4AA093/2FundamentalTheoremofCalaculus-MoreAreas_files/slide0134.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Given = a 2D figure R and we would like to find its area.  
Take a ruler x and measure R from one end to another, say from x =3D a to x =3D b.
Let L(x) =3D cross–sectional length at x.
a=3Dx0
b=3Dx
Let the sequence {a=3Dx0, x<= /span>1<= /span>, x2, ..= xn<= /span>=3Db} be a regular partition of [a, b] and select an arbitrary point xi<= /span> in = each sub-interval
[x i–1, xi].
*
x1
x2
x1
*
xi–1
xi
R<= /b> <= /span>
More= on Area
L(x= )
------=_NextPart_01C97FE2.59D46D00 Content-Location: file:///C:/8E4AA093/2FundamentalTheoremofCalaculus-MoreAreas_files/slide0164.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Given = a 2D figure R and we would like to find its area.  
Take a ruler x and measure R from one end to another, say from x =3D a to x =3D b.
Let L(x) =3D cross–sectional length at x.
a=3Dx0
b=3Dx
Let the sequence {a=3Dx0, x<= /span>1<= /span>, x2, ..= xn<= /span>=3Db} be a regular partition of [a, b] and select an arbitrary point xi<= /span> in = each sub-interval
[x i–1, xi].
*
x1
x2
x1
*
x2
*
xi–1
xi
R<= /b> <= /span>
More= on Area
L(x= )
------=_NextPart_01C97FE2.59D46D00 Content-Location: file:///C:/8E4AA093/2FundamentalTheoremofCalaculus-MoreAreas_files/slide0165.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Given = a 2D figure R and we would like to find its area.  
Take a ruler x and measure R from one end to another, say from x =3D a to x =3D b.
Let L(x) =3D cross–sectional length at x.
a=3Dx0
b=3Dx
Let the sequence {a=3Dx0, x<= /span>1<= /span>, x2, ..= xn<= /span>=3Db} be a regular partition of [a, b] and select an arbitrary point xi<= /span> in = each sub-interval
[x i–1, xi].
*
x1
x2
x1
*
x2
*
x3
*
xi–1
xi
xi
*
R<= /b> <= /span>
More= on Area
L(x= )
------=_NextPart_01C97FE2.59D46D00 Content-Location: file:///C:/8E4AA093/2FundamentalTheoremofCalaculus-MoreAreas_files/slide0161.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Given = a 2D figure R and we would like to find its area.  
Take a ruler x and measure R from one end to another, say from x =3D a to x =3D b.
Let L(x) =3D cross–sectional length at x.
a=3Dx0
b=3Dx
Let the sequence {a=3Dx0, x<= /span>1<= /span>, x2, ..= xn<= /span>=3Db} be a regular partition of [a, b] and select an arbitrary point xi<= /span> in = each sub-interval
[x i–1, xi]. Let Δx =3D the width of each subint= erval.
*
x1
x2
x1
*
x2
*
x3
*
xi–1
xi
xi
*
R<= /b> <= /span>
More= on Area
L(x= )
------=_NextPart_01C97FE2.59D46D00 Content-Location: file:///C:/8E4AA093/2FundamentalTheoremofCalaculus-MoreAreas_files/slide0163.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Given = a 2D figure R and we would like to find its area.  
Take a ruler x and measure R from one end to another, say from x =3D a to x =3D b.
Let L(x) =3D cross–sectional length at x.
a=3Dx0
b=3Dx
Let the sequence {a=3Dx0, x<= /span>1<= /span>, x2, ..= xn<= /span>=3Db} be a regular partition of [a, b] and select an arbitrary point xi<= /span> in = each sub-interval [x i–1, xi]. Let = 16;x =3D the width of each subinterval. T= he rectangle with L(x<= span style=3D'font-size:89%;position:relative;top:.37em;mso-text-raise:-25%'>i<= /span>) as the height and Δx as = the width approximate the area s= pan from xi–1 to xi.
*
x1
x2
x1
*
x2
*
x3
*
xi–1
xi
xi
*
R<= /b> <= /span>
More= on Area
L(x= )
------=_NextPart_01C97FE2.59D46D00 Content-Location: file:///C:/8E4AA093/2FundamentalTheoremofCalaculus-MoreAreas_files/slide0135.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
Given = a 2D figure R and we would like to find its area.  
Take a ruler x and measure R from one end to another, say from x =3D a to x =3D b.
Let L(x) =3D cross–sectional length at x.
a=3Dx0
b=3Dx
Let the sequence {a=3Dx0, x<= /span>1<= /span>, x2, ..= xn<= /span>=3Db} be a regular partition of [a, b] and select an arbitrary point xi<= /span> in = each sub-interval [x i–1, xi]. Let = 16;x =3D the width of each subinterval. T= he rectangle with L(x= i<= /span>) as height and Δx as width approximates = the area spanned from xi&= #8211;1 to = xi<= /span>.
*
x1
x2
xi–1
xi
x1
*
x2
*
x3
*
xi
*
L(x= i)
Δx
R<= /b> <= /span>
More= on Area
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a=3Dx0
b=3Dx
x1
x2
xi–1
xi
x1
*
x2
*
x3
*
xi
*
L(x= i)
Δx
The Riemann sum of all such rectangles approximates the area of R
R<= /b> <= /span>
More= on Area
------=_NextPart_01C97FE2.59D46D00 Content-Location: file:///C:/8E4AA093/2FundamentalTheoremofCalaculus-MoreAreas_files/slide0136.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
More= on Area
a=3Dx0
b=3Dx
x1
x2
xi–1
xi
x1
*
x2
*
x3
*
xi
*
L(x= i)
Δx
The Riemann sum of all such rectangles approximates the area of R
=
=
=
=
R<= /b> <= /span>
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More= on Area
a=3Dx0
b=3Dx
x1
x2
xi–1
xi
x1
*
x2
*
x3
*
xi
*
L(x= i)
Δx
The Riemann sum of all such rectangles approximates the area of R, i.e.
=
=
=
=
R<= /b> <= /span>
L(x1<= /span>) = 16;x+  
*
i=3D= 1
n
W= 21; L(xi)Δx  
*
L(x<= /span>2<= /span>) = 16;x+ …  
*
L(x<= /span>n<= /span>) = 16;x =3D 
*=
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More= on Area
a=3Dx0
b=3Dx
x1
x2
xi–1
xi
x1
*
x2
*
x3
*
xi
*
L(x= i)
Δx
=
=
=
=
In fa= ct the area is its limit as Δ= ;xà0= or n &agrav= e;∞.
R<= /b> <= /span>
L(x1<= /span>) = 16;x+  
*
i=3D= 1
n
W= 21; L(xi)Δx  
*
L(x<= /span>2<= /span>) = 16;x+ …  
*
L(x<= /span>n<= /span>) = 16;x =3D 
*=
The Riemann sum of all such rectangles approximates the area of R, i.e.
------=_NextPart_01C97FE2.59D46D00 Content-Location: file:///C:/8E4AA093/2FundamentalTheoremofCalaculus-MoreAreas_files/slide0167.xml Content-Transfer-Encoding: quoted-printable Content-Type: text/xml; charset="utf-8" ------=_NextPart_01C97FE2.59D46D00 Content-Location: file:///C:/8E4AA093/2FundamentalTheoremofCalaculus-MoreAreas_files/slide0169.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
More= on Area
a=3Dx0
b=3Dx
x1
x2
xi–1
xi
x1
*
x2
*
x3
*
xi
*
L(x= i)
Δx
=
=
=
=
That = is, the area of R =3D lim
i=3D= 1
n
W= 21; L(xi)Δx.  
*
&= #916;x à0
R<= /b> <= /span>
L(x1<= /span>) = 16;x+  
*
i=3D= 1
n
W= 21; L(xi)Δx  
*
L(x<= /span>2<= /span>) = 16;x+ …  
*
L(x<= /span>n<= /span>) = 16;x =3D 
*=
The Riemann sum of all such rectangles approximates the area of R. i.e.
In fa= ct the area is its limit as Δ= ;xà0= or n &agrav= e;∞.
------=_NextPart_01C97FE2.59D46D00 Content-Location: file:///C:/8E4AA093/2FundamentalTheoremofCalaculus-MoreAreas_files/slide0169.xml Content-Transfer-Encoding: quoted-printable Content-Type: text/xml; charset="utf-8" ------=_NextPart_01C97FE2.59D46D00 Content-Location: file:///C:/8E4AA093/2FundamentalTheoremofCalaculus-MoreAreas_files/slide0170.htm Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset="us-ascii" Slide 1
More= on Area
a=3Dx0
b=3Dx
x1
x2
xi–1
xi
x1
*
x2
*
x3
*
xi
*
L(x= i)
Δx
=
=
=
=
That = is, the area of R =3D lim
i=3D= 1
n
W= 21; L(xi)Δx.  
*
&= #916;x à0
By th= e FTC, we’ve the area of R = =3D= 3;
∫
x=3Da
b
L(x) dx= 3;
the cross–length function
R<= /b> <= /span>
L(x1<= /span>) = 16;x+  
*
i=3D= 1
n
W= 21; L(xi)Δx  
*
L(x<= /span>2<= /span>) = 16;x+ …  
*
L(x<= /span>n<= /span>) = 16;x =3D 
*=
The Riemann sum of all such rectangles approximates the area of R. i.e.
In fa= ct the area is its limit as Δ= ;xà0= or n &agrav= e;∞.
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Example A.
a. Find the area bounded by y =3D &= #8211;x2 + 2x and y =3D x2
We need to find the span of the area. They are the = x–coordinates of the <= /span>= intersections of the curves.
More= on Area
y =3D –x2 + 2x
y =3D x2
Set the equations equal to = find the intersection points.
x2 =3D –x2 + 2x
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Example A.
a. Find the area bounded by y =3D &= #8211;x2 + 2x and y =3D x2
We need to find the span of the area. They are the = x–coordinates of the <= /span>= intersections of the curves.
More= on Area
y =3D –x2 + 2x
y =3D x2
Set the equations equal to = find the intersection points.
x2 =3D –x2 + 2x
2x2 =3D 2x
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Example A.
a. Find the area bounded by y =3D &= #8211;x2 + 2x and y =3D x2
We need to find the span of the area. They are the = x–coordinates of the <= /span>= intersections of the curves.
More= on Area
y =3D –x2 + 2x
y =3D x2
Set the equations equal to = find the intersection points.
x2 =3D –x2 + 2x
2x2 =3D 2x
2x(x – 1) =3D 0 so x =3D 0, 1.
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Example A.
a. Find the area bounded by y =3D &= #8211;x2 + 2x and y =3D x2
We need to find the span of the area. They are the = x–coordinates of the <= /span>= intersections of the curves.
More= on Area
y =3D –x2 + 2x
y =3D x2
Set the equations equal to = find the intersection points.
x2 =3D –x2 + 2x
2x2 =3D 2x
2x(x – 1) =3D 0 so x =3D 0, 1.
Hence the area is
∫
x=3D0
1
(–x2 + 2x) – x2 dx
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Example A.
a. Find the area bounded by y =3D &= #8211;x2 + 2x and y =3D x2
We need to find the span of the area. They are the = x–coordinates of the <= /span>= intersections of the curves.
More= on Area
y =3D –x2 + 2x
y =3D x2
Set the equations equal to = find the intersection points.
x2 =3D –x2 + 2x
2x2 =3D 2x
2x(x – 1) =3D 0 so x =3D 0, 1.
Hence the area is
∫
x=3D0
1
(–x2 + 2x) – x2 dx =3D
∫
1
–2x2 + 2x dx
0
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Example A.
a. Find the area bounded by y =3D &= #8211;x2 + 2x and y =3D x2
We need to find the span of the area. They are the = x–coordinates of the <= /span>= intersections of the curves.
More= on Area
y =3D –x2 + 2x
y =3D x2
Set the equations equal to = find the intersection points.
x2 =3D –x2 + 2x
2x2 =3D 2x
2x(x – 1) =3D 0 so x =3D 0, 1.
Hence the area is
∫
x=3D0
1
(–x2 + 2x) – x2 dx =3D
∫
1
–2x2 + 2x dx
=3D
–2
3
x3 + x2
|
0
0
1
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Example A.
a. Find the area bounded by y =3D &= #8211;x2 + 2x and y =3D x2
We need to find the span of the area. They are the = x–coordinates of the <= /span>= intersections of the curves.
More= on Area
y =3D –x2 + 2x
y =3D x2