SubInterval
[x
i1 ,
x
i
]
Sample Point
Left
Endpoint
x
i
*
= x
i1
Height of
Rectangle i
f(x
i
*)
Length of
Rectangle i
Δx = x
i1
 x
i
Area of
Rectangle i
f(x
i
*) Δx
[ 2, 5 ]
x
1
*
=
2
f(2) = 2
2
= 4
5 – 2 = 3
4 x 3 =
12
[ 5, 8 ]
x
2
*
=
5
f(5) = 5
2
= 25
8 – 5 = 3
25 x 3 =
75
[ 8,
11 ]
x
3
*
=
8
f(8) = 8
2
= 64
11 – 8 = 3
64 x 3 =
192
[ 11, 14 ]
x
4
*
=
11
f(11) = 11
2
= 121
14 – 11 = 3
121 x 3 = 363
TOTAL AREA
=
642
Example of Riemann Sums for f(x) = x
2
on [2,14] with n = 4 subintervals
with equal length given by (ba)/n = (14 – 2)/4 = 3. Sample pts: Left Endpt
b
14
4
∫
a
f(x)dx
=
∫
2
x
2
dx
≈
Σ
i=1
f(x
i
*)Δx
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentSubInterval
[x
i1
, x
i
]
Sample Point
Right
Endpt
x
i
*
= x
i
Height of
Rectangle i
f(x
i
*)
Length of
Rectangle i
Δx = x
i1
 x
i
Area of
Rectangle i
f(x
i
*) Δx
[ 2, 5 ]
x
1
*
=
5
f(5) = 5
2
= 25
5 – 2 = 3
25 x 3 =
75
[ 5, 8 ]
x
2
*
=
8
f(8) = 8
2
= 64
8 – 5 = 3
64 x 3 =
192
[ 8, 11 ]
x
3
*
=
11
f(11) = 11
2
= 121
11 – 8 = 3
121 x 3 = 363
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '11
 Augustarainsford
 Riemann Sums, Rectangle, Riemann integral, Riemann sum, Bernhard Riemann, sample pts

Click to edit the document details