341
Answers
501–600
Answers and Explanations
You want to use four rectangles of equal width to estimate the area under f
(x) = 1 + 2x
over the interval 0 ≤ x ≤ 4. Begin by dividing the length of the interval by 4 to find the
width of each rectangle:
∆ = − =
x 4 0
4
1
Then divide the interval [0, 4] into 4 equal pieces, each with a width of 1, to get the
intervals [0, 1], [1, 2], [2, 3], and [3, 4]. Using the right endpoint of each interval to
calculate the heights of the rectangles gives you the values x = 1, x = 2, x = 3, and x = 4.
The height of each rectangle is f
(x). To approximate the area under the curve, multiply
the width of each rectangle by f
(x) and add the areas:
1 1 1 2 1 3 1 4
1 2 1
1 2 2
1 2 3
1 2 4
f
f
f
f
( )
( )
( )
( )
( )
( )
( )
( )
+
+
+
= +
+ +
+ +
+ +
[
] [
] [
] [
] ]
= 24
524.
–8.89
In this example, the given function has values that are both positive and negative on
the given interval, so don’t make the mistake of thinking the Riemann sum approximates the area between the curve and the x-axis. In this case, the Riemann sum
approximates the value of
the area of the region above the -axis
that s bounded abov
x
’
e e by the function
the area of the region below the
−
x x -axis
that s bounded below by the function
’
You want to use five rectangles of equal width over the interval 2 ≤ x ≤ 6. Begin by
dividing the length of the interval by 5 to find the width of each rectangle:
∆ = − =
x 6 2
5
4
5
Then divide the interval [2, 6] into 5 equal pieces, each with a width of 4
5
, to get the
intervals 2 14
5
,
, 14
5
18
5
,
, 18
5
22
5
,
, 22
5
26
5
,
, and
26
5
6
,
. Using the right endpoint of
each interval to calculate the heights of the rectangles gives you the values x = 14
5
,
x = 18
5
, x = 22
5
, x = 26
5
, and x = 6. The height of each rectangle is f
(x). (Note that a
“negative height” indicates that the rectangle is below the x-axis.) To find the Riemann
sum, multiply the width of each rectangle by f
(x) and add the areas:
4
5
14
5
4
5
18
5
4
5
22
5
4
5
26
5
4
5
6
4
5
14
5
14
5
f
f
f
f
f
( ) ( ) ( ) ( ) ( )
+
+
+
+
=
sin
+
+
+
+
4
5
18
5
18
5
4
5
22
5
22
5
4
5
26
5
26
5
sin
s in
sin
4 4
5
6
6
8 89
sin
.
[
]
≈ −
Answers
501–600
Answers and Explanations
You want to use four rectangles of equal width to estimate the area under f
(x) = 1 + 2x
over the interval 0 ≤ x ≤ 4. Begin by dividing the length of the interval by 4 to find the
width of each rectangle:
∆ = − =
x 4 0
4
1
Then divide the interval [0, 4] into 4 equal pieces, each with a width of 1, to get the
intervals [0, 1], [1, 2], [2, 3], and [3, 4]. Using the right endpoint of each interval to
calculate the heights of the rectangles gives you the values x = 1, x = 2, x = 3, and x = 4.
The height of each rectangle is f
(x). To approximate the area under the curve, multiply
the width of each rectangle by f
(x) and add the areas:
1 1 1 2 1 3 1 4
1 2 1
1 2 2
1 2 3
1 2 4
f
f
f
f
( )
( )
( )
( )
( )
( )
( )
( )
+
+
+
= +
+ +
+ +
+ +
[
] [
] [
] [
] ]
= 24
524.
–8.89
In this example, the given function has values that are both positive and negative on
the given interval, so don’t make the mistake of thinking the Riemann sum approximates the area between the curve and the x-axis. In this case, the Riemann sum
approximates the value of
the area of the region above the -axis
that s bounded abov
x
’
e e by the function
the area of the region below the
−
x x -axis
that s bounded below by the function
’
You want to use five rectangles of equal width over the interval 2 ≤ x ≤ 6. Begin by
dividing the length of the interval by 5 to find the width of each rectangle:
∆ = − =
x 6 2
5
4
5
Then divide the interval [2, 6] into 5 equal pieces, each with a width of 4
5
, to get the
intervals 2 14
5
,
, 14
5
18
5
,
, 18
5
22
5
,
, 22
5
26
5
,
, and
26
5
6
,
. Using the right endpoint of
each interval to calculate the heights of the rectangles gives you the values x = 14
5
,
x = 18
5
, x = 22
5
, x = 26
5
, and x = 6. The height of each rectangle is f
(x). (Note that a
“negative height” indicates that the rectangle is below the x-axis.) To find the Riemann
sum, multiply the width of each rectangle by f
(x) and add the areas:
4
5
14
5
4
5
18
5
4
5
22
5
4
5
26
5
4
5
6
4
5
14
5
14
5
f
f
f
f
f
( ) ( ) ( ) ( ) ( )
+
+
+
+
=
sin
+
+
+
+
4
5
18
5
18
5
4
5
22
5
22
5
4
5
26
5
26
5
sin
s in
sin
4 4
5
6
6
8 89
sin
.
[
]
≈ −