343
Answers
501–600
Answers and Explanations
Then divide the interval [1, 3] into eight equal pieces, each with a width of 1
4
, to get
the intervals 1 5
4
,
, 5
4
6
4
,
, 6
4
7
4
,
, 7
4
2
,
, 2 9
4
,
, 9
4
10
4
,
, 10
4
11
4
,
, and 11
4
3
,
. Using
the right endpoint of each interval to calculate the height of the rectangles gives you
the values x = 5
4
, x = 6
4
, x = 7
4
, x = 2, x = 9
4
, x = 10
4
, x = 11
4
, and x = 3. 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:
527.
0.29
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 four rectangles of equal width over the interval 0 ≤ x ≤ 3. Begin by
dividing the length of the interval by 4 to find the width of each rectangle:
∆ x =
=
3 0
4
3
4
−
Then divide the interval [0, 3] into 4 equal pieces, each with a width of 3
4
, to get the
intervals 0 3
4
,
, 3
4
6
4
,
, 6
4
9
4
,
, and 9
4
3
,
. Recall that to find the midpoint of an interval, you simply add the left and the right endpoints together and then divide by 2 (that
is, you average the two values). In this case, the midpoints are x = 3
8
, x = 9
8
, x = 15
8
, and
x = 21
8
. 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:
3
4
3
8
3
4
9
8
3
4
15
8
3
4
21
8
3
4
2
3
8
3
4
2
9
f
f
f
f
( ) + ( ) + ( ) + ( )
=
( )
+
cos
c os 8 8
3
4
2
15
8
3
4
2
21
8
0 29
( )
+
( )
+
( )
≈
cos
c os
.
1
4
5
4
1
4
6
4
1
4
7
4
1
4
2 1
4
9
4
1
4
10
4
1
4
11
4
1
f
f
f
f
f
f
f
( ) + ( ) + ( ) + ( )+ ( ) + ( ) + ( ) + 4 4
3
1
4
5 4
5 4
1
1
4
6 4
6 4
1
1
4
7 4
7 4
1
f ( )
=
+
+
+
+
+
+ 1 1
4
2
2 1
1
4
9 4
9 4
1
1
4
10 4
10 4
1
1
4
11 4
11 4
+
+
+
+
+
+
+ +
+
+
≈
1
1
4
3
3 1
1 34
.
Answers
501–600
Answers and Explanations
Then divide the interval [1, 3] into eight equal pieces, each with a width of 1
4
, to get
the intervals 1 5
4
,
, 5
4
6
4
,
, 6
4
7
4
,
, 7
4
2
,
, 2 9
4
,
, 9
4
10
4
,
, 10
4
11
4
,
, and 11
4
3
,
. Using
the right endpoint of each interval to calculate the height of the rectangles gives you
the values x = 5
4
, x = 6
4
, x = 7
4
, x = 2, x = 9
4
, x = 10
4
, x = 11
4
, and x = 3. 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:
527.
0.29
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 four rectangles of equal width over the interval 0 ≤ x ≤ 3. Begin by
dividing the length of the interval by 4 to find the width of each rectangle:
∆ x =
=
3 0
4
3
4
−
Then divide the interval [0, 3] into 4 equal pieces, each with a width of 3
4
, to get the
intervals 0 3
4
,
, 3
4
6
4
,
, 6
4
9
4
,
, and 9
4
3
,
. Recall that to find the midpoint of an interval, you simply add the left and the right endpoints together and then divide by 2 (that
is, you average the two values). In this case, the midpoints are x = 3
8
, x = 9
8
, x = 15
8
, and
x = 21
8
. 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:
3
4
3
8
3
4
9
8
3
4
15
8
3
4
21
8
3
4
2
3
8
3
4
2
9
f
f
f
f
( ) + ( ) + ( ) + ( )
=
( )
+
cos
c os 8 8
3
4
2
15
8
3
4
2
21
8
0 29
( )
+
( )
+
( )
≈
cos
c os
.
1
4
5
4
1
4
6
4
1
4
7
4
1
4
2 1
4
9
4
1
4
10
4
1
4
11
4
1
f
f
f
f
f
f
f
( ) + ( ) + ( ) + ( )+ ( ) + ( ) + ( ) + 4 4
3
1
4
5 4
5 4
1
1
4
6 4
6 4
1
1
4
7 4
7 4
1
f ( )
=
+
+
+
+
+
+ 1 1
4
2
2 1
1
4
9 4
9 4
1
1
4
10 4
10 4
1
1
4
11 4
11 4
+
+
+
+
+
+
+ +
+
+
≈
1
1
4
3
3 1
1 34
.