Part II: The Answers
344
Answers
501–600
528.
0.32
In this example, the given function has values that are both positive and negative on
the given interval, so 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 1 ≤ x ≤ 5. Begin by
dividing the length of the interval by 5 to find the width of each rectangle:
∆ x = − =
5 1
5
4
5
Divide the interval [1, 5] into 5 equal pieces, each with a width of 4
5
, to get the intervals
1 9
5
,
, 9
5
13
5
,
, 13
5
17
5
,
, 17
5
21
5
,
, and 21
5
5
,
. 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 = 7
5
, x = 11
5
,
x =
=
15
5
3, x = 19
5
, and x = 23
5
. 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
7
5
4
5
11
5
4
5
3 4
5
19
5
4
5
23
5
4
5
7 5
7 5 1
f
f
f
f
f
( ) + ( ) + + ( ) + ( )
=
+
( )
sin
+
+
+
+
+
+
4
5
11 5
11 5 1
4
5
3
3 1
4
5
19 5
19 5 1
sin
sin
sin
+
+
≈
4
5
23 5
23 5 1
0 32
sin
.
529.
160.03
Because the given function has values that are strictly greater than or equal to zero on
the given interval, you can interpret the Riemann sum as approximating the area that’s
underneath the curve and bounded below by the x-axis.
You want to use six rectangles of equal width over the interval 1 ≤ x ≤ 3. Begin by dividing the length of the interval by 6 to find the width of each rectangle:
∆ x = − =
4 1
6
1
2
Then divide the interval [1, 4] into 6 equal pieces, each with a width of 1
2
, to get the
intervals 1 3
2
,
, 3
2
2
,
, 2 5
2
,
, 5
2
3
,
, 3 7
2
,
, and 7
2
4
,
. 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 = 5
4
, x = 7
4
,
344
Answers
501–600
528.
0.32
In this example, the given function has values that are both positive and negative on
the given interval, so 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 1 ≤ x ≤ 5. Begin by
dividing the length of the interval by 5 to find the width of each rectangle:
∆ x = − =
5 1
5
4
5
Divide the interval [1, 5] into 5 equal pieces, each with a width of 4
5
, to get the intervals
1 9
5
,
, 9
5
13
5
,
, 13
5
17
5
,
, 17
5
21
5
,
, and 21
5
5
,
. 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 = 7
5
, x = 11
5
,
x =
=
15
5
3, x = 19
5
, and x = 23
5
. 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
7
5
4
5
11
5
4
5
3 4
5
19
5
4
5
23
5
4
5
7 5
7 5 1
f
f
f
f
f
( ) + ( ) + + ( ) + ( )
=
+
( )
sin
+
+
+
+
+
+
4
5
11 5
11 5 1
4
5
3
3 1
4
5
19 5
19 5 1
sin
sin
sin
+
+
≈
4
5
23 5
23 5 1
0 32
sin
.
529.
160.03
Because the given function has values that are strictly greater than or equal to zero on
the given interval, you can interpret the Riemann sum as approximating the area that’s
underneath the curve and bounded below by the x-axis.
You want to use six rectangles of equal width over the interval 1 ≤ x ≤ 3. Begin by dividing the length of the interval by 6 to find the width of each rectangle:
∆ x = − =
4 1
6
1
2
Then divide the interval [1, 4] into 6 equal pieces, each with a width of 1
2
, to get the
intervals 1 3
2
,
, 3
2
2
,
, 2 5
2
,
, 5
2
3
,
, 3 7
2
,
, and 7
2
4
,
. 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 = 5
4
, x = 7
4
,