Part II: The Answers
340
Answers
501–600
Then divide the interval [1, 4] into 7 equal pieces, each with a width of 3
7
, to get the
intervals 1 10
7
,
, 10
7
13
7
,
, 13
7
16
7
,
, 16
7
19
7
,
, 19
7
22
7
,
, 22
7
25
7
,
, and 25
7
4
,
. Using
the left endpoint of each interval to calculate the heights of the rectangles gives you
the values x = 1, x = 10
7
, x = 13
7
, x = 16
7
, x = 19
7
, x = 22
7
, and x = 25
7
. 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:
3
7
1 3
7
10
7
3
7
13
7
3
7
16
7
3
7
19
7
3
7
22
7
3
7
25
7
f
f
f
f
f
f
f
( )
( ) ( ) ( ) ( ) ( )
+
+
+
+
+
+
( ( )
+ ( )
( )
( )
=
+
+
+
+
3
7
4 1 2 1
3
7
4 10
7
2 10
7
3
7
4 13
7
2 13
7
ln
ln
ln
( )
( )
+
+
+
+
+
3
7
4 16
7
2 16
7
3
7
4 19
7
2 19
7
3
7
4 22
ln
ln
ln 7 7
2 22
7
3
7
4 25
7
2 25
7
22 66
+
+
+
≈
( )
( )
ln
.
522.
2.788 × 10
10
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 eight rectangles of equal width to estimate the area under f
(x) = e
3x
+ 4
over the interval 1 ≤ x ≤ 9. Begin by dividing the length of the interval by 8 to find the
width of each rectangle:
∆ = − =
x 9 1
8
1
Then divide the interval [1, 9] into 8 equal pieces, each with a width of 1, to get the intervals [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], and [8, 9]. Using the left endpoint of each
interval to calculate the heights of the rectangles gives you the values x = 1, x = 2, x = 3,
x = 4, x = 5, x = 6, x = 7, and x = 8. 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:
523.
24
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.
1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8
4
3 1
f
f
f
f
f
f
f
f
e
( )+ ( )+ ( )+ ( )+ ( )+ ( )+ ( )+ ( )
=
+
( )
+
+
+
+
+
+
+
+
+
+
( )
( )
( )
( )
( )
e
e
e
e
e
3 2
3 3
3 4
3 5
3 6
4
4
4
4
4
+
+
+
+
≈
×
( )
( )
e
e
3 7
3 8
10
4
4
2 788 10
.
340
Answers
501–600
Then divide the interval [1, 4] into 7 equal pieces, each with a width of 3
7
, to get the
intervals 1 10
7
,
, 10
7
13
7
,
, 13
7
16
7
,
, 16
7
19
7
,
, 19
7
22
7
,
, 22
7
25
7
,
, and 25
7
4
,
. Using
the left endpoint of each interval to calculate the heights of the rectangles gives you
the values x = 1, x = 10
7
, x = 13
7
, x = 16
7
, x = 19
7
, x = 22
7
, and x = 25
7
. 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:
3
7
1 3
7
10
7
3
7
13
7
3
7
16
7
3
7
19
7
3
7
22
7
3
7
25
7
f
f
f
f
f
f
f
( )
( ) ( ) ( ) ( ) ( )
+
+
+
+
+
+
( ( )
+ ( )
( )
( )
=
+
+
+
+
3
7
4 1 2 1
3
7
4 10
7
2 10
7
3
7
4 13
7
2 13
7
ln
ln
ln
( )
( )
+
+
+
+
+
3
7
4 16
7
2 16
7
3
7
4 19
7
2 19
7
3
7
4 22
ln
ln
ln 7 7
2 22
7
3
7
4 25
7
2 25
7
22 66
+
+
+
≈
( )
( )
ln
.
522.
2.788 × 10
10
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 eight rectangles of equal width to estimate the area under f
(x) = e
3x
+ 4
over the interval 1 ≤ x ≤ 9. Begin by dividing the length of the interval by 8 to find the
width of each rectangle:
∆ = − =
x 9 1
8
1
Then divide the interval [1, 9] into 8 equal pieces, each with a width of 1, to get the intervals [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], and [8, 9]. Using the left endpoint of each
interval to calculate the heights of the rectangles gives you the values x = 1, x = 2, x = 3,
x = 4, x = 5, x = 6, x = 7, and x = 8. 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:
523.
24
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.
1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8
4
3 1
f
f
f
f
f
f
f
f
e
( )+ ( )+ ( )+ ( )+ ( )+ ( )+ ( )+ ( )
=
+
( )
+
+
+
+
+
+
+
+
+
+
( )
( )
( )
( )
( )
e
e
e
e
e
3 2
3 3
3 4
3 5
3 6
4
4
4
4
4
+
+
+
+
≈
×
( )
( )
e
e
3 7
3 8
10
4
4
2 788 10
.