Use right and left endpoints and the given number of rectangles to find two approximations of the area of the region between the graph of the function and the x-axis over the given interval. f(x) = 2x + 2, [0, 2], 4 rectangles Step 1 To find two approximations of the area of the region between the graph of the function f(x) = 2x + 2 and the x-axis over the interval [0, 2] using 4 rectangles, first partition the interval [0, 2 ] into n = | Then the width of each rectangle is given by using the following formula. b-a Ax= Step 2 Therefore, the width of each rectangle is b-a n Ax = n 4 -0.5 2 - 0 Step 3 Consider the right endpoints approximation of the area of the region. Observe the region between the graph of the function f(x) = 2x + 2 and the x-axis over the interval [0, 2] with four circumscribed rectangles, which is shown below. 1 6 4 0.5 1.0 1.5 The right endpoints of the n intervals are Ax(i) where i = 1 to 4 Thus, the right end points of four intervals are Ax(i) = 2 2.0 Then substitute 4x ==-- and n = 4 to find the left end points of four intervals. 1 That is, the four right end points of the intervals are Therefore, the four intervals are given as follows. ·[글·ㄷ 2 2.5 1 (i), wherei 1 to 4. 1, 2 2 X and 2. (222) 4 subintervals.
Use right and left endpoints and the given number of rectangles to find two approximations of the area of the region between the graph of the function and the x-axis over the given interval. f(x) = 2x + 2, [0, 2], 4 rectangles Step 1 To find two approximations of the area of the region between the graph of the function f(x) = 2x + 2 and the x-axis over the interval [0, 2] using 4 rectangles, first partition the interval [0, 2 ] into n = | Then the width of each rectangle is given by using the following formula. b-a Ax= Step 2 Therefore, the width of each rectangle is b-a n Ax = n 4 -0.5 2 - 0 Step 3 Consider the right endpoints approximation of the area of the region. Observe the region between the graph of the function f(x) = 2x + 2 and the x-axis over the interval [0, 2] with four circumscribed rectangles, which is shown below. 1 6 4 0.5 1.0 1.5 The right endpoints of the n intervals are Ax(i) where i = 1 to 4 Thus, the right end points of four intervals are Ax(i) = 2 2.0 Then substitute 4x ==-- and n = 4 to find the left end points of four intervals. 1 That is, the four right end points of the intervals are Therefore, the four intervals are given as follows. ·[글·ㄷ 2 2.5 1 (i), wherei 1 to 4. 1, 2 2 X and 2. (222) 4 subintervals.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![Use right and left endpoints and the given number of rectangles to find two approximations of the area of the region between the graph of the function and the x-axis over the given interval.
f(x) = 2x + 2, [0, 2], 4 rectangles
Step 1
To find two approximations of the area of the region between the graph of the function f(x) = 2x + 2 and the x-axis over the interval [0, 2] using 4 rectangles, first partition the interval [0, 2 ] into n =
Then the width of each rectangle is given by using the following formula.
b - a
Step 3
Ax =
Step 2
Therefore, the width of each rectangle is
b - a
n
Ax =
n
-0.5
4
Consider the right endpoints approximation of the area of the region.
Then substitute Ax =
2
Observe the region between the graph of the function f(x) = 2x + 2 and the x-axis over the interval [0, 2] with four circumscribed rectangles, which is shown below.
Submit
- 0
6
5
3
0.5
1.0
The right endpoints of the n intervals are Ax(i) where i = 1 to 4
1.5
Thus, the right end points of four intervals are Ax(i)
1
That is, the four right end points of the intervals are
Therefore, the four intervals are given as follows.
1
2
and n = 4 to find the left end points of four intervals.
oppot come book!
2.0
]][0
2
2
2.5
1
(i), wherei = 1 to 4.
1,
2
2
X
and 2.
2
2]
4 subintervals.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff9b6a03f-2298-4441-ab19-fb50f28ec42f%2F104747fc-8d79-4720-a78e-ee28d91dda62%2Fpk8tacu_processed.png&w=3840&q=75)
Transcribed Image Text:Use right and left endpoints and the given number of rectangles to find two approximations of the area of the region between the graph of the function and the x-axis over the given interval.
f(x) = 2x + 2, [0, 2], 4 rectangles
Step 1
To find two approximations of the area of the region between the graph of the function f(x) = 2x + 2 and the x-axis over the interval [0, 2] using 4 rectangles, first partition the interval [0, 2 ] into n =
Then the width of each rectangle is given by using the following formula.
b - a
Step 3
Ax =
Step 2
Therefore, the width of each rectangle is
b - a
n
Ax =
n
-0.5
4
Consider the right endpoints approximation of the area of the region.
Then substitute Ax =
2
Observe the region between the graph of the function f(x) = 2x + 2 and the x-axis over the interval [0, 2] with four circumscribed rectangles, which is shown below.
Submit
- 0
6
5
3
0.5
1.0
The right endpoints of the n intervals are Ax(i) where i = 1 to 4
1.5
Thus, the right end points of four intervals are Ax(i)
1
That is, the four right end points of the intervals are
Therefore, the four intervals are given as follows.
1
2
and n = 4 to find the left end points of four intervals.
oppot come book!
2.0
]][0
2
2
2.5
1
(i), wherei = 1 to 4.
1,
2
2
X
and 2.
2
2]
4 subintervals.
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Follow-up Questions
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Follow-up Question
Σ[(21)](3) - Σ | 2
4
-
4
=
-Σ||
i=1
=
1
Step 10
Therefore, using the left end points, the sum of the areas of four rectangles is
¡- 1
KB)
i=1
2
- (-/-)
= 5 +
2
+
4
2
i=1
(¹=¹)
2
1
to approximate the sum of area using the left end points.
1
2
3
NH
1, and
+2
1), wherei = 1 to 4.
3
+ 1)
2
+
2
3
3
(¹)
4](https://content.bartleby.com/qna-images/question/f9b6a03f-2298-4441-ab19-fb50f28ec42f/c264455c-9669-4730-9949-f912f35b7881/ctnopv_processed.png)
Transcribed Image Text:Thus, the left end points of four intervals are Ax(i – 1) =
That is, the four left end points of the intervals are 0,
Therefore, the four intervals are given as follows.
where i = 1 to 4.
Step 9
Using the left end points the height of each rectangle is
1
The width of each rectangle is
4
0
i=1
-
4
i=1
1
1
Step 11
simplify Σ{f(/z1)](3)
Σ[(21)](3) - Σ | 2
4
-
4
=
-Σ||
i=1
=
1
Step 10
Therefore, using the left end points, the sum of the areas of four rectangles is
¡- 1
KB)
i=1
2
- (-/-)
= 5 +
2
+
4
2
i=1
(¹=¹)
2
1
to approximate the sum of area using the left end points.
1
2
3
NH
1, and
+2
1), wherei = 1 to 4.
3
+ 1)
2
+
2
3
3
(¹)
4
Solution
by Bartleby ExpertFollow-up Question
i=1
i +2 (²²)
Step 7
i=1
4
i=1
=
-0.5
to approximate the sum of the area using right end points.
=
=
Then substitute Ax =
i=1
1
6
= 5+
5
i=1
NH
S
+
i + 2
i=1
0.5
4
Thus, the approximate sum of the area using right end points is 9
Step 8
Consider the left endpoints approximation of the area of the region.
2
2
Observe the region between the graph of the function f(x) = 2x + 2 and the x-axis over the interval [0, 2
Submit Skip (you cannot come back)
1.0
+ 1)
1.5
The left endpoints of the n intervals are Ax(i − 1) where i = 1 to
+
Thus, the left end points of four intervals are Ax(i – 1)
That is, the four left end points of the intervals are 0,
Therefore, the four intervals are given as follows.
2.0
11/13
and n = 4 to find the left end points of four intervals.
2
]], [[
2
2
1
9
2.5
4
(i − 1), wherei = 1 to 4.
1, and
2
2
2
2]
with four inscribed rectangles shown in the following figure.](https://content.bartleby.com/qna-images/question/f9b6a03f-2298-4441-ab19-fb50f28ec42f/c4eeeca2-6701-47b4-9f7f-135ecff8537e/e676kge_processed.png)
Transcribed Image Text:Step 6
Simplify
(16)
4
Σ{(J) -Σ[• L®2 (2)+ E2](;)
i=1
i +2 (²²)
Step 7
i=1
4
i=1
=
-0.5
to approximate the sum of the area using right end points.
=
=
Then substitute Ax =
i=1
1
6
= 5+
5
i=1
NH
S
+
i + 2
i=1
0.5
4
Thus, the approximate sum of the area using right end points is 9
Step 8
Consider the left endpoints approximation of the area of the region.
2
2
Observe the region between the graph of the function f(x) = 2x + 2 and the x-axis over the interval [0, 2
Submit Skip (you cannot come back)
1.0
+ 1)
1.5
The left endpoints of the n intervals are Ax(i − 1) where i = 1 to
+
Thus, the left end points of four intervals are Ax(i – 1)
That is, the four left end points of the intervals are 0,
Therefore, the four intervals are given as follows.
2.0
11/13
and n = 4 to find the left end points of four intervals.
2
]], [[
2
2
1
9
2.5
4
(i − 1), wherei = 1 to 4.
1, and
2
2
2
2]
with four inscribed rectangles shown in the following figure.
Solution
by Bartleby ExpertFollow-up Question
![Then substitute Ax =
Thus, the right end points of four intervals are Ax(i)
That is, the four right end points of the intervals are
Therefore, the four intervals are given as follows.
0
where i = 1 to 4.
11/123
and n = 4 to find the left end points of four intervals.
Step 6
0
Step 4
Using the right end points, the height of each rectangle is
1
4
ΣΤΗ
i=1
2
The width of each rectangle is
i=1
120
i=1
¯],
4
Σ[()](3) - Σ
i=1
i=1
1
1
= 5 +
Step 5
Therefore, using the right end points, the sum of the areas of four rectangles is
=
- Σ Σ
+
-(-)
Submit Skip (you cannot come back)
=
2
2
1× (-;-).
1(²)
2
Simplify (¹) to approximate the sum of the area using right end points.
(²)
1
2
+
2
(i), wherei = 1 to 4.
3
2
+ 1)
2
+
and 2.
2
2]](https://content.bartleby.com/qna-images/question/f9b6a03f-2298-4441-ab19-fb50f28ec42f/4feecbfa-cf81-446a-9e90-85108afd4440/1edumla_processed.png)
Transcribed Image Text:Then substitute Ax =
Thus, the right end points of four intervals are Ax(i)
That is, the four right end points of the intervals are
Therefore, the four intervals are given as follows.
0
where i = 1 to 4.
11/123
and n = 4 to find the left end points of four intervals.
Step 6
0
Step 4
Using the right end points, the height of each rectangle is
1
4
ΣΤΗ
i=1
2
The width of each rectangle is
i=1
120
i=1
¯],
4
Σ[()](3) - Σ
i=1
i=1
1
1
= 5 +
Step 5
Therefore, using the right end points, the sum of the areas of four rectangles is
=
- Σ Σ
+
-(-)
Submit Skip (you cannot come back)
=
2
2
1× (-;-).
1(²)
2
Simplify (¹) to approximate the sum of the area using right end points.
(²)
1
2
+
2
(i), wherei = 1 to 4.
3
2
+ 1)
2
+
and 2.
2
2]
Solution
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