Find the Exact Area Under the Curve Using Integration Questions

find the area and answer questions.

Find the approximate area under the curve by dividing the intervals into n subintervals and then adding up the areas of the inscribed rectangles. The height of each rectangle may be found by evaluating the function for each value of x. Your instructor will assign you n 1and n 2.

n1=6 n2=12

y = 2 x x 2 + 1 between x = 0 and x = 6 for n 1and n 2

Find the exact area under the curve using integration y = 2 x x 2 + 1 between x = 0 and x = 6

Explain the reason for the difference in your answers.

  = 1475+
3/13
5729
753
985
~ 74.2247
4
4
4
A+B ~ 93.8572 + 74.2247 – 168.0819
A = Sº 2x V x2 + 1dx
2(x2 +1) 7
3
0
= 149.3748
Reply
Quashawn Turner
O
:
Yesterday
Luyu,
It seems we got different answers for the first 2 questions. You used 7 and 13 interval points instead of 6 and 12. Then, you need to state why we
obtain different results for part A and B.
Reply
:
O
Alfred Lee
Yesterday
Luyu,
I will accept your work for n=6 and 12. NOTE: For n=6 and n=12, you went too far. For n=6, should have stopped at f(5), and for n=12 you should have
stopped at f(11).
You need to provide a conclusion on why you believe n=6 is different from n=12.
Reply
a) y = 2x V x2 + 1 x = 0, x = 6 n1 = 6, 12 = 12
=
For ni = 6
b-a
The length of each interval is given by: A x=
6-0
= 1
6
n
We then divide the interval [0, 6) into 6 sub intervals. The length of each interval is 1.
0,1,2,3,4,5
We will find the area of each rectangle and then add them up to find the total area.
An 1.f (an)
Aj = 1.f(0) = 0
A2 = 1.f(1) = 2V2
Az = 1. f(2) = 415
A4 = 1. f(3) = 6V10
Az = 1.f (4) = 8/17
A6 = 1.f(5) = 10/26
Ar = 0+2/2+4/5+6V10+8V17 + 10/26 = 114.721
=
=
For n2 = 12
b-a
The length of each interval is given by: Ax=
=
6-0
12
=
1
2
n
We then divide the interval (0,6) into 12 sub intervals. The length of each interval is 1.
0,1,1, 3,2,1,3,2,4, 3, 5,
5
11
9
,
We will find the area of each rectangle and then add them up to find the total area.
A1 = 1.f(0) = 0
5
2
A2 = }. f(1) =
Az = }. f(1) = V2
=
We will find the area of each rectangle and then add them up to find the total area.
5
N NG
=
2
–
.
.
A1 =1.f(0) = 0
A2 = 1 . f (1)
{=
Az = 1 . f(1) = V2
Aq = 1.1 (1) = {/13
As = 1 . f (2) = 275
A6 = }.() = 1/29
A7 3 . f (3) = 3/10
As = }. () = {/53
f3
Ag = 1 . f (4) = 4v17
A10 = {f(t) = /85
1. x{
A11 = 1 . f (5) = 5/26
A12 = _ . f ( 1 ) = 5 V5
AT
=
2
.
=
2
–
.
· 131.585
b) Integration
S (2x V x2 +1) dx
V
Let u =
22 +1
du = ** (x2 +1) 7 dx
2du = x (x2 +1)
ſ 2u’du
– 1 / 3 dx
=
2
A11 = 1 . f (5) = 5/26
A12 = f(1) = 5/5
2. V5
AT 131.585
–
b) Integration
S (2x V x2 +1) dx
V
Let u =
x2 + 1
V
– du
du = x (x2 +1) 7
2du = x(x2 +1) ? da
ſ 2u’du
=
Replace u with
V
22 +1
1[(V22 +1)]
를 22 +1) “) from a = 0 to x = 6
(v@*+1)* – }(vo +1)* = 2 (87787 – 1) = 149.375
Vo?
3
3
= 2
37/37
3
=
3
For part a the difference occurs because the area approaches the exact value as you increase the value of n.
The difference between part a and b occurs because when estimating the area under the curve, some parts of the rectangles are
Reply
O
Alfred Lee
Yesterday
Good job, Quashawn. Your work is correct.
Reply
2
A11 = 1 . f (5) = 5/26
A12 = f(1) = 5/5
2. V5
AT 131.585
–
b) Integration
S (2x V x2 +1) dx
V
Let u =
x2 + 1
V
– du
du = x (x2 +1) 7
2du = x(x2 +1) ? da
ſ 2u’du
=
Replace u with
V
22 +1
1[(V22 +1)]
를 22 +1) “) from a = 0 to x = 6
(v@*+1)* – }(vo +1)* = 2 (87787 – 1) = 149.375
Vo?
3
3
= 2
37/37
3
=
3
For part a the difference occurs because the area approaches the exact value as you increase the value of n.
The difference between part a and b occurs because when estimating the area under the curve, some parts of the rectangles are
Reply
O
Alfred Lee
Yesterday
Good job, Quashawn. Your work is correct.
Reply
a) y = 2x V x2 + 1 x = 0, x = 6 n1 = 6, 12 = 12
=
For ni = 6
b-a
The length of each interval is given by: A x=
6-0
= 1
6
n
We then divide the interval [0, 6) into 6 sub intervals. The length of each interval is 1.
0,1,2,3,4,5
We will find the area of each rectangle and then add them up to find the total area.
An 1.f (an)
Aj = 1.f(0) = 0
A2 = 1.f(1) = 2V2
Az = 1. f(2) = 415
A4 = 1. f(3) = 6V10
Az = 1.f (4) = 8/17
A6 = 1.f(5) = 10/26
Ar = 0+2/2+4/5+6V10+8V17 + 10/26 = 114.721
=
=
For n2 = 12
b-a
The length of each interval is given by: Ax=
=
6-0
12
=
1
2
n
We then divide the interval (0,6) into 12 sub intervals. The length of each interval is 1.
0,1,1, 3,2,1,3,2,4, 3, 5,
5
11
9
,
We will find the area of each rectangle and then add them up to find the total area.
A1 = 1.f(0) = 0
5
2
A2 = }. f(1) =
Az = }. f(1) = V2
=
We will find the area of each rectangle and then add them up to find the total area.
5
N NG
=
2
–
.
.
A1 =1.f(0) = 0
A2 = 1 . f (1)
{=
Az = 1 . f(1) = V2
Aq = 1.1 (1) = {/13
As = 1 . f (2) = 275
A6 = }.() = 1/29
A7 3 . f (3) = 3/10
As = }. () = {/53
f3
Ag = 1 . f (4) = 4v17
A10 = {f(t) = /85
1. x{
A11 = 1 . f (5) = 5/26
A12 = _ . f ( 1 ) = 5 V5
AT
=
2
.
=
2
–
.
· 131.585
b) Integration
S (2x V x2 +1) dx
V
Let u =
22 +1
du = ** (x2 +1) 7 dx
2du = x (x2 +1)
ſ 2u’du
– 1 / 3 dx
=
2
A11 = 1 . f (5) = 5/26
A12 = f(1) = 5/5
2. V5
AT 131.585
–
b) Integration
S (2x V x2 +1) dx
V
Let u =
x2 + 1
V
– du
du = x (x2 +1) 7
2du = x(x2 +1) ? da
ſ 2u’du
=
Replace u with
V
22 +1
1[(V22 +1)]
를 22 +1) “) from a = 0 to x = 6
(v@*+1)* – }(vo +1)* = 2 (87787 – 1) = 149.375
Vo?
3
3
= 2
37/37
3
=
3
For part a the difference occurs because the area approaches the exact value as you increase the value of n.
The difference between part a and b occurs because when estimating the area under the curve, some parts of the rectangles are
Reply
O
Alfred Lee
Yesterday
Good job, Quashawn. Your work is correct.
Reply