Scribd
Upload
76 views13 pages

Solution: Your Input: Approximate The Integral Gles

The document provides the steps to approximate an integral using the Simpson's rule with 6 rectangles. Specifically: - The integral is evaluated from 0 to √23, and is divided into 6 subintervals of length √3. - The function is evaluated at the endpoints of each subinterval. - These function values are summed using the Simpson's rule weighting and multiplied by the interval length cubed, giving an approximation of the integral of 3.1622770.

Uploaded by

Armando Lios
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
76 views13 pages

Solution: Your Input: Approximate The Integral Gles

The document provides the steps to approximate an integral using the Simpson's rule with 6 rectangles. Specifically: - The integral is evaluated from 0 to √23, and is divided into 6 subintervals of length √3. - The function is evaluated at the endpoints of each subinterval. - These function values are summed using the Simpson's rule weighting and multiplied by the interval length cubed, giving an approximation of the integral of 3.1622770.

Uploaded by

Armando Lios
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
You are on page 1/ 13

SOLUTION

Your input: approximate the

integral ∫‾‾‾‾‾‾‾‾‾‾√∫0310x2+189x2+18 dx using n=6 rectan
gles.
The Simpson's rule states

that ∫≈−−∫abf(x)dx≈Δx3(f(x
0)+4f(x1)+2f(x2)+4f(x3)+2f(x4)+...+2f(xn−2)+4f(xn−1)+f(xn))
where −Δx=b−an.

We have that a=0, b=3, n=6.

Therefore, −Δx=3−06=12.

Divide the interval [0,3] into n=6 subintervals of length Δx=12, with the


following
endpoints: a=0,12,1,32,2
,52,3=b.
Now, we just evaluate the function at these endpoints:
f(x0)=f(a)=f(0)=1=1

‾‾‾√4f(x1
)=4f(12)=4829=4.02461561694996
‾‾‾√2f(x2)=2f(1)=4219=2.03670030886926

‾‾‾√
4f(x3)=4f(32)=123417=4.11596604342021
‾‾‾√2f(x4)=2f
(2)=2879=2.0727509006864

‾‾‾‾‾‾√
4f(x5)=4f(52)=41062699=4.16494914096215
‾‾‾√
f(x6)=f(b)=f(3)=23311=1.04446593573419
Finally, just sum up the above values and multiply
by Δx3=16: 



16(1+
4.02461561694996+2.03670030886926+...
+4.16494914096215+1.04446593573419)=3.07657465777036
Answer: 3.07657465777036.
(1+(((9*x^2)/4)^(1/3))^2)^0.5
Your input: approximate the

integral ∫‾√‾√‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√∫0233422333(x2)23+1 

dx using n=6 rectangles.
The Simpson's rule states

that ∫≈−−∫abf(x)dx≈Δx3(f(x
0)+4f(x1)+2f(x2)+4f(x3)+2f(x4)+...+2f(xn−2)+4f(xn−1)+f(xn))
where −Δx=b−an.

We have that a=0, ‾√b=23, n=6.

Therefore, ‾√−‾√Δx=23−06=33.

Divide the interval ‾√[0,23] into n=6 subintervals of length ‾√Δx=33,


with the following
endpoints: ‾√‾√‾√‾√‾√‾√a=0,33,233,3,433,533,23
=b.
Now, we just evaluate the function at these endpoints:
f(x0)=f(a)=f(0)=1=1

‾√‾‾‾‾‾‾‾√
4f(x1)=4f(33)=46234+1=5.40441569418735

‾√‾‾‾‾‾‾‾√
2f(x2)=2f(233)=21+323=3.51003351724846

‾√‾‾‾‾‾‾‾‾‾√
4f(x3)=4f(3)=41+94223=8.55256908015686

‾√‾√⋅‾‾‾‾‾‾‾‾‾‾‾‾√
2f(x4)=2f(433)=21+223⋅323=4.99659195388929
‾√‾‾‾√‾‾‾‾‾‾‾‾‾‾‾‾√
4f(x5)=4f(533)=41+54223453=11.35448472
92039

‾√‾‾‾√f(x6)=f(b)=f(23)=10=3.16227766016838
Finally, just sum up the above values and multiply
by ‾√Δx3=39: ‾√


39(1+5.40
441569418735+3.51003351724846+...
+11.3544847292039+3.16227766016838)=7.30932612155179
Answer: 7.30932612155179.

3)
Pregunta 4
Pregunta5

You might also like