Evaluate | dx approximately, using (a) the trapezodial rule, and (b) Simpson's rule, where the interval '1+x² [0, 1] is divided into n = 4 equal parts.
Evaluate | dx approximately, using (a) the trapezodial rule, and (b) Simpson's rule, where the interval '1+x² [0, 1] is divided into n = 4 equal parts.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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3.24) my professor says I have to explain the steps in the solved problems in the picture. Not just copy eveything down from the text.
![Numerical methods for evaluating definite integrals
1
dx
5.24.
Evaluate
z approximately, using (a) the trapezodial rule, and (b) Simpson's rule, where the interval
.2
1+ x
[0, 1] is divided into n = 4 equal parts.
Let f(x) = 1/(1 + x²). Using the notation on Page 104, we find Ar = (b – a)/n = (1 – 0)/4 = 0.25. Then,
keeping four decimal places, we have yo = f(0) = 1.0000, y = f(0.25) = 0.9412, y2 = f(0.50) = 0.8000, y3 =
f(0.75) = 0.6400, and y, = f(1) = 0.50000.
%3D
%3D
(a) The trapezoidal rule gives
Ar
{Yo +2y, + 2y, + 2y; + y4} =
2
0.25
{1.0000 + 2(0.9412)+ 2(0.8000) + 2(0.6400) + 0.500}
= 0.7828.
(b) Simpson's rule gives
Ar
· {Yy, + 4y, + 2y, + 4y, + y,} =
3
0.25
{1.0000 + 4(0.9412) + 2(0.8000) + 4(0.6400) + 0.500}
2
= 0.7854.
The true value is t/4 - 0.7854.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad530511-9758-42d9-9994-894bc4eea3f3%2Fff975bc6-67a1-4be2-bc1b-bcb66a3efee7%2F3rr74q_processed.png&w=3840&q=75)
Transcribed Image Text:Numerical methods for evaluating definite integrals
1
dx
5.24.
Evaluate
z approximately, using (a) the trapezodial rule, and (b) Simpson's rule, where the interval
.2
1+ x
[0, 1] is divided into n = 4 equal parts.
Let f(x) = 1/(1 + x²). Using the notation on Page 104, we find Ar = (b – a)/n = (1 – 0)/4 = 0.25. Then,
keeping four decimal places, we have yo = f(0) = 1.0000, y = f(0.25) = 0.9412, y2 = f(0.50) = 0.8000, y3 =
f(0.75) = 0.6400, and y, = f(1) = 0.50000.
%3D
%3D
(a) The trapezoidal rule gives
Ar
{Yo +2y, + 2y, + 2y; + y4} =
2
0.25
{1.0000 + 2(0.9412)+ 2(0.8000) + 2(0.6400) + 0.500}
= 0.7828.
(b) Simpson's rule gives
Ar
· {Yy, + 4y, + 2y, + 4y, + y,} =
3
0.25
{1.0000 + 4(0.9412) + 2(0.8000) + 4(0.6400) + 0.500}
2
= 0.7854.
The true value is t/4 - 0.7854.
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