Numerical Integration - Problem

Hurly
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Homework Statement



Hey I Need Help With Numerical Integration:

I Have Data Sets Which is Shown In The Picture Below

http://imageshack.us/photo/my-images/69/numericalintegration.png/

Could Someone Use One of The Data Sets to Show Hows it Done Then i'll Do The Rest of Them

Needed Answers:

The H Value-
The N Value-
The Total Area-

Homework Equations



Areai = h * (F(Xi)+F(Xi+1) / 2)

With Xi+1 = Xi+hN = (XFinal - XInitial) / h

Total Area = Sum of The Areai values from 0 - N-1
 
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Hey Hurly and welcome to the forums.

The data says b,c,d but doesn't define what they are. Let's assuming that they are 2nd, 3rd and 4th observations corresponding to f(x1), f(x2) and f(x3).

Then as an example the first line corresponds to f(x1) = 1, f(x2) = 4, f(x3) = -6, f(x0) = 3 and f(x4) = 4.

You will have to use some defined or calculated value for h. Using this information, can you find the approximate integral between x0 and x4 with your formula?
 
\frac{Xf-Xi}{2*n}* (f(x0)+2*f(x1)+2*f(x2)+2*f(x3)+f(x4))

\frac{4-3}{2*n}* (f(3)+2*f(1)+2*f(4)+2*f(-6)+f(4))

0.13*(f(3)+2*f(1)+2*f(4)+2*f(-6)+f(4))
 
Excellent! Now just replace those "f"s with the correct value from your data set and do the arithmetic.
 
Data Set Example

x0 = -1 ( X initial)
x1 = 0
x2 = 1
x3 = 2
x4 = 3 ( X Final)

5 data Sets so 4 Traps

N = 4

H = \frac{xFinal - xInitial}{n}

H = \frac{3 - (-1)}{4} = 1

H = 1

Area = \frac{1}{2}*H*(x0+2(x1+x2+x3)+x4)

Area = \frac{1}{2}*1*(-1+2(0+1+2)+3)

Area = 4

- Is This The Correct Way?
 
Is This Correct?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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