To find the area inside the curve.

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Hi all,

I want to find the area inside the curve present in my pdf attachment. (2nd graph).

I got this graph using the data points present in the excel worksheet.

Position column will be always same, but the force column varies.

I need a program in excel to do this for all my other force values. Can anyone help me in this..

Thnx in advance
Rsh
 

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You say "area inside the curve" but position is given only by a single number. In order to have an "area" you have to have a two dimensional situation with position given by two numbers. Also you have a column labeled "force". What does that have to do with area?
 
rsh said:
Hi all,

I want to find the area inside the curve present in my pdf attachment. (2nd graph).

I got this graph using the data points present in the excel worksheet.

Position column will be always same, but the force column varies.

I need a program in excel to do this for all my other force values. Can anyone help me in this..

Thnx in advance
Rsh

In it simplest form the area would simply be the sum of all forces (if the position increments are the same each time).
Otherwise you need to calculate the difference of each positional increment, multiply it by the corresponding force, and sum all results (integration by rectangular interpolation).
If you need more accurate results, you'd need to apply the trapezoidal rule.
Or an even more advanced method of numerical integration, which can be found with google.
 
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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