Area under graph when log-log scale gives straight line

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SUMMARY

The discussion focuses on calculating the area under a logarithmic graph representing power density against frequency, specifically between 110Hz and 200Hz. The slope of the line is determined to be 0.0001/3, leading to the equation log(y) = (0.0001/3)(x - 110) + 0.001. The exponential form of the function is derived as y = (e^0.001)e^(0.0001(x - 110)/3). The method of integrating this function to find the area under the curve is confirmed as valid.

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robsmith82
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Hi,

I have a graph with a frequency on the x-axis and power density on the y axis. Both scales are logarithmic, and the graph shows a straight line between the points (110Hz, 0.001V^2/Hz) and (200Hz, 0.004V^2/Hz). I need to work out the area under this portion of the graph as it relates to total power.

So, my understanding is I need to find the function of the line, then set up and evaluate a definite intergral between 110Hz and 200Hz. My real question is, how do I work out the function of the line, and is this a correct method?

Thanks

Rob
 
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That straight line has slope .0001/3 and passes through 0(110Hz, 0.001) so has equation log(y)= (0.0001/3)(x- 110)+ 0.001. Taking exponentials of both sides,
y= (e^{0.001})e^{\frac{0.0001(x-110)}{3}}

That should not be difficult to integrate.
 
ok, i can integrate that, my real question is about how you found the function of y... How did you find that slope?
 

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