Understanding the Point-to-Point Equation in a Book

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I'm just having trouble seeing how the book is getting from point-to-point. I understand the area=height X Width part of the equation, but I don't see how in the seconds step (2i/5)(2/5) turns into [-(2i/5)^2+5](2/5). Smae goes for the second step in the second equation circled in red.
 
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height is given by f(x) = f(2i/5) so you plug in x=2i/5:
f\left(\frac{2i}{5}\right)=-\left(\frac{2i}{5}\right)^2+5

same goes for the second equation

you have an expression for x as a function of i that you plug into find f as a function of i.
 
Ahh, now I see. I totally forgot about the f(x)=-x^2 equation. Thanks!
 
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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