Finding the Area Under a Parabola Using Archimedes' Method

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I've started with Apostol's first volume of calculus and I'm already having a problem with the first exercise in the book. Weak. It asks to use Archimedes' method to find the area under a parabola from 0<x<b for a parabola where y = a*b^2 + c. Using the summation of upper/lower rectangles of width b/n and height a(k*b/n)^2 +c. Now each rectangle has an area of a(k^2)(b/n)^3 + (bc)/n. k goes from 0 to n-1 for the lower rectangles and from 1 to n for the upper rectangles. My problem comes when summing up all of the rectangles using the arithmetic summation. I get the first portion correct, but end up with (bc)/n instead of bc as the second term in the area formula. Any way that I'm approaching this incorrectly or should approach this?
 
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The sum bc/n+bc/n+...+bc/n (n times) is bc.
 
Ah, right. Stupid of me to miss that result.
 
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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