Find the area of [tex] y = 2x^{2} [/tex] using the method of exhaustion. I know the area to be [tex] \frac{2x^{3}}{3} [/tex] from the rules. However, in Apostol's book, it shows you that the area under of [tex] y = x^{2} [/tex] is [tex] \frac{x^{3}}{3} [/tex]. So I will consider that a given. That means, intuitively, the area of [tex] y = 2x^{2} [/tex] is [tex] \frac{2x^{3}}{3} [/tex].(adsbygoogle = window.adsbygoogle || []).push({});

I drew the graph, and found that the area of the rectangle was: [tex] (\frac{b}{n})(\frac{2k^{2}b^{2}}{n^{2}}) [/tex] = [tex] \frac{b^{3}}{n^{3}} 2k^{2} [/tex]. So, [tex] S_{n} = \frac{2b^{3}}{n^{3}}(1^{2}+2^{2} + .. + n^{2}) [/tex] and [tex] s_{n} \frac{2b^{3}}{n^{3}}(1^{2}+ 2^{2} + ... + (n-1)^{2}) [/tex]

We know that [tex] 1^{2} + 2^{2} + ... + n^{2} = \frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6} [/tex].

So [tex] 1^{2} + 2^{2} + ... + (n-1)^{2}} < \frac{2n^{3}}{3}< 1^{2}+2^{2}+...+n^{2} [/tex]. Multiplying both sides by [tex] \frac{2b^{3}}{n^{3}} [/tex] we get [tex] s_{n} < \frac{4b^{3}}{3} < S_{n} [/tex]. But I know the area to be [tex] \frac{2b^{3}}{3} [/tex]. Where did I make my mistake?

Thanks

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# Homework Help: Integrals: Method of Exhaustion

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