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I need to get a formula for area under $y=x^{-2}$ for $x \in (1,a)$, where $a \in (1, +\infty)$, WITHOUT using integrals. I tried following:

  1. Let $h=\frac{a}{n}$, where $n$ is natural number of sections you got by dividing abscissa from $1$ to $a$
  2. Then $S = h\cdot(1h)^{-2}+h\cdot (2h)^{-2}+...+h \cdot(nh)^{-2}= (\frac{1}{h})\cdot(1+\frac{1}{4}+\frac{1}{9}+...+\frac{1}{n^2})=???$
What am i doing wrong? Maybe, there is another way to get this formula (without integrals)?
Levon Minasian
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1 Answers1

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You have taken your lower limit wrong in your solution, If I define $$f(x) = \frac{1}{x^2}$$ $n$ be index of $n^{\text{th}}$ rectangle, and $h$ be the thickness of each rectangle, then $h = \cfrac{a-1}{n}$

Then required area would be $$A = \lim_{n \to \infty} \frac{a-1}{n} \sum^{n-1}_{k=0} f(1 + kh) = (a-1) \lim_{n \to \infty} \sum^{n-1}_{k =0} \cfrac{n}{(n + k(a-1))^2} $$

Hope you can take from here, now.

xrfxlp
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