2 unit integration

[88siege]

Need help with part (iii) please. I'm not sure where to start?

[RuiAce]

[88siege]

I've gotten this so far. I keep getting 99/100 on the RHS and not 100/101  ? ://

[RuiAce]

My bad, you're meant to go to 101 not 100.

Make sure the LHS is still correct though. Right now your LHS actually only goes up to \( \frac1{99^2} \), not \( \frac{1}{100^2} \). That is to say,
\[ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots + \frac{1}{99^2} > \frac{99}{100} \]
After fixing my mistake, it should correctly work out.
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Also it's not really Simpson's rule here, because we're estimating using rectangles and not parabolas. The reasoning for part ii) is simply because the rectangles, as we have drawn them, overestimate the area.

[88siege]

Oh ok, would this be correct??? I don't get why I was going up by 1/99^2 and not 1/100^2 ??

[RuiAce]

Think about how you did the earlier one then. You had \( \int_1^5 \frac{1}{x^2}\,dx \), but why was it that you only went up to \( \frac1{4^2} \) and not \( \frac1{5^2} \)?

It's the exact same reasoning here.

Anyway, it's correct (apart from the Simpson's rule statement which you probably missed - see my earlier edit)

[88siege]

Ohhhhh I get it never mind, I didn't see the edit. Thanks for the help!!