Porisma

Worked solution

From four rectangles to an exact area

The diagram shows y=x2y=x^2 between x=0x=0 and x=1x=1, with four rectangles of equal width.

(i) Show that their total area is 143∑r=14r2\frac1{4^3}\sum_{r=1}^4r^2.

(ii) If nn rectangles are used instead, deduce an expression for their total area.

(iii) Show that lim⁡n→∞1n3∑r=1nr2=13\lim_{n\to\infty}\frac1{n^3}\sum_{r=1}^nr^2=\frac13. What does this limit represent?

(iv) Give a reason why 1n3∑r=1nr2>13\frac1{n^3}\sum_{r=1}^nr^2>\frac13. You may use ∑r=1nr2=n6(n+1)(2n+1)\sum_{r=1}^nr^2=\frac n6(n+1)(2n+1) without proof.

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Four equal-width right-endpoint rectangles above the increasing curve y=x squared on [0,1].
Four equal-width right-endpoint rectangles above the increasing curve y=x squared on [0,1].
Four equal-width right-endpoint rectangles above the increasing curve y=x squared on [0,1].

CJC · Lecture · Example 1

(i)

Each width is 1/41/4. The right-endpoint heights are (1/4)2,(2/4)2,(3/4)2,(4/4)2(1/4)^2,(2/4)^2,(3/4)^2,(4/4)^2. Thus A4=14[(14)2+(24)2+(34)2+(44)2]=143∑r=14r2=1532A_4=\frac14\left[(\frac14)^2+(\frac24)^2+(\frac34)^2+(\frac44)^2\right]=\frac1{4^3}\sum_{r=1}^4r^2=\frac{15}{32}.

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(ii)

Replace four by nn: width 1/n1/n, right endpoint r/nr/n and height (r/n)2(r/n)^2. Therefore An=∑r=1n1n(rn)2=1n3∑r=1nr2A_n=\sum_{r=1}^n\frac1n(\frac rn)^2=\frac1{n^3}\sum_{r=1}^nr^2.

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(iii)

Use the supplied identity: An=n(n+1)(2n+1)6n3=16(2+3n+1n2)→13A_n=\frac{n(n+1)(2n+1)}{6n^3}=\frac16(2+\frac3n+\frac1{n^2})\to\frac13. This is the area under y=x2y=x^2 from 00 to 11, namely ∫01x2 dx\int_0^1x^2\,dx.

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(iv)

The curve is strictly increasing, so each right-endpoint rectangle lies above the curve except at its right edge. Their total is a strict overestimate of the actual area 1/31/3 for every positive integer nn.

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Common mistakes

A rectangle needs height times width; omitting the width loses a factor of n.

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