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Worked solution

Recognise a right-endpoint sum and a cube-root limit

(i) For continuous ff, explain with a sketch why

lim⁡n→∞1n[f(1n)+f(2n)+⋯+f(nn)]=∫01f(x) dx.\lim_{n\to\infty}\frac1n\left[f\left(\frac1n\right)+f\left(\frac2n\right)+\cdots+f\left(\frac nn\right)\right]=\int_0^1f(x)\,dx.
marks

(ii) Hence evaluate lim⁡n→∞1n(13+23+⋯+n3n3)\displaystyle\lim_{n\to\infty}\frac1n\left(\frac{\sqrt[3]1+\sqrt[3]2+\cdots+\sqrt[3]n}{\sqrt[3]n}\right). marks

[N2015/I/3]

CJC · Review problems · Question 2

Answer

3/43/4.

(i) Identify width, sample point and height

Partition [0,1][0,1] into nn strips of width Δx=1/n\Delta x=1/n. In strip [(r−1)/n,r/n][(r-1)/n,r/n], the right-endpoint height is f(r/n)f(r/n) and the signed rectangle contribution is f(r/n)/nf(r/n)/n. Summing gives the displayed expression. Continuity on the closed interval ensures its Riemann sums converge to ∫01f(x) dx\int_0^1f(x)\,dx. For positive ff this is ordinary area; where f<0f<0 the contributions are signed, so continuity does not imply the integral is a nonnegative geometric area.

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(ii) Rewrite before identifying the function

Distribute the cube-root denominator: r3/n3=(r/n)1/3\sqrt[3]r/\sqrt[3]n=(r/n)^{1/3}. The limit is the right-endpoint sum for f(x)=x1/3f(x)=x^{1/3}, hence ∫01x1/3 dx=[34x4/3]01=3/4\int_0^1x^{1/3}\,dx=[\tfrac34x^{4/3}]_0^1=3/4. The source method is: rewrite to match (i), identify ff, then apply the integral result.

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Illustrative right-endpoint rectangles on [0,1]. The r-th rectangle has width 1/n, horizontal interval [(r−1)/n,r/n], and height f(r/n); the limit is the integral for any continuous f.
Illustrative right-endpoint rectangles on [0,1]. The r-th rectangle has width 1/n, horizontal interval [(r−1)/n,r/n], and height f(r/n); the limit is the integral for any continuous f.
Illustrative right-endpoint rectangles on [0,1]. The r-th rectangle has width 1/n, horizontal interval [(r−1)/n,r/n], and height f(r/n); the limit is the integral for any continuous f.

Common mistakes

For a general continuous function, rectangle contributions represent signed integral values rather than unsigned area.

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