2015 H2 Mathematics Paper 1 Question 3

Definite Integrals: Areas and Volumes

Answers

(i)

1n{f(1n)+…+f(nn)}\frac{1}{n} \Big\{ f\left( \frac{1}{n} \right) + \ldots + f\left( \frac{n}{n} \right) \Big\}is the total area of n{n} rectangles shown in the sketch.
This is an approximation of ∫01f(x)  dx,{\displaystyle \int_0^1 f(x) \; \mathrm{d}x, } which is the area under the curve.
As n→∞,{n \to \infty, } the total area of the rectangles approach the area under the curve.
Hence lim⁡n→∞1n{f(1n)+f(2n)+…+f(nn)}\displaystyle \lim_{n \to \infty} \frac{1}{n} \Bigg\{ f\left( \frac{1}{n} \right) + f\left( \frac{2}{n} \right) + \ldots + f\left( \frac{n}{n} \right) \Bigg\} is ∫01f(x)  dx.{\displaystyle \int_0^1 f(x) \; \mathrm{d}x.}

(ii)

34{\frac{3}{4}}

Full solutions

(i)

1n{f(1n)+…+f(nn)}\frac{1}{n} \Big\{ f\left( \frac{1}{n} \right) + \ldots + f\left( \frac{n}{n} \right) \Big\}is the total area of n{n} rectangles shown in the sketch.
This is an approximation of ∫01f(x)  dx,{\displaystyle \int_0^1 f(x) \; \mathrm{d}x, } which is the area under the curve.
As n→∞,{n \to \infty, } the total area of the rectangles approach the area under the curve.
Hence lim⁡n→∞1n{f(1n)+f(2n)+…+f(nn)}\displaystyle \lim_{n \to \infty} \frac{1}{n} \Bigg\{ f\left( \frac{1}{n} \right) + f\left( \frac{2}{n} \right) + \ldots + f\left( \frac{n}{n} \right) \Bigg\} is ∫01f(x)  dx.{\displaystyle \int_0^1 f(x) \; \mathrm{d}x.}

(ii)

lim⁡n→∞(13+23+…+n3n3)=∫01x3  dx=[34x43]01=34  ■\begin{align*} & \lim_{n\to \infty} \left( \frac{\sqrt[3]{1} + \sqrt[3]{2} + \ldots + \sqrt[3]{n} }{\sqrt[3]{n}} \right) \\ & = \int_0^1 \sqrt[3] x \; \mathrm{d}x \\ & = \left[ \frac{3}{4} x^{\frac{4}{3}} \right]_0^1 \\ & = \frac{3}{4} \; \blacksquare \end{align*}