June Revision Exercise 3 Q4

By KS Teng

(i)
A = 2\{ \frac{1}{n} (1) + \frac{1}{n}e^{\frac{1}{4n}} + \frac{1}{n}e^{\frac{2}{4n}} + \ldots + \frac{1}{n}e^{\frac{n-1}{4n}}\}

= \frac{2}{n} \{e^{\frac{0}{4n}} + e^{\frac{1}{4n}} + e^{\frac{2}{4n}} + \ldots + e^{\frac{n-1}{4n}} \}

= \frac{2}{n}\sum_{r=0}^{n-1} e^{\frac{r}{4n}}

(ii)
Area of R 2 \int_0^1 e^{\frac{y}{4}} ~dy = 8(e^{\frac{1}{4}}-1)

As n \rightarrow \infty, A \rightarrow \text{Actual Area} = 8(e^{\frac{1}{4}}-1)

Back to June Revision Exercise 3

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