The diagram shows part of the graph of $y={{x}^{2}}+2$ from $x=1$ to $x=2$. The area under the

Question

The diagram shows part of the graph of $y={{x}^{2}}+2$ from $x=1$ to $x=2$. The area under the curve in this interval may be approximated by the total area of $n$ rectangles, $A$, as shown. The width of each rectangle is $\frac{1}{n}$. Given that $\underset{r=1}{\overset{n}{\mathop \sum }}\,{{r}^{2}}=\frac{n}{6}\left( n+1 \right)\left( 2n+1 \right)$, show that $A=\frac{13}{3}+\frac{3}{2n}+\frac{1}{6{{n}^{2}}}$. Deduce the exact value of $\mathop{\displaystyle\displaystyle \displaystyle\int }_{1}^{2}\left( 2+{{x}^{2}} \right)\,\mathrm{d}x$, justifying your answer. [6] Region $R$ is bounded by the curve $y=\ln x$, $x=1$, and $y=\ln 5$. Find the area of $R$.[2] Find the exact volume of the solid generated when $R$ is rotated through $4$ right angles about the $y$-axis.[4]

Accepted Answer

(a) $A=\dfrac{13}{3}+\dfrac{3}{2n}+\dfrac{1}{6n^2}$, $\displaystyle\int_1^2(x^2+2)\,\mathrm{d}x=\dfrac{13}{3}$ (b)(i) $4-\ln5$ square units $\approx2.39$ (b)(ii) $(12-\ln5)\pi$ cubic units

2024 CJC P1 Q9

2024 CJC P1 Q9

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