Find $\displaystyle\int{\frac{x}{\sqrt{2x-1}}}\,\mathrm{d}x$. [3] Using the substitution $t= an x$,

Question

Find $\displaystyle\int{\frac{x}{\sqrt{2x-1}}}\,\mathrm{d}x$. [3] Using the substitution $t=	an x$, find $\displaystyle\int{\frac{1}{4{{\cos }^{2}}x+9{{\sin }^{2}}x}\,\mathrm{d}x}$. [4] The diagram above shows part of the graph of $y={{x}^{2}}+3x$ , with rectangles approximating the area under the curve from $x=0$ to $x=1$. The area under the curve may be approximated by the total area, $A$, of $\left( n-1 ight)$ rectangles each of width $\frac{1}{n}$. Given that $\underset{r=1}{\overset{n}{\mathop \sum }}\,{{r}^{2}}=\frac{1}{6}n\left( n+1 ight)\left( 2n+1) ight)$, show that $A=\frac{\left( n-1 ight)\left( 11n-1 ight)}{6{{n}^{2}}}$. Explain briefly how the value of $\mathop{\displaystyle\int }_{0}^{1}{{x}^{2}}+3x\,\mathrm{d}x$ can be deduced from this expression, and hence find this value exactly without integration. [6]

Accepted Answer

(a) $x\sqrt{2x-1} - \frac{1}{3}\sqrt{2x-1}^{\,3} + C$; (b) $\frac{1}{6}\tan^{-1}\left(\frac{3}{2}\tan x\right) + C$; (c) $\frac{11}{6}$ units²

2019 DHS P1 Q10

2019 DHS P1 Q10

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