Using the substitution $u = \sqrt{2x - 1}$, show that \(\displaystyle\int \frac{x}{\sqrt{2x - 1}}

Question

Using the substitution $u = \sqrt{2x - 1}$, show that $\displaystyle\int \frac{x}{\sqrt{2x - 1}} \, \mathrm{d}x = \displaystyle\int \frac{u^2 + 1}{2} \, \mathrm{d}u$.[2] The diagram below shows the graph of $y = 2 - \frac{x}{\sqrt{2x - 1}}$. Indicate on the same diagram in the Printed Answer Booklet, the equation of the asymptote of the curve and the coordinates of its turning point and points of intersection with the $x$-axis.[2] The region $R$ is bounded by the curve $y = 2 - \frac{x}{\sqrt{2x - 1}}$ and the lines $x = 1$, $x = 3$ and $y = \frac{1}{2}$. Find the exact area of $R$.[5] (c) The region $S$ is bounded by the curves $y = 2 - \frac{x}{\sqrt{2x - 1}}$, $x = 2(y - 1)^2 + 1$, the lines $x = 1$, $x = 4$ and the $x$-axis. Express $x = 2(y - 1)^2 + 1$ in the form $y = \mathrm{f}(x)$.[1] On the same diagram as in part (b)(i), sketch the graph of $x = 2(y - 1)^2 + 1$.[2] Find the volume of the solid generated when $S$ is rotated through $2\pi$ radians about the $x$-axis. Give your answer correct to $3$ decimal places.[3]

Accepted Answer

(b)(ii)$\frac{11-4\sqrt 5}{3}$ (c)(i)$y=1 \pm \sqrt {\frac{x-1}{2}}$ (c)(iii)4.518 units$^3$

2025 RI P1 Q10

2025 RI P1 Q10