Find $\displaystyle\int \frac{1}{\sqrt{3-2x}} \, \mathrm{d}x.$[2] \(\displaystyle\int \frac{x}{x-1

Question

Find $\displaystyle\int \frac{1}{\sqrt{3-2x}} \, \mathrm{d}x.$[2] $\displaystyle\int \frac{x}{x-1} \, \mathrm{d}x, \text{ where}\hspace{0.5em} x > 1.$[2] A curve $C$ has equation $y = \ln x - \frac{1}{2x}$. Sketch the graph of $C$, stating the equation of the asymptote and the coordinates of the point where $C$ crosses the $x$-axis.[2] Find the equation of the tangent to $C$ at the point where $x = 3$, giving your answer in the form $y = mx + c$ where $m$ and $c$ are exact values to be determined.[3] Find the numerical value of the area of the region bounded by $C$, the tangent in (b)(ii) and the $x$-axis.[4]

Accepted Answer

(a)(i)$-(3-2x)^{\frac{1}{2}}+c$ (a)(ii)$x+\ln (x -1)+c$ (b)(ii)$y=\frac{7}{18}x+\ln 3-\frac{4}{3}$ (b)(iii)0.273 units$^2$

2024 TJC H1 Q4

2024 TJC H1 Q4