Question

The function $f\left(x\right)$ whose graph passes through the point $(0,\frac{7}{3})$ and whose derivatives is $x\sqrt{1-x^2}$ is given by.

Answer 1

$\frac{dy}{dx}=x\sqrt{1-x^2}\Rightarrow dy=x\sqrt{1-x^2}dx\int dy=\int x\sqrt{1-x^2}dxLetu=1-x^2\Rightarrow du=-2xdx\Rightarrow dx=-\frac{1}{2x}du\therefore \int x\sqrt{1-x^2}dx=\int x(-\frac{1}{2x})du\sqrt{u}=-\frac{1}{2}\int \sqrt{u}duNow,\int \sqrt{u}du=\frac{2}{3}{u}^{3/2}\therefore -\frac{1}{2}\int \sqrt{u}du=-\frac{1}{3}{u}^{3/2}=\frac{1}{3}{(1-x^2)}^{3/2}\therefore y=-\frac{1}{3}{(1-x^2)}^{3/2}+c$

Now, this curve passes through $(0,\frac{7}{3})$

$\therefore \frac{7}{3}=-\frac{1}{3}{(1-0)}^{3/2}+c\Rightarrow \frac{7}{3}=-\frac{1}{3}+c\Rightarrow c=\frac{8}{3}\therefore y=-\frac{1}{3}{(1-x^2)}^{3/2}+\frac{8}{3}$