Mathematics - Area Under Curves Question with Solution | TestHub

$f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x+0)}{h}\Rightarrow {\lim }_{h\to 0}\frac{f(x)+f(h)+h\sqrt{f(x)}-f(x)-f(0)-0\sqrt{f(x)}}{h}$

$\Rightarrow {\lim }_{\mathrm{h}\to 0}\left(\frac{\mathrm{f}(\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}-0}\right)+\sqrt{\mathrm{f}(\mathrm{x})}$

$f^{\prime }(x)=\sqrt{f(x)}$

$\int \frac{{\mathrm{f}}^{\prime }(\mathrm{x})}{\sqrt{\mathrm{f}(\mathrm{x})}}\mathrm{dx}=\int \mathrm{dx}$

$2\sqrt{f(x)}=x+c$

$f(x)=\frac{x^2}{4}$ when $\alpha =0$.

The area is minimum, so the

required minimum area is:

$$=2{\int }_0^{9}2\sqrt{y}dy$$

$$\Rightarrow 4{\left(\frac{{\mathrm{y}}^{3/2}}{3/2}\right)}_0^9=72sq.unit.$$