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Question Number 108652 by mohammad17 last updated on 18/Aug/20

Commented by mohammad17 last updated on 18/Aug/20

$$sirareyoucanhelpmepleas?$$

Answered by Aziztisffola last updated on 18/Aug/20

$$\mathrm{Q}2)\frac{\mathrm{dw}}{\mathrm{dy}}=\frac{1}{1+{\mathrm{y}}^2}$$$$\frac{\mathrm{dw}}{\mathrm{ds}}=\frac{1}{1+{\mathrm{s}}^2}$$$$\mathrm{Q}3)\frac{1}{1-0}{\int }_0^1\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^1{(\mathrm{sx}-\mathrm{r})}^2\mathrm{dx}$$$$={\left[\frac{1}{3\mathrm{s}}{(\mathrm{sx}-\mathrm{r})}^3\right]}_0^1=\frac{1}{3\mathrm{s}}({(\mathrm{s}-\mathrm{r})}^3-{(-\mathrm{r})}^3)$$$$=\frac{{(\mathrm{s}-\mathrm{r})}^3+{\mathrm{r}}^3}{3\mathrm{s}}$$

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