Question

Integrate (0 to 8) root (10 - x )+ (root x + root (10 - x))

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{I}={\int }_0^8\left[\sqrt{10-\mathrm{x}}+\left\{\sqrt{\mathrm{x}}+\sqrt{10-\mathrm{x}}\right\}\right]\mathrm{dx} \\ ={\int }_0^8\left[\sqrt{10-\mathrm{x}}+\sqrt{\mathrm{x}}+\sqrt{10-\mathrm{x}}\right]\mathrm{dx} \\ ={\int }_0^8\left[2\sqrt{10-\mathrm{x}}+\sqrt{\mathrm{x}}\right]\mathrm{dx} \\ ={\int }_0^8\left[2{\left(10-\mathrm{x}\right)}^{1/2}+{\mathrm{x}}^{1/2}\right]\mathrm{dx} \\ ={\left[-\frac{4}{3}{\left(10-\mathrm{x}\right)}^{3/2}+\frac{2}{3}\left({\mathrm{x}}^{3/2}\right)\right]}_0^8 \\ =\left[\left\{-\frac{4}{3}{\left(10-8\right)}^{3/2}+\frac{2}{3}{\left(8\right)}^{3/2}\right\}-\left\{-\frac{4}{3}{\left(10-0\right)}^{3/2}+\frac{2}{3}{\left(0\right)}^{3/2}\right\}\right] \\ =\left[\left\{-\frac{4}{3}\times 2\sqrt{2}+\frac{2}{3}\times 16\sqrt{2}\right\}+\frac{4}{3}\times {10}^{3/2}\right] \\ =-\frac{8\sqrt{2}}{3}+\frac{32\sqrt{2}}{3}+\frac{4\times {10}^{3/2}}{3} \\ =\frac{24\sqrt{2}+4\times {10}^{3/2}}{3} \end{gathered}$$