Application of Integrals

$$\begin{gathered} \mathrm{The} \mathrm{point} \mathrm{of} \mathrm{intersection} \mathrm{of} \mathrm{these} \mathrm{curves} \mathrm{will} \mathrm{be} \\ \sqrt{\mathrm{x}}={\mathrm{x}}^3 \\ \Rightarrow \mathrm{x}=0,1 \end{gathered}$$

Then, the area is given by

$$\begin{gathered} A={\int }_0^1\left(\sqrt{x}-x^3\right)dx \\ ={\left[\frac{2}{3}{x}^{\frac{3}{2}}-\frac{x^4}{4}\right]}_0^1 \\ =\frac{2}{3}\left(1\right)-\frac{1}{4} \\ =\frac{5}{12} \end{gathered}$$

$$\begin{array}{l}\text{Required Area}={\int }_{-6}^4(4-\frac{x^2}{4})-\left(\frac{x-4}{2}\right)dx\\ \alpha ={\int }_{-6}^4(6-\frac{x^2}{4}-\frac{x}{2})dx\\ ={(6-\frac{x^3}{12}-\frac{x^2}{4})}^{-4}\\ \alpha =\frac{125}{3}\text{ Sq. units}\\ 6\alpha =250\end{array}$$

(B)

$$\begin{gathered} \mathrm{The} \mathrm{point} \mathrm{of} \mathrm{intersection} \mathrm{of} \mathrm{these} \mathrm{curves} \mathrm{will} \mathrm{be} \\ \sqrt{\mathrm{x}}={\mathrm{x}}^3 \\ \Rightarrow \mathrm{x}=0,1 \end{gathered}$$

Then, the area is given by

$$\begin{gathered} A={\int }_0^1\left(\sqrt{x}-x^3\right)dx \\ ={\left[\frac{2}{3}{x}^{\frac{3}{2}}-\frac{x^4}{4}\right]}_0^1 \\ =\frac{2}{3}\left(1\right)-\frac{1}{4} \\ =\frac{5}{12} \end{gathered}$$

\(\begin{aligned}
& \mathrm{e}^x=1-\mathrm{e}^x \\
& \mathrm{e}^x=\frac{1}{2} \\
& \mathrm{x}=-\ln 2 \\
& \text { required area }=\int_{-\ln 2}^0\left(e^x-\left(1-e^x\right)\right) d x=\left[2 e^x-x\right]_{-\ln 2}^0 \\
& =2-\left(2 e^{-\ln 2}+\ln 2\right)=2-1-\ln 2=1-\ln 2
\end{aligned}\)

$$\begin{gathered} \text{Given, } \\ {x}^2+4x+2\le y\le |x+2| \\ y={(x+2)}^2-2 \& y= |x+2| \end{gathered}$$

$$\begin{gathered} \text{Required area} = {\int }_{-2}^2\left|x\right|dx -{\int }_{-2}^2\left(x^2-2\right)dx \\ = 2{\int }_0^2\left(x\right)dx -2{\int }_0^2\left(x^2-2\right)dx \\ ={\left[\left(x^2\right)-2\left(\frac{{\left(x\right)}^3}{3}-2x\right)\right]}_0^2 \\ =\left[4-2\left(\frac{8}{3}-4\right)\right]=\left[4+\frac{8}{3}\right]=\frac{20}{3} \text{sq. units} \end{gathered}$$

\(\begin{aligned}
& \mathrm{e}^x=1-\mathrm{e}^x \\
& \mathrm{e}^x=\frac{1}{2} \\
& \mathrm{x}=-\ln 2 \\
& \text { required area }=\int_{-\ln 2}^0\left(e^x-\left(1-e^x\right)\right) d x=\left[2 e^x-x\right]_{-\ln 2}^0 \\
& =2-\left(2 e^{-\ln 2}+\ln 2\right)=2-1-\ln 2=1-\ln 2
\end{aligned}\)