Question

The area bounded by the curve y = x |x| and the ordinates x = −1 and x = 1 is given by
(a) 0

(b) $\frac{1}{3}$

(c) $\frac{2}{3}$

(d) $\frac{4}{3}$

Answer 1

$\left(\mathrm{c}\right)\frac{2}{3}$



The given equation of the curve is
$$\begin{gathered} y=x\left|x\right| \\ \Rightarrow y=\left\{\begin{array}{l}{x}^{2}x\ge 0\\ -x^{2}x<0\end{array}\right. \\ \mathrm{Now},\mathrm{solving}x=1\mathrm{and}y=x\left|x\right|\mathrm{we}\mathrm{get} \\ x=1\Rightarrow y=1 \\ \Rightarrow A\left(1,1\right)\mathrm{is}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{cuve}y=x\left|x\right|\mathrm{and}x=1 \\ \mathrm{Also},\mathrm{solving}x=-1\mathrm{and}y=x\left|x\right|\mathrm{we}\mathrm{get} \\ x=-1\Rightarrow y=-1 \\ \Rightarrow A\text{'}\left(-1,-1\right)\mathrm{is}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{cuve}y=x\left|x\right|\mathrm{and}x=-1 \\ \mathrm{If}P\left(x,y_1\right),x>0\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{on}y=x\left|x\right|\mathrm{then}{y}_1>0\Rightarrow \left|y_1\right|=y_1 \\ \mathrm{And}Q\left(x,y_2\right),x<0\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{on}y=x\left|x\right|\mathrm{then}{y}_2<0\Rightarrow \left|y_2\right|=-y_2 \\ \mathrm{Required}\mathrm{area}={\int }_{-1}^0\left|y_2\right|dx+{\int }_0^1\left|y_1\right|dx \\ ={\int }_{-1}^0-y_{2}dx+{\int }_0^1{y}_{1}dx \\ ={\int }_{-1}^0-\left(-x^2\right)dx+{\int }_0^1{x}^{2}dx \\ ={\int }_{-1}^0{x}^{2}dx+{\int }_0^1{x}^{2}dx \\ ={\left[\frac{x^3}{3}\right]}_{-1}^0+{\left[\frac{x^3}{3}\right]}_0^1 \\ =\left[0-{\frac{\left(-1\right)}{3}}^3\right]+\left(\frac{1^3}{3}-0\right) \\ =\frac{1}{3}+\frac{1}{3} \\ =\frac{2}{3}\mathrm{sq}\mathrm{units} \end{gathered}$$