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

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$\frac{2}{3}$ The given equation of the curve is y = x \left| x \right| \Rightarrow y = \begin{cases} x^2 & x \geq 0\ - x^2 & x < 0 \end{cases} \text{ Now, solving }x = 1\text{ and }y = x\left| x \right|\text{ we get } x = 1 \Rightarrow y = 1 \Rightarrow A\left( 1, 1 \right)\text{ is point of intersection of the cuve }y = x\left| x \right|\text{ and }x = 1 \text{ Also, solving }x = - 1\text{ and }y = x\left| x \right|\text{ we get } x = - 1 \Rightarrow y = - 1 \Rightarrow A'\left( - 1, - 1 \right)\text{ is point of intersection of the cuve }y = x\left| x \right|\text{ and }x = - 1 \text{ If P}\left( x, y_1 \right) , x > 0\text{ is a point on }y = x \left| x \right|\text{ then }y_1 > 0 \Rightarrow \left| y_1 \right| = y_1 \text{ And Q}\left( x, y_2 \right) , x < 0\text{ is a point on }y = x \left| x \right|\text{ then }y_2 < 0 \Rightarrow \left| y_2 \right| = - y_2 \text{ Required 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[ x^$\frac{3}{3}$ \right]_{- 1}^0 + \left[ x^$\frac{3}{3}$ \right]_0^1 = \left[ 0 - \frac{\left( - 1 \right)}{3}^3 \right] + \left( 1^$\frac{3}{3}$ - 0 \right) = $\frac{1}{3}$ + $\frac{1}{3}$ = $\frac{2}{3}$\text{ sq units } [Solution diagram: 44135_solution_1.png]

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