Question

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

पर्याय

• 0

• \[\frac{1}{3}\]
• \[\frac{2}{3}\]
• \[\frac{4}{3}\]

Answer 1

\[\frac{2}{3}\]

The given equation of the curve is
\[y = x \left| x \right|\]
\[ \Rightarrow y = \begin{cases} x^2 & x \geq 0\\ - x^2 & x < 0 \end{cases}\]
\[\text{ Now, solving }x = 1\text{ and }y = x\left| x \right|\text{ we get }\]
\[x = 1 \Rightarrow y = 1 \]
\[ \Rightarrow A\left( 1, 1 \right)\text{ is point of intersection of the cuve }y = x\left| x \right|\text{ and }x = 1\]
\[\text{ Also, solving }x = - 1\text{ and }y = x\left| x \right|\text{ we get }\]
\[x = - 1 \Rightarrow y = - 1\]
\[ \Rightarrow A'\left( - 1, - 1 \right)\text{ is point of intersection of the cuve }y = x\left| x \right|\text{ and }x = - 1\]
\[\text{ If P}\left( x, y_1 \right) , x > 0\text{ is a point on }y = x \left| x \right|\text{ then }y_1 > 0 \Rightarrow \left| y_1 \right| = y_1 \]
\[ \text{ And Q}\left( x, y_2 \right) , x < 0\text{ is a point on }y = x \left| x \right|\text{ then }y_2 < 0 \Rightarrow \left| y_2 \right| = - y_2 \]
\[\text{ Required area }= \int_{- 1}^0 \left| y_2 \right| dx + \int_0^1 \left| y_1 \right| dx\]
\[ = \int_{- 1}^0 - y_2 dx + \int_0^1 y_1 dx\]
\[ = \int_{- 1}^0 - \left( - x^2 \right) dx + \int_0^1 x^2 dx\]
\[ = \int_{- 1}^0 x^2 dx + \int_0^1 x^2 dx\]
\[ = \left[ \frac{x^3}{3} \right]_{- 1}^0 + \left[ \frac{x^3}{3} \right]_0^1 \]
\[ = \left[ 0 - \frac{\left( - 1 \right)}{3}^3 \right] + \left( \frac{1^3}{3} - 0 \right)\]
\[ = \frac{1}{3} + \frac{1}{3}\]
\[ = \frac{2}{3}\text{ sq units }\]