Practicing Success
The area of the region $\left\{(x, y): y \geq x^2\right.$ and $\left.y \leq|x|\right\}$ is |
2 1 $\frac{1}{2}$ $\frac{1}{3}$ |
$\frac{1}{3}$ |
region → $y \geq x^2$ and $y \leq x$ so plotting regions on cartesion plane so intersecting point of $y=x^2$ and $y=|x|$ $\Rightarrow x^2=|x|$ So x = -1, 0, 1 y= 1, 0, 1 Corresponding as for parabola in first quadrant $y = x^2 ⇒ x = \sqrt{2}$ for line $y = x ⇒ x = y$ → area of region × 2 = Total area area of 1 region = $\int\limits_0^1 \sqrt{y}-y d y$ So total area = $2\int\limits_0^1 \sqrt{y}-y d y$ $=2\left[\frac{y^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{y^{1+1}}{1+1}\right]_0^1$ $=2\left[\frac{2}{3} y^{3 / 2}-\frac{y^2}{2}\right]_0^1= 2\left[\frac{2}{3}-\frac{1}{2}\right] =\frac{4}{3}-1=\frac{4-3}{3}$ $=\frac{1}{3}$ |