CUET Preparation Today
The area of the region bounded by the curves y2 = x and y = x is: |
$\frac{1}{2}$ $\frac{2}{3}$ $\frac{1}{6}$ $\frac{5}{12}$ |
$\frac{1}{6}$ |
$y^2=x,\;y=x.$ $x=y^2.$ $\text{Points of intersection:}$ $y=y^2.$ $y(y-1)=0.$ $y=0,1.$ $\text{Area}=\int_{0}^{1}(y-y^2)\,dy.$ $=\left[\frac{y^2}{2}-\frac{y^3}{3}\right]_{0}^{1}.$ $=\frac12-\frac13.$ $=\frac16.$ $\text{Area}=\frac16.$ |
