Question

Find the area of the region bounded by y =\[\sqrt{x}\] and y = x.

Answer 1

\[y = \sqrt{x} . . . \left( 1 \right)\text{ is a parabola opening side ways, with vertex at O(0, 0) and + ve }x - \text{ axis as axis of symmetry }\]
\[x = y . . . \left( 2 \right)\text{ is a straight line passsing through O(0, 0) and at angle }{45}^o\text{ with the }x - \text{ axis }\]
\[\text{ Solving }\left( 1 \right)\text{ and }\left( 2 \right) \]
\[ y^2 = x = y \]
\[ \Rightarrow y^2 = y \]
\[ \Rightarrow y(y - 1) = 0 \]
\[ \Rightarrow y = 0\text{ or }y = 1\text{ and }x = 0\text{ or }x = 1\]
\[\text{ Thus, the line intersects the parabola at O(0, 0 ) and A(1, 1) }\]
\[\text{ Consider a approximating rectangle of length }= \left| y_2 - y_1 \right|\text{ and width }= dx \]
\[\text{ Area of approximating rectangle }= \left| y_2 - y_1 \right| dx\]
\[\text{ Approximating rectangle moves from }x = 0\text{ to }x = 1 \]
\[ \therefore\text{ Area of the shaded region }= \int_0^1 \left| y_2 - y_1 \right| dx = \int_0^1 \left( y_2 - y_1 \right) dx ...............\left[ As, y_2 > y_1 , \left| y_2 - y_1 \right| = y_2 - y_1 \right] \]
\[ \Rightarrow A = \int_0^1 \left( \sqrt{x} - x \right) dx \]
\[ \Rightarrow A = \left[ \frac{x^\frac{3}{2}}{\frac{3}{2}} - \frac{x^2}{2} \right]_0^1 \]
\[ \Rightarrow A = \left[ \frac{1^\frac{3}{2}}{\frac{3}{2}} - \frac{1^2}{2} - 0 \right]\]
\[ \Rightarrow A = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}\text{ sq . units }\]
\[ \therefore\text{ Area bound by the parabola and straight line }= \frac{1}{6}\text{ sq . units }\]