Question

Sketch the region bounded by the curves y = x² + 2, y = x, x = 0 and x = 1. Also, find the area of this region.

Answer 1

We have,
\[y = x^2 + 2\] and \[y = x\]
We see that parabola and the line \[y = x\] do not intersect \[x = 1\]  is a line parallel to y axis Point of intersection between parabola and \[x = 1\] is \[\text{ Putting }x = 1\text{ in }y = x^2 + 2,\text{ we get, }\]
\[y = 1 + 2 = 3\]
\[\text{ Point of intersection of two lines is given by }\]
\[\text{ Putting }x = 1\text{ in }y = x, \text{ we get, }\]
\[y = 1 \]
\[\text{ Consider a vetical strip of length }\left| y_2 - y_1 \right| \text{ and width }= dx \text{ such that }P\left( x, y_2 \right)\text{ lies on parabola and Q }\left( x , y_1 \right)\text{ lies on }y = x\]
\[\text{ Shaded area }= \int_0^1 \left| y_2 - y_1 \right| dx\]
\[ = \int_0^1 \left( y_2 - y_1 \right) dx ..............\left\{ \because \left| y_2 - y_1 \right| \Rightarrow y_2 - y_1 \text{ as }y_2 > y_1 , \right\}\]
\[ = \int_0^1 \left\{ \left( x^2 + 2 \right) - \left( x \right) \right\} dx\]
\[ = \int_0^1 \left( x^2 + 2 - x \right)dx\]
\[ = \left[ \frac{x^3}{3} - \frac{x^2}{2} + 2x \right]_0^1 \]
\[ = \frac{1}{3} - \frac{1}{2} + 2\]
\[ = \frac{2 - 3 + 12}{6}\]
\[ = \frac{11}{6}\text{ sq units }\]
\[\text{ Thus, area enclosed by parabola and given two lines }= \frac{11}{6}\text{ sq units }\]