Question

Find the area of the region bounded by the parabola y² = 2x + 1 and the line x − y − 1 = 0.

Answer 1

We have,
\[y^2 = 2x + 1\] and \[x - y - 1 = 0\]
To find the intersecting points of the curves ,we solve both the equations.
\[y^2 = 2\left( 1 + y \right) + 1\]
\[ \Rightarrow y^2 - 2 - 2y - 1 = 0\]
\[ \Rightarrow y^2 - 2y - 3 = 0\]
\[ \Rightarrow \left( y - 3 \right)\left( y + 1 \right) = 0\]
\[ \Rightarrow y = 3\text{ or }y = - 1\]
\[ \therefore x = 4\text{ or }0\]
\[\text{ Consider a horizantal strip of length }\left| x_2 - x_1 \right|\text{ and width dy where }P\left( x_2 , y \right)\text{ lies on straight line and Q }\left( x_1 , y \right)\text{ lies on the parabola . }\]
\[\text{ Area of approximating rectangle }= \left| x_2 - x_1 \right| dy , \text{ and it moves from }y = - 1\text{ to }y = 3\]
\[\text{ Required area = area }\left( OADO \right) = \int_{- 1}^3 \left| x_2 - x_1 \right| dy\]
\[ = \int_{- 1}^3 \left| x_2 - x_1 \right| dy ...........\left\{ \because \left| x_2 - x_1 \right| = x_2 - x_1 as x_2 > x_1 \right\}\]
\[ = \int_{- 1}^3 \left\{ \left( 1 + y \right) - \frac{1}{2}\left( y^2 - 1 \right) \right\}dy\]
\[ = \int_{- 1}^3 \left\{ 1 + y - \frac{1}{2} y^2 + \frac{1}{2} \right\}dy\]
\[ = \int_{- 1}^3 \left\{ \frac{3}{2} + y - \frac{1}{2} y^2 \right\}dy\]
\[ = \left[ \frac{3}{2}y + \frac{y^2}{2} - \frac{1}{6} y^3 \right]_{- 1}^3 \]
\[ = \left[ \frac{9}{2} + \frac{9}{2} - \frac{27}{6} \right] - \left[ \frac{- 3}{2} + \frac{1}{2} + \frac{1}{6} \right]\]
\[ = \left[ \frac{9}{2} \right] + \left[ \frac{5}{6} \right]\]
\[ = \frac{16}{3}\text{ sq units }\]
\[\text{ Area enclosed by the line and given parabola }= \frac{16}{3}\text{ sq units }\]