Question

Find the area of the region bounded by the curves y = x − 1 and (y − 1)² = 4 (x + 1).

Answer 1

We have, y = x − 1 and (y − 1)² = 4 (x + 1)
\[\therefore \left( x - 1 - 1 \right)^2 = 4\left( x + 1 \right)\]
\[ \Rightarrow \left( x - 2 \right)^2 = 4\left( x + 1 \right)\]
\[ \Rightarrow x^2 + 4 - 4x = 4x + 4\]
\[ \Rightarrow x^2 + 4 - 4x - 4x - 4 = 0\]
\[ \Rightarrow x^2 - 8x = 0\]
\[ \Rightarrow x = 0\text{ or }x = 8\]
\[ \therefore y = - 1\text{ or }7\]
\[\text{ Consider a horizantal strip of length }\left| x_2 - x_1 \right| \text{ and width dy where }P\left( x_2 , y \right)\text{ lies on straight line and Q }\left( x_1 , y \right)\text{ lies on the parabola .} \]
\[\text{ Area of approximating rectangle }= \left| x_2 - x_1 \right| dy ,\text{ and it moves from }y = - 1\text{ to }y = 7\]
\[\text{ Required area = area }\left( OADO \right) = \int_{- 1}^7 \left| x_2 - x_1 \right| dy\]
\[ = \int_{- 1}^7 \left| x_2 - x_1 \right| dy ...........\left\{ \because \left| x_2 - x_1 \right| = x_2 - x_1\text{ as }x_2 > x_1 \right\}\]
\[ = \int_{- 1}^7 \left[ \left( 1 + y \right) - \frac{1}{4}\left\{ \left( y - 1 \right)^2 - 4 \right\} \right]dy\]
\[ = \int_{- 1}^7 \left\{ 1 + y - \frac{1}{4} \left( y - 1 \right)^2 + 1 \right\}dy\]
\[ = \int_{- 1}^7 \left\{ 2 + y - \frac{1}{4} \left( y - 1 \right)^2 \right\}dy\]
\[ = \left[ 2y + \frac{y^2}{2} - \frac{1}{12} \left( y - 1 \right)^3 \right]_{- 1}^7 \]
\[ = \left[ 14 + \frac{49}{2} - \frac{1}{12} \times 6 \times 6 \times 6 \right] - \left[ - 2 + \frac{1}{2} + \frac{1}{12} \times 2 \times 2 \times 2 \right]\]
\[ = \left[ 14 + \frac{49}{2} - 18 \right] - \left[ - 2 + \frac{1}{2} + \frac{2}{3} \right]\]
\[ = \left[ \frac{41}{2} \right] + \left[ \frac{5}{6} \right]\]
\[ = \frac{64}{3}\text{ sq units }\]
\[\text{ Area enclosed by the line and given parabola }= \frac{64}{3}\text{ sq units }\]