Question

Find the area of the region included between the parabola y² = x and the line x + y = 2.

Find the area enclosed by the parabola y² = x and the line y + x = 2.

Answer 1

We have,
\[y^2 = x\] and \[x + y = 2\] 

To find the intersecting points of the curves ,we solve both the equations.
\[y^2 + y - 2 = 0\]
\[ \Rightarrow \left( y + 2 \right)\left( y - 1 \right) = 0\]
\[ \Rightarrow y = - 2\text{ or }y = 1\]
\[ \therefore x = 4 \text{ or }1\]
\[\text{ Consider a horizantal strip of length }\left| x_2 - x_1 \right|\text{ and width }dy\text{ where }P\left( x_2 , y \right)\text{ lies on straight line and Q}\left( x_1 , y \right)\text{ lies on the parabola }. \]
\[\text{ Area of approximating rectangle }= \left| x_2 - x_1 \right| dy ,\text{ and it moves from }y = - 2\text{ to }y = 1\]
\[\text{ Required area = area }\left(\text{ OADO }\right) = \int_{- 2}^1 \left| x_2 - x_1 \right| dy\]
\[ = \int_{- 2}^1 \left| x_2 - x_1 \right| dy .............\left\{ \because \left| x_2 - x_1 \right| = x_2 - x_1 as x_2 > x_1 \right\}\]
\[ = \int_{- 2}^1 \left\{ \left( 2 - y \right) - y^2 dy \right\}\]
\[ = \left[ 2y - \frac{y^2}{2} - \frac{y^3}{3} \right]_{- 2}^1 \]
\[ = \left[ 2 - \frac{1}{2} - \frac{1}{3} \right] - \left[ - 4 - 2 + \frac{8}{3} \right]\]
\[ = 2 - \frac{1}{2} - \frac{1}{3} + 6 - \frac{8}{3}\]
\[ = \frac{9}{2}\text{ sq units }\]
\[\text{ Area enclosed by the line and given parabola }= \frac{9}{2}\text{ sq units }\]

Answer 2

y² = x,

x + y = 2

i.e. x = 2 – y

Solving the equations:

y² = 2 – y,

i.e. y² + y – 2 = 0

i.e. (y + 2)(y – 1) = 0

∴  y = 1, – 2

∴ x = 1, 4 (respectively)

Points of intersection of parabola and line

= (1, 1) and (4, – 2)

Also the line cuts X-axis at (2, 0)

Area above X-axis

= `int_0^1 sqrt(x)  dx + int_1^2(2 - x)dx`

= `2/3[x^(3/2)]_0^1 + [2x - x^2/2]_1^2`

= `2/3 + (4 - 2) - (2 - 1/2)`

= `2/3 + 2 - 3/2`

= `(4 + 12 - 9)/6`

= `7/6`   ...(1)

Area below X-axis

= `|int_0^4 sqrt(x) dx| - |int_2^4(2 - x)dx|`

= `2/3[x^(3/2)]_0^4 - [2x - x^2/2]_2^4`

= `2/3(8) - |[(8 - 8) - (4 - 2)]|`

= `16/3 - |(-2)|`

= `16/3 - 2`

= `10/3`   ...(2)

From (1) and (2)

Required area = `7/6 + 10/3`

= `(7 + 20)/6`

= `27/6`

= `9/2` sq. units