The area bounded by y = 2 - x^2 and x + y = 0 is: - Vidyadip

MCQ

The area bounded by $y = 2 - x^2$ and $x + y = 0$ is:

A
$\frac{7}{2}\text{ sq. units}$
✓
$\frac{9}{2}\text{ sq. units}$
C
$9\text{ sq. units}$
D
$\text{none of these}$

✓

Answer

Correct option: B.

$\frac{9}{2}\text{ sq. units}$

To find the points of intersection of $x + y = 0$ and $y = 2 - x^2.$
We put $x = -y$ in $y = 2 - x^2,$
We get $y = 2 - y^2$
$\Rightarrow y^2 + y - 2 = 0$
$\Rightarrow y - 1, y + 2 = 0$
$\Rightarrow y = 1, -2$
$\Rightarrow x = -1, 2$
Therefore, the points of intersection are $A(-1, 1)$ and $C(2, -2).$
The area of the required region $\ce{ABCD},$
$\text{A} = \int\limits^2_{-1}(\text{y}_1-\text{y}_{2})\text{dx} ($Where, $y_1 = 2 - x^2$ and $y_2 = -x)$
$=\int\limits^2_{-1}(2-\text{x}^{2}+\text{x})\text{dx}$
$ = \Big[2\text{x}-\frac{\text{x}^{3}}{3}+\frac{\text{x}^{2}}{2}\Big]^2_{-1}$
$= \bigg\{2(2)-\frac{(2)^{3}}{3}+\frac{(2)^{2}}{2}\bigg\}-\bigg\{2(-1)-\frac{(-1)^{3}}{3}+\frac{(-1)^{2}}{2}\bigg\}$
$= \Big(4-\frac{8}{3}+2\Big)-\Big(-2+\frac{1}{3}+\frac{1}{2}\Big)$
$=6-\frac{8}{3}+2-\frac{1}{3}-\frac{1}{2}$
$= 8 - \frac{9}{3}-\frac{1}{2}$
$= 5 -\frac{1}{2}$
$\frac{9}{2}\text{ sq. units}$

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