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Application of Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplication of Integrals1 Mark
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
    None of these
✓

Answer

Correct option: B.
$\frac{9}{2}\text{ sq. units}$
$\text{(IMAGE)}$
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 $\text{ABCD},$
$\text{A} = \int\limits^2_{-1}(\text{y}_1-\text{y}_{2})\text{dx}$ (Where, $\mathrm{y}_1=2-\mathrm{x}^2$ and $\mathrm{y}_2=-\mathrm{x}$ )
$=\int\limits^2_{-1}(2-\text{x}^{2}+\text{x})\text{dx}$
$ = \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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