Question

The area bounded by y = 2 − x² and x + y = 0 is
(a) $\frac{7}{2}$sq. units

(b) $\frac{9}{2}$sq. units

(c) 9 sq. units

(d) none of these

Answer 1

$\left(\mathrm{b}\right)\frac{9}{2}\mathrm{sq}.\mathrm{units}$



To find the points of intersection of x + y = 0 and y = 2 − x².
We put x = − y in y = 2 − x², we get

$$\begin{gathered} y=2-y^2 \\ \Rightarrow y^2+y-2=0 \\ \Rightarrow \left(y-1\right)\left(y+2\right)=0 \\ \Rightarrow y=1,-2 \end{gathered}$$
$\Rightarrow x=-1,2$
Therefore, the points of intersection are A(−1, 1) and C(2, −2).

The area of the required region ABCD,

$$\begin{gathered} A={\int }_{-1}^2\left(y_1-y_2\right)\mathit{d}x\left(\mathrm{Where},y_1=2-x^2\mathrm{and}{y}_2=-x\right) \\ ={\int }_{-1}^2\left(2-x^2+x\right)\mathit{d}x \\ ={\left[2x-\frac{x^3}{3}+\frac{x^2}{2}\right]}_{-1}^2 \\ =\left\{2\left(2\right)-\frac{{\left(2\right)}^3}{3}+\frac{{\left(2\right)}^2}{2}\right\}-\left\{2\left(-1\right)-\frac{{\left(-1\right)}^3}{3}+\frac{{\left(-1\right)}^2}{2}\right\} \\ =\left(4-\frac{8}{3}+2\right)-\left(-2+\frac{1}{3}+\frac{1}{2}\right) \\ =6-\frac{8}{3}+2-\frac{1}{3}-\frac{1}{2} \\ =8-\frac{9}{3}-\frac{1}{2} \\ =5-\frac{1}{2} \\ =\frac{9}{2}\mathrm{square}\mathrm{units}. \end{gathered}$$