The area bounded by y = 2 − x2 and x + y = 0 is _________ .

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KollegeApply · Accepted Answer

$\frac{9}{2}$\text{ sq . 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 y = 2 - y^2 \Rightarrow y^2 + y - 2 = 0 \Rightarrow \left( y - 1 \right)\left( y + 2 \right) = 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 ABCD, A = \int_{- 1}^2 \left( y_1 - y_2 \right) d x ...........\left(\text{Where, }y_1 = 2 - x^2\text{ and }y_2 = - x \right) = \int_{- 1}^2 \left( 2 - x^2 + x \right) d x = \left[ 2x - x^$\frac{3}{3}$ + x^$\frac{2}{2}$ \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 - $\frac{1}{2}$ = $\frac{9}{2}$\text{ square units }. [Solution diagram: 43933_solution_1.png]

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