The area enclosed by the curve y = -x2 and the straight line x + y + 2 = 0 is:

Question

The area enclosed by the curve $y = - x^{2}$ and the straight line $x + y + 2 = 0$ is:

Smart Achievers · Accepted Answer

(1)

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

Explanation:

We have $y = - x^{2}$ and $x + y + 2 = 0$

$\Rightarrow$ $- x - 2 = - x^{2}$

$\Rightarrow$ $x^{2} - x - 2 = 0$

$\Rightarrow$ $x^{2} + x - 2 x - 2 = 0$

$\Rightarrow$ $x (x + 1) - 2 (x + 1) = 0$

$\Rightarrow$ $(x - 2) (x + 1) = 0$

$\Rightarrow$ $x = 2, - 1$

Hence, the area of the shaded region is

$A =$ $| \int_{- 1}^{2} (- x - 2 + x^{2}) d x |$

$=$ $| \int_{- 1}^{2} (x^{2} - x - 2) d x |$

$=$ $| {[\frac{x^{3}}{3} - \frac{x^{2}}{2} - 2 x]}_{- 1}^{2} |$

$=$ $| [\frac{8}{3} - \frac{4}{2} - 4 + \frac{1}{3} + \frac{1}{2} - 2] |$

$=$ $| \frac{16 - 12 - 24 + 2 + 3 - 12}{6} |$

$=$ $| - \frac{27}{6} |$

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

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