The area bounded by the curve x = 2 - y - y^2 | vedclass

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(C) The curve is given by $x = 2 - y - y^2$. The $Y$-axis is the line $x = 0$. To find the points of intersection with the $Y$-axis,set $x = 0$: $2 -

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The curve is given by $x = 2 - y - y^2$. The $Y$-axis is the line $x = 0$. To find the points of intersection with the $Y$-axis,set $x = 0$: $2 - y - y^2 = 0 \implies y^2 + y - 2 = 0$. Factoring the quadratic: $(y + 2)(y - 1) = 0$,so $y = -2$ and $y = 1$. The area $A$ is given by the integral $\int_{-2}^{1} |x| dy = \int_{-2}^{1} (2 - y - y^2) dy$. Evaluating the integral: $A = [2y - \frac{y^2}{2} - \frac{y^3}{3}]_{-2}^{1}$. At $y = 1$: $2(1) - \frac{1}{2} - \frac{1}{3} = 2 - \frac{5}{6} = \frac{7}{6}$. At $y = -2$: $2(-2) - \frac{(-2)^2}{2} - \frac{(-2)^3}{3} = -4 - 2 + \frac{8}{3} = -6 + \frac{8}{3} = -\frac{10}{3}$. $A = \frac{7}{6} - (-\frac{10}{3}) = \frac{7}{6} + \frac{20}{6} = \frac{27}{6} = \frac{9}{2}$ sq. units.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

The area bounded by the curve $x = 2 - y - y^2$ and the $Y$-axis is

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