The area of the region bounded by the curve y | vedclass

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(D) The given curve is $y = 2 - x - 3x^2$. The region is bounded by the $X$-axis $(y = 0)$, the $Y$-axis $(x = 0)$, and the line $x = -2$.First, find

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) The given curve is $y = 2 - x - 3x^2$. The region is bounded by the $X$-axis $(y = 0)$, the $Y$-axis $(x = 0)$, and the line $x = -2$.First, find the roots of the curve by setting $y = 0$:$2 - x - 3x^2 = 0 \implies 3x^2 + x - 2 = 0 \implies (3x - 2)(x + 1) = 0$.The roots are $x = -1$ and $x = 2/3$.For $x \in [-2, -1]$, $y = 2 - x - 3x^2$ is negative (e.g.,at $x = -1.5$, $y = 2 + 1.5 - 3(2.25) = 3.5 - 6.75 = -3.25 For $x \in [-1, 0]$, $y = 2 - x - 3x^2$ is positive (e.g.,at $x = -0.5$, $y = 2 + 0.5 - 3(0.25) = 2.5 - 0.75 = 1.75 > 0$).The total area is given by:Area $= \int_{-2}^{-1} -(2 - x - 3x^2) dx + \int_{-1}^{0} (2 - x - 3x^2) dx$$= \int_{-2}^{-1} (3x^2 + x - 2) dx + \int_{-1}^{0} (2 - x - 3x^2) dx$$= [x^3 + \frac{x^2}{2} - 2x]_{-2}^{-1} + [2x - \frac{x^2}{2} - x^3]_{-1}^{0}$$= [(-1 + 0.5 + 2) - (-8 + 2 + 4)] + [(0) - (-2 - 0.5 + 1)]$$= [1.5 - (-2)] + [0 - (-1.5)]$$= 3.5 + 1.5 = 5$.Thus, the correct option is $D$.

The area of the region bounded by the curve $y = 2 - x - 3x^2$, the $X$-axis, the $Y$-axis, and the line $x = -2$ is

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