The area bounded by the curves y = (x + 1)^2, | vedclass

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(D) The curves are $y = (x + 1)^2$ and $y = (x - 1)^2$. The line is $y = \frac{1}{4}$.By symmetry,the area is twice the area bounded by $y = (x - 1)^2

Vedclass · Accepted Answer

$1/3$

Vedclass · Answer

(D) The curves are $y = (x + 1)^2$ and $y = (x - 1)^2$. The line is $y = \frac{1}{4}$.By symmetry,the area is twice the area bounded by $y = (x - 1)^2$,the $y$-axis $(x=0)$,and the line $y = \frac{1}{4}$ in the region $x \ge 0$.For $y = (x - 1)^2$,we have $x - 1 = \pm \sqrt{y}$,so $x = 1 \pm \sqrt{y}$. Since we are considering the right branch,$x = 1 - \sqrt{y}$ for $x \in [0, 1]$.The intersection of $y = (x - 1)^2$ and $y = \frac{1}{4}$ gives $(x - 1)^2 = \frac{1}{4}$,so $x - 1 = -1/2$ (as $x The area $A = 2 \int_{1/4}^{1} (1 - \sqrt{y}) dy$ is incorrect based on the graph. The correct integral for the region bounded by $y = (x - 1)^2$,$x=0$,and $y=1/4$ is $\int_{0}^{1/2} (1/4 - (x-1)^2) dx + \int_{1/2}^{1} (1/4 - 0) dx$ is not right. Let's use horizontal strips: The area is $2 \int_{1/4}^{1} (1 - \sqrt{y}) dy = 2 [y - \frac{2}{3} y^{3/2}]_{1/4}^{1} = 2 [(1 - 2/3) - (1/4 - \frac{2}{3} \cdot \frac{1}{8})] = 2 [1/3 - (1/4 - 1/12)] = 2 [1/3 - 2/12] = 2 [1/3 - 1/6] = 2 [1/6] = 1/3$ sq. units.

The area bounded by the curves $y = (x + 1)^2$,$y = (x - 1)^2$ and the line $y = \frac{1}{4}$ is

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