The area of the region enclosed by the curves | vedclass

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(D) Given curves are $y^2 = 4(x+1)$ and $y^2 = 5(x-4)$.To find the intersection points, equate the $x$ values:$x = \frac{y^2}{4} - 1$ and $x = \frac{y

Vedclass · Accepted Answer

$200/3$

Vedclass · Answer

(D) Given curves are $y^2 = 4(x+1)$ and $y^2 = 5(x-4)$.To find the intersection points, equate the $x$ values:$x = \frac{y^2}{4} - 1$ and $x = \frac{y^2}{5} + 4$.$\frac{y^2}{4} - 1 = \frac{y^2}{5} + 4$$\frac{y^2}{4} - \frac{y^2}{5} = 5$$\frac{y^2}{20} = 5 \implies y^2 = 100 \implies y = \pm 10$.When $y = 10$, $x = \frac{100}{4} - 1 = 24$. So, intersection points are $(24, 10)$ and $(24, -10)$.The area is given by $\int_{-10}^{10} (x_{	ext{right}} - x_{	ext{left}}) dy$.$x_{	ext{right}} = \frac{y^2}{4} + 1$ is incorrect; let's re-evaluate: $y^2 = 4(x+1) \implies x = \frac{y^2}{4} - 1$ (left curve) and $y^2 = 5(x-4) \implies x = \frac{y^2}{5} + 4$ (right curve).Area $= \int_{-10}^{10} [(\frac{y^2}{4} - 1) - (\frac{y^2}{5} + 4)] dy$ is wrong. The region is bounded by $x = \frac{y^2}{5} + 4$ on the left and $x = \frac{y^2}{4} - 1$ on the right? No, looking at the graph, $y^2=4(x+1)$ is the outer curve. Let's use $x = \frac{y^2}{4} - 1$ and $x = \frac{y^2}{5} + 4$.Area $= \int_{-10}^{10} [(\frac{y^2}{4} - 1) - (\frac{y^2}{5} + 4)] dy = \int_{-10}^{10} (\frac{y^2}{20} - 5) dy = 2 \int_{0}^{10} (\frac{y^2}{20} - 5) dy = 2 [\frac{y^3}{60} - 5y]_0^{10} = 2 [\frac{1000}{60} - 50] = 2 [\frac{50}{3} - 50] = 2 [-\frac{100}{3}] = -200/3$. Taking absolute value, Area $= 200/3$.

The area of the region enclosed by the curves $y^2=4(x+1)$ and $y^2=5(x-4)$ is

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