The area of the region {(x, y) : 0 le y le 6 | vedclass

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(B) The curves are $y = 6-x$ and $y^2 = 4x-3$. First, find the intersection points: $(6-x)^2 = 4x-3 \implies 36 - 12x + x^2 = 4x - 3 \implies x^2 - 16

Vedclass · Accepted Answer

$9$

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(B) The curves are $y = 6-x$ and $y^2 = 4x-3$. First, find the intersection points: $(6-x)^2 = 4x-3 \implies 36 - 12x + x^2 = 4x - 3 \implies x^2 - 16x + 39 = 0$. Factoring gives $(x-3)(x-13) = 0$, so $x=3$ or $x=13$. At $x=3$, $y=3$. The region is bounded by $x=0$, $y=6-x$, and $y^2=4x-3$. The area is given by $\int_0^3 (6-x) dx - \int_{3/4}^3 2\sqrt{x-3/4} dx$. Evaluating the first integral: $[6x - x^2/2]_0^3 = 18 - 4.5 = 13.5$. Evaluating the second integral: $\int_{3/4}^3 2(x-3/4)^{1/2} dx = [2 \cdot \frac{2}{3} (x-3/4)^{3/2}]_{3/4}^3 = \frac{4}{3} (3-0.75)^{3/2} = \frac{4}{3} (2.25)^{3/2} = \frac{4}{3} (3.375) = 4.5$. Total Area $= 13.5 - 4.5 = 9$.

The area of the region $\{(x, y) : 0 \le y \le 6 - x, y^2 \ge 4x - 3, x \ge 0\}$ is:

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