The area of the region bounded by the parabol | vedclass

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(C) The given parabola is $y^2 = 27x$.
Since the region is bounded by the parabola and the line $x = 1$,the area $A$ is given by the integral of $y$

Vedclass · Accepted Answer

$4\sqrt{3}$

Vedclass · Answer

(C) The given parabola is $y^2 = 27x$.
Since the region is bounded by the parabola and the line $x = 1$,the area $A$ is given by the integral of $y$ with respect to $x$ from $x = 0$ to $x = 1$.
Since $y^2 = 27x$,we have $y = \sqrt{27x} = 3\sqrt{3}\sqrt{x}$.
The area is symmetric about the $x$-axis,so the total area is $2 	imes \int_{0}^{1} y \, dx$.
$A = 2 \int_{0}^{1} 3\sqrt{3} \sqrt{x} \, dx = 6\sqrt{3} \int_{0}^{1} x^{1/2} \, dx$.
$A = 6\sqrt{3} \left[ \frac{x^{3/2}}{3/2} ight]_{0}^{1} = 6\sqrt{3} 	imes \frac{2}{3} [x^{3/2}]_{0}^{1}$.
$A = 4\sqrt{3} (1 - 0) = 4\sqrt{3}$ sq. units.

The area of the region bounded by the parabola $y^2 = 27x$ and the line $x = 1$ is . . . . . . sq. units. (in $sqrt{3}$)

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