The area of the region enclosed by the lines | vedclass

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Question

(C) The lines are $y = 2x$, $y = \frac{x}{2}$, and $x = 2$.Intersection points:$1$. $y = 2x$ and $y = \frac{x}{2}$ intersect at $(0, 0)$.$2$. $y = 2x$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The lines are $y = 2x$, $y = \frac{x}{2}$, and $x = 2$.Intersection points:$1$. $y = 2x$ and $y = \frac{x}{2}$ intersect at $(0, 0)$.$2$. $y = 2x$ and $x = 2$ intersect at $(2, 4)$.$3$. $y = \frac{x}{2}$ and $x = 2$ intersect at $(2, 1)$.The area $A$ is given by the integral $\int_{0}^{2} (2x - \frac{x}{2}) \, dx$.$A = \int_{0}^{2} \frac{3x}{2} \, dx$.$A = \frac{3}{2} [\frac{x^2}{2}]_{0}^{2} = \frac{3}{4} [4 - 0] = 3$ sq. units.

The area of the region enclosed by the lines $y = 2x$, $2y = x$ and $x = 2$ (in sq. units) is...

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