The area of the region enclosed by the lines | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The lines are $y = 2x$, $y = x/2$, and $x = 2$.The intersection of $y = 2x$ and $y = x/2$ is at $(0, 0)$.The intersection of $y = 2x$ and $x = 2$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The lines are $y = 2x$, $y = x/2$, and $x = 2$.The intersection of $y = 2x$ and $y = x/2$ is at $(0, 0)$.The intersection of $y = 2x$ and $x = 2$ is at $(2, 4)$.The intersection of $y = x/2$ and $x = 2$ is at $(2, 1)$.The area $A$ is given by the integral $\int_{0}^{2} (2x - x/2) \, dx$.$A = \int_{0}^{2} (3x/2) \, dx$.$A = \frac{3}{2} \left[ \frac{x^2}{2} ight]_{0}^{2}$.$A = \frac{3}{4} [4 - 0] = 3 	ext{ sq. units}$.

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

Similar Questions

Find the area of the region bounded by $x^{2}=4y$,$y=2$,$y=4$ and the $y-$axis in the first quadrant.

The area of the shaded region in the given figure of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ is . . . . . . sq. units.

The volume of the solid generated by revolving the region bounded by the parabolas $y=x^{2}$ and $x=y^{2}$ about the $y$-axis is:

The area of the region (in sq. units) enclosed by the curve $y=x^3-19x+30$ and the $x$-axis is

The area bounded by the lines $y = 2 + x$,$y = 2 - x$,and $x = 2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module