The area bounded by the lines y = 2 + x,y = 2 | vedclass

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

Question

(B) The lines are $y = 2 + x$,$y = 2 - x$,and $x = 2$.First,find the intersection points:$1$. Intersection of $y = 2 + x$ and $y = 2 - x$: $2 + x = 2

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The lines are $y = 2 + x$,$y = 2 - x$,and $x = 2$.First,find the intersection points:$1$. Intersection of $y = 2 + x$ and $y = 2 - x$: $2 + x = 2 - x \implies 2x = 0 \implies x = 0$. Thus,$y = 2$. Point $C$ is $(0, 2)$.$2$. Intersection of $y = 2 + x$ and $x = 2$: $y = 2 + 2 = 4$. Point $B$ is $(2, 4)$.$3$. Intersection of $y = 2 - x$ and $x = 2$: $y = 2 - 2 = 0$. Point $A$ is $(2, 0)$.The region is a triangle $ABC$ with vertices $A(2, 0)$,$B(2, 4)$,and $C(0, 2)$.The length of the base $AB$ (vertical line) is $|4 - 0| = 4$.The height of the triangle from vertex $C$ to the line $x = 2$ is the horizontal distance from $x = 0$ to $x = 2$,which is $2$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4 	imes 2 = 4 	ext{ sq. units}$.Alternatively,using integration: Area $= \int_{0}^{2} ((2 + x) - (2 - x)) dx = \int_{0}^{2} 2x dx = [x^2]_{0}^{2} = 4 - 0 = 4 	ext{ sq. units}$.

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

Similar Questions

The area (in square units) of the region bounded by $x^2=8y$,$x=4$ and the $X$-axis is:

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

The area bounded by the $y-$axis,$y=\cos x$ and $y=\sin x$ when $0 \leq x \leq \frac{\pi}{2}$ is

The area (in sq. units) in the first quadrant bounded by the curve $y=x^2+2$ and the lines $y=x+1$,$x=0$,and $x=2$ is:

The area bounded by the parabola ${y^2} = 4ax$ and its latus rectum is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module