Area of the region bounded by the curve |x| + | vedclass

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

Question

(C) The given equation is $|x| + y = 1$,which can be written as $y = 1 - |x|$.This equation represents two lines:$1$) For $x \ge 0$,$y = 1 - x$.$2$) F

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) The given equation is $|x| + y = 1$,which can be written as $y = 1 - |x|$.This equation represents two lines:$1$) For $x \ge 0$,$y = 1 - x$.$2$) For $x The region is bounded by these lines and the $x$-axis $(y = 0)$.The vertices of the region are $(0, 1)$,$(1, 0)$,and $(-1, 0)$.This forms a triangle with base length $b = 1 - (-1) = 2$ and height $h = 1$.The area of the triangle is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 2 	imes 1 = 1$ sq. units.

Area of the region bounded by the curve $|x| + y = 1$ is . . . . . . sq. units.

Similar Questions

The area bounded by the curve $x=\log (|y|)$,the lines $x=-1$ and $x=0$ is

The area of the region bounded by the line $y=2x+1$,the $X$-axis,and the ordinates $x=-1$ and $x=1$ is

If the tangent to the curve $y = 1 - x^2$ at $x = \alpha,$ where $0 < \alpha < 1,$ meets the axes at $P$ and $Q.$ Also $\alpha$ varies,the minimum value of the area of the triangle $OPQ$ is $k$ times the area bounded by the axes and the part of the curve for which $0 < x < 1,$ then $k$ is equal to

The area bounded by the curve $y^2 = x$,the line $y = 4$,and the $y$-axis is

Area bounded by the curve $y = \sin x$ between $x = 0$ and $x = 2\pi$ is ......... $sq. \text{ unit}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module