The area (in sq. units) enclosed between the | vedclass

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

Question

(C) Given curves are $y=x^2$ $(i)$ and $y=|x|$ $(ii)$.Since both curves are symmetric about the $y$-axis,the total area is twice the area in the first

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(C) Given curves are $y=x^2$ $(i)$ and $y=|x|$ $(ii)$.Since both curves are symmetric about the $y$-axis,the total area is twice the area in the first quadrant where $x \ge 0$.For $x \ge 0$,$y=|x|=x$.Solving $y=x^2$ and $y=x$,we get $x^2=x$,which implies $x(x-1)=0$,so $x=0$ or $x=1$.The intersection points in the first quadrant are $(0,0)$ and $(1,1)$.The area in the first quadrant is $\int_0^1 (x - x^2) dx$.Total area $= 2 \int_0^1 (x - x^2) dx$$= 2 \left[ \frac{x^2}{2} - \frac{x^3}{3} ight]_0^1$$= 2 \left( \frac{1}{2} - \frac{1}{3} ight) = 2 \left( \frac{1}{6} ight) = \frac{1}{3} 	ext{ sq. units}$.

The area (in sq. units) enclosed between the curves $y=x^2$ and $y=|x|$ is

Similar Questions

If the area of the larger portion bounded between the curves $x^2+y^2=25$ and $y=|x-1|$ is $\frac{1}{4}(b \pi+c)$,where $b, c \in N$,then $b+c$ is equal to $ . . .. .. $

The area bounded by the curves ${y^2} = 8x$ and $y = x$ is

Find the area of the region bounded by the two parabolas $y=x^{2}$ and $y^{2}=x$.

The area bounded by the curves $y = |x| - 1$ and $y = -|x| + 1$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module