The area enclosed between the curves y^2 = x | vedclass

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

Question

(A) The curves are $y^2 = x$ and $y = |x|$.Since $y = |x|$,the curves are $y^2 = x$ and $y = x$ (for $x \ge 0$) and $y = -x$ (for $x < 0$).However,$y^

Vedclass · Accepted Answer

$\frac{1}{6}$

Vedclass · Answer

(A) The curves are $y^2 = x$ and $y = |x|$.Since $y = |x|$,the curves are $y^2 = x$ and $y = x$ (for $x \ge 0$) and $y = -x$ (for $x However,$y^2 = x$ implies $x \ge 0$,so we only consider the region where $x \ge 0$ and $y = x$.The intersection points are found by $x^2 = x$,which gives $x(x-1) = 0$,so $x = 0$ and $x = 1$.In the interval $[0, 1]$,$\sqrt{x} \ge x$.The required area is given by $\int_{0}^{1} (\sqrt{x} - x) dx$.$= \left[ \frac{x^{3/2}}{3/2} - \frac{x^2}{2} \right]_{0}^{1}$.$= \left( \frac{2}{3} - \frac{1}{2} \right) - (0 - 0) = \frac{4-3}{6} = \frac{1}{6}$.

The area enclosed between the curves $y^2 = x$ and $y = |x|$ is

Similar Questions

The area enclosed by the curves $y = \sin^{-1}(\cos x)$ and $y = \cos^{-1}(\sin x)$ for $x \in \left[ \frac{\pi}{2}, \frac{3\pi}{2} \right]$ is:

Area of the region (in sq units) bounded by the curves $y=\sqrt{x}$,$x=\sqrt{y}$ and the lines $x=1$,$x=4$ is

If the area of the region $\{(x, y): |x^2-2| \leq y \leq x\}$ is $A$,then $6A + 16\sqrt{2}$ is equal to $...........$.

The area of the region included between the parabolas $y^2=8x$ and $x^2=8y$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module