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

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

Question

(D) To find the area enclosed between the curves $y^2=x$ and $y=x$,we first find their points of intersection by solving the equations simultaneously:

Vedclass · Accepted Answer

$\frac{1}{6}$ sq. units

Vedclass · Answer

(D) To find the area enclosed between the curves $y^2=x$ and $y=x$,we first find their points of intersection by solving the equations simultaneously:$y^2 = y \implies y^2 - y = 0 \implies y(y-1) = 0$.Thus,the points of intersection are at $y=0$ and $y=1$. Correspondingly,$x=0$ and $x=1$.The area $A$ is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=1$:$A = \int_0^1 (\sqrt{x} - x) \, dx$$A = \left[ \frac{x^{3/2}}{3/2} - \frac{x^2}{2} ight]_0^1$$A = \left[ \frac{2}{3}x^{3/2} - \frac{1}{2}x^2 ight]_0^1$$A = \left( \frac{2}{3}(1)^{3/2} - \frac{1}{2}(1)^2 ight) - (0 - 0)$$A = \frac{2}{3} - \frac{1}{2} = \frac{4-3}{6} = \frac{1}{6} 	ext{ sq. units}$.

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

Similar Questions

The area of the region bounded by the hyperbola $x^2-y^2=9$ and its latus rectum is

The area enclosed by $y^{2}=8x$ and $y=\sqrt{2}x$ that lies outside the triangle formed by $y=\sqrt{2}x$,$x=1$,and $y=2\sqrt{2}$ is equal to

The area enclosed by the curves $y=\ln(x+e^{2})$,$x=\ln(2/y)$ (which is $y=2e^{-x}$) and $x=\ln 2$,above the line $y=1$ is:

If $f(x)$ is a continuous,increasing,and odd function such that $\int_{-1}^{4} f(x) \,dx = 10$ and $\int_{0}^{1} f(x) \,dx = \frac{3}{2}$,then the area bounded by $y = f(x)$,the $x$-axis,and the ordinates $x = -4$ and $x = 4$ is:

The area bounded by the curve ${x^2} = 4y$ and the straight line $x = 4y - 2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module