The area of the region bounded by the curves | vedclass

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

Question

(A) Given curves are $y=x^{2}$ and $x=y^{2}$, which are parabolas.To find the points of intersection, substitute $y=x^{2}$ into $x=y^{2}$:$x=(x^{2})^{

Vedclass · Accepted Answer

$1/3$

Vedclass · Answer

(A) Given curves are $y=x^{2}$ and $x=y^{2}$, which are parabolas.To find the points of intersection, substitute $y=x^{2}$ into $x=y^{2}$:$x=(x^{2})^{2}$$x=x^{4}$$x^{4}-x=0$$x(x^{3}-1)=0$$x(x-1)(x^{2}+x+1)=0$Since $x^{2}+x+1=0$ has no real roots, we have $x=0$ and $x=1$.When $x=0$, $y=0$. When $x=1$, $y=1$.Thus, the points of intersection are $(0,0)$ and $(1,1)$.The area of the bounded region is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=1$:$	ext{Area} = \int_{0}^{1} (\sqrt{x} - x^{2}) \, dx$$= \left[ \frac{x^{3/2}}{3/2} - \frac{x^{3}}{3} ight]_{0}^{1}$$= \left[ \frac{2}{3}x^{3/2} - \frac{1}{3}x^{3} ight]_{0}^{1}$$= (\frac{2}{3}(1) - \frac{1}{3}(1)) - (0 - 0)$$= \frac{2}{3} - \frac{1}{3} = \frac{1}{3} 	ext{ sq units}$.

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

Similar Questions

Let $f(\alpha)$ denote the area of the region in the first quadrant bounded by $x=0, x=1, y^{2}=x$ and $y=|\alpha x-5|-|1-\alpha x|+\alpha x^{2}$. Then $f(0)+f(1)$ is equal to

The area of the region bounded by the curves $y = \sin x$,$y = \cos x$ and $x = 0$ is

The area of the region bounded by the curve $y = 2^{kx}$ and the lines $x = 0$ and $x = 2$ in the first quadrant is $\frac{2}{\log_e 2}$. Find the value of $k$.

The area bounded by the curve $y = x^2 + 2$, the $x$-axis, and the lines $x = 1$ and $x = 2$ is:

The area bounded by the curve $y = x^2 + 4x + 5$,the coordinate axes,and the minimum ordinate is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module