The area bounded by the curves y^2 - x = 0 an | vedclass

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

Question

(B) The given curves are $y^2 = x$ and $y = x^2$. To find the points of intersection,substitute $x = y^2$ into $y = x^2$,which gives $y = (y^2)^2 = y^

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(B) The given curves are $y^2 = x$ and $y = x^2$. To find the points of intersection,substitute $x = y^2$ into $y = x^2$,which gives $y = (y^2)^2 = y^4$. This implies $y^4 - y = 0$,so $y(y^3 - 1) = 0$. The intersection points are at $y = 0$ and $y = 1$. For $y = 0$,$x = 0$,and for $y = 1$,$x = 1$. The area $A$ bounded by the curves is given by the integral: $A = \int_0^1 (\sqrt{x} - x^2) dx$ $A = \left[ \frac{x^{3/2}}{3/2} - \frac{x^3}{3} \right]_0^1$ $A = \left( \frac{2}{3}(1)^{3/2} - \frac{1}{3}(1)^3 \right) - (0 - 0)$ $A = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$ Thus,the area is $\frac{1}{3}$ square units.

The area bounded by the curves $y^2 - x = 0$ and $y - x^2 = 0$ is

Similar Questions

The area (in sq. units) bounded by the curves $y=\frac{8}{x}$,$y=2x$ and $x=4$ is

The area of the region inside the circle $(x-2 \sqrt{3})^2+y^2=12$ and outside the parabola $y^2=2 \sqrt{3} x$ is

Area enclosed by the curves $y = \ln x$,$y = \ln|x|$,$y = |\ln x|$,and $y = |\ln|x||$ is equal to

Area of the region enclosed between the curves $x = y^2 - 1$ and $x = |y| \sqrt{1 - y^2}$ is

The area (in sq. units) of the region enclosed between the parabola $y^{2}=2x$ and the line $x+y=4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module