The area (in square units) of the region boun | vedclass

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

Question

(C) Given curves are $y = -2x^2$ and $y = 1 - 3x^2$.To find the points of intersection,set the equations equal to each other:$-2x^2 = 1 - 3x^2$$x^2 =

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(C) Given curves are $y = -2x^2$ and $y = 1 - 3x^2$.To find the points of intersection,set the equations equal to each other:$-2x^2 = 1 - 3x^2$$x^2 = 1$$x = \pm 1$.Since the curves are symmetric about the $y$-axis,the area $A$ is given by:$A = 2 \int_{0}^{1} (y_{upper} - y_{lower}) dx$Here,$y_{upper} = 1 - 3x^2$ and $y_{lower} = -2x^2$.$A = 2 \int_{0}^{1} ((1 - 3x^2) - (-2x^2)) dx$$A = 2 \int_{0}^{1} (1 - x^2) dx$$A = 2 [x - \frac{x^3}{3}]_{0}^{1}$$A = 2 (1 - \frac{1}{3}) = 2 (\frac{2}{3}) = \frac{4}{3}$ square units.

The area (in square units) of the region bounded by the curves $y + 2x^2 = 0$ and $y + 3x^2 = 1$ is equal to

Similar Questions

The area of the region bounded by the curve $y = x^2 + 2$ and the lines $y = x$,$x = 0$,and $x = 3$ is

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

Find the area of the region bounded by the parabola $(y-2)^2 = x-1$,the tangent to the parabola at the point $(2,3)$,and the $x$-axis.

The area enclosed by the curves $y^2+4x=4$ and $y-2x=2$ is:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module