The area bounded between the parabola y^{2}=4 | vedclass

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

Question

(C) To find the area bounded between the parabola $y^{2}=4x$ and the line $y=2x-4$,we first find their points of intersection by substituting $x = \fr

Vedclass · Accepted Answer

$9$ sq unit

Vedclass · Answer

(C) To find the area bounded between the parabola $y^{2}=4x$ and the line $y=2x-4$,we first find their points of intersection by substituting $x = \frac{y+4}{2}$ into the parabola equation:$y^{2} = 4\left(\frac{y+4}{2}ight)$$y^{2} = 2(y+4)$$y^{2} - 2y - 8 = 0$$(y-4)(y+2) = 0$Thus,the intersection points occur at $y = 4$ and $y = -2$.The required area is given by the integral of the difference between the line and the parabola with respect to $y$:$	ext{Area} = \int_{-2}^{4} \left( \frac{y+4}{2} - \frac{y^{2}}{4} ight) dy$$= \left[ \frac{y^{2}}{4} + 2y - \frac{y^{3}}{12} ight]_{-2}^{4}$$= \left( \frac{16}{4} + 8 - \frac{64}{12} ight) - \left( \frac{4}{4} - 4 - \frac{-8}{12} ight)$$= \left( 4 + 8 - \frac{16}{3} ight) - \left( 1 - 4 + \frac{2}{3} ight)$$= \left( 12 - \frac{16}{3} ight) - \left( -3 + \frac{2}{3} ight)$$= \frac{20}{3} - \left( -\frac{7}{3} ight) = \frac{27}{3} = 9 	ext{ sq unit}$.

The area bounded between the parabola $y^{2}=4x$ and the line $y=2x-4$ is equal to

Similar Questions

The area enclosed between the curve $y = \log_e(x + e)$ and the coordinate axes is

The area between the curve $y = \sin^2 x$,the $x$-axis,and the ordinates $x = 0$ and $x = \frac{\pi}{2}$ is:

The area bounded by the curve $y = x \sin x$ and the $x$-axis between $x = 0$ and $x = 2\pi$ is:

The area bounded by the curve $y=x^3-3x^2+2x$ and the $X$-axis is (in square units)

Find the area bounded by the curves $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module