The area inside the parabola y^2 = 4ax,betwee | vedclass

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

Question

(C) The area $A$ bounded by the parabola $y^2 = 4ax$ and the lines $x = a$ and $x = 4a$ is given by the integral:$A = 2 \int_{a}^{4a} y \, dx$Since $y

Vedclass · Accepted Answer

$\frac{56a^2}{3}$ sq. units

Vedclass · Answer

(C) The area $A$ bounded by the parabola $y^2 = 4ax$ and the lines $x = a$ and $x = 4a$ is given by the integral:$A = 2 \int_{a}^{4a} y \, dx$Since $y^2 = 4ax$,we have $y = 2\sqrt{a}\sqrt{x}$ (considering the area above the $x$-axis and multiplying by $2$ for symmetry).$A = 2 \int_{a}^{4a} 2\sqrt{a} \sqrt{x} \, dx = 4\sqrt{a} \int_{a}^{4a} x^{1/2} \, dx$$A = 4\sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_{a}^{4a} = 4\sqrt{a} \cdot \frac{2}{3} \left[ x^{3/2} \right]_{a}^{4a}$$A = \frac{8\sqrt{a}}{3} \left[ (4a)^{3/2} - a^{3/2} \right]$$A = \frac{8\sqrt{a}}{3} \left[ 8a\sqrt{a} - a\sqrt{a} \right]$$A = \frac{8\sqrt{a}}{3} \left[ 7a\sqrt{a} \right] = \frac{56a^2}{3}$ sq. units.

The area inside the parabola $y^2 = 4ax$,between the lines $x = a$ and $x = 4a$ is equal to

Similar Questions

The area (in sq units) of the region bounded by $x=-1$, $x=2$, $y=x^2+1$ and $y=2x-2$ is

The area of the region bounded by the lines $2x - y + 1 = 0$, $y = -1$, $y = 3$ and the y-axis is

Find the area bounded by the curves $\{(x, y): y \geq x^{2} \text{ and } y=|x|\}$.

The area of the region bounded by the curve $y = \tan x$,the $X$-axis,and the line $x = \frac{\pi}{3}$ is

Area of the region bounded by the curve $y=\cos x$,$x=\frac{\pi}{2}$ and $x=\frac{3 \pi}{2}$ is . . . . . . sq. units.

Vedclass Test Series

Exam Paper Generator

Online Exam Module