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 bounded by $y=x^{3}$,$y=8$,and $x=0$ is

The area (in $sq. \ units$) bounded by the curve $y = x^2 + 2x + 1$,the tangent to it at $(1, 4)$,and the $Y$-axis is

The area enclosed by $y=3x-5$, $y=0$, $x=3$, and $x=5$ is

Find the area under the curve $y=x^{4}$ bounded by the lines $x=1$,$x=5$ and the $x$-axis.

The area of the region bounded by the curve $y = |x - 5|$,$y = 0$,$x = 0$,and $x = 2$ is . . . . . . sq. units.

Vedclass Test Series

Exam Paper Generator

Online Exam Module