Let S(alpha) = {(x,y) : y^2 leq x, 0 leq x le | vedclass

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

Question

(C) The region $S(\alpha)$ is bounded by the parabola $y^2 = x$ and the vertical line $x = \alpha$.The area $A(\alpha)$ is given by:$A(\alpha) = \int_

Vedclass · Accepted Answer

$4\left(\frac{4}{25}\right)^{\frac{1}{3}}$

Vedclass · Answer

(C) The region $S(\alpha)$ is bounded by the parabola $y^2 = x$ and the vertical line $x = \alpha$.The area $A(\alpha)$ is given by:$A(\alpha) = \int_{0}^{\alpha} 2\sqrt{x} \, dx = 2 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{\alpha} = \frac{4}{3} \alpha^{3/2}$.Given the ratio $A(\lambda) : A(4) = 2 : 5$,we have:$\frac{\frac{4}{3} \lambda^{3/2}}{\frac{4}{3} 4^{3/2}} = \frac{2}{5}$$\frac{\lambda^{3/2}}{8} = \frac{2}{5}$$\lambda^{3/2} = \frac{16}{5}$$\lambda = \left( \frac{16}{5} \right)^{2/3} = \left( \frac{16^2}{5^2} \right)^{1/3} = \left( \frac{256}{25} \right)^{1/3} = 4 \left( \frac{4}{25} \right)^{1/3}$.Thus,$\lambda = 4\left(\frac{4}{25}\right)^{1/3}$.

Let $S(\alpha) = \{(x,y) : y^2 \leq x, 0 \leq x \leq \alpha\}$ and $A(\alpha)$ be the area of the region $S(\alpha)$. If for a $\lambda, 0 < \lambda < 4, A(\lambda) : A(4) = 2 : 5$,then $\lambda$ equals:

Similar Questions

The area of the region bounded by the curve $y=x^2$ and the line $y=16$ is

The area of the region bounded by the curve $y=4x-x^{2}$ and the $x$-axis is

The value of $a$ $(a > 0)$ for which the area bounded by the curves $y = \frac{x}{6} + \frac{1}{x^2}$,$y = 0$,$x = a$,and $x = 2a$ has the least value,is

The area bounded by the curve $y = f(x)$,$x-$ axis and ordinates $x = 1$ and $x = b$ is $(b - 1)\sin(3b + 4)$. Then $f(x)$ is:

The area of the region bounded by the $y$-axis,$y=\cos x$,and $y=\sin x$,when $0 \leq x \leq \frac{\pi}{4}$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module