The area of the region lying in the first qua | vedclass

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

Question

(A) Given the curve $y = 4x^2$, we express $x$ in terms of $y$ as $x = \sqrt{\frac{y}{4}} = \frac{\sqrt{y}}{2}$.The area $A$ in the first quadrant bou

Vedclass · Accepted Answer

$\frac{1}{3}[8 - 2\sqrt{2}]$ sq. units

Vedclass · Answer

(A) Given the curve $y = 4x^2$, we express $x$ in terms of $y$ as $x = \sqrt{\frac{y}{4}} = \frac{\sqrt{y}}{2}$.The area $A$ in the first quadrant bounded by the $Y$-axis and the lines $y = 2$ and $y = 4$ is given by the integral:$A = \int_{2}^{4} x \, dy = \int_{2}^{4} \frac{\sqrt{y}}{2} \, dy$.$A = \frac{1}{2} \int_{2}^{4} y^{1/2} \, dy = \frac{1}{2} \left[ \frac{y^{3/2}}{3/2} \right]_{2}^{4} = \frac{1}{2} \cdot \frac{2}{3} [y^{3/2}]_{2}^{4}$.$A = \frac{1}{3} [4^{3/2} - 2^{3/2}] = \frac{1}{3} [8 - 2\sqrt{2}]$ sq. units.

The area of the region lying in the first quadrant and bounded by the curve $y = 4x^2$, the $Y$-axis, and the lines $y = 2$ and $y = 4$ is:

Similar Questions

Area inside the parabola ${y^2} = 4ax$ between the lines $x = a$ and $x = 4a$ is equal to

If $S$ be the area of the region enclosed by $y=e^{-x^2}, y=0, x=0$,and $x=1$. Then
$(A) S \geq \frac{1}{e}$
$(B) S \geq 1-\frac{1}{e}$
$(C) S \leq \frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$
$(D) S \leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$

The area of the region bounded by the curves $y = \sin x$,$y = \cos x$ and $x = 0$ is

The area of the region $\{(x, y):|x-y| \leq y \leq 4 \sqrt{x}\}$ is

The graphs of $f(x) = x^2$ and $g(x) = cx^3$ (where $c > 0$) intersect at the points $(0, 0)$ and $\left( \frac{1}{c}, \frac{1}{c^2} \right)$. If the area of the region lying between these graphs over the interval $[0, 1/c]$ is equal to $2/3$,then the value of $c$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module