The area bounded by the parabola y = 4x^2,the | vedclass

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

Question

(C) Given the equation of the parabola is $y = 4x^2$. Expressing $x$ in terms of $y$,we get $x^2 = \frac{y}{4}$,which implies $x = \frac{\sqrt{y}}{2}$

Vedclass · Accepted Answer

$\frac{7}{3} 	ext{ sq. units}$

Vedclass · Answer

(C) Given the equation of the parabola is $y = 4x^2$. Expressing $x$ in terms of $y$,we get $x^2 = \frac{y}{4}$,which implies $x = \frac{\sqrt{y}}{2}$ (since the area is in the first quadrant,we take the positive root). The area bounded by the curve,the $y$-axis,and the lines $y = 1$ and $y = 4$ is given by the integral: $	ext{Area} = \int_{1}^{4} x \, dy = \int_{1}^{4} \frac{\sqrt{y}}{2} \, dy$. $= \frac{1}{2} \int_{1}^{4} y^{1/2} \, dy$. $= \frac{1}{2} \left[ \frac{y^{3/2}}{3/2} ight]_{1}^{4} = \frac{1}{2} 	imes \frac{2}{3} \left[ y^{3/2} ight]_{1}^{4}$. $= \frac{1}{3} [4^{3/2} - 1^{3/2}] = \frac{1}{3} [8 - 1] = \frac{7}{3} 	ext{ sq. units}$.

The area bounded by the parabola $y = 4x^2$,the $y$-axis,and the lines $y = 1$ and $y = 4$ is:

Similar Questions

Find the area lying above the $x-$ axis and included between the circle $x^{2}+y^{2}=8x$ and inside the parabola $y^{2}=4x$.

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

Let $y$ be the function which passes through $(1, 2)$ having slope $(2x + 1)$. The area bounded between the curve and $x$-axis is

The area of the region enclosed between the parabola $y^{2}=x$ and the line $y=mx$ is $\frac{1}{48}$. Then, the value of $m$ is

The area bounded by the $y-$axis,$y=\cos x$ and $y=\sin x$ when $0 \leq x \leq \frac{\pi}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module