The area (in sq. units) of the region bounded | vedclass

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

Question

(A) Given curves are $x^2=4y$ and $x=4y-2$. From the second equation,$4y = x+2$. Substituting this into the first equation,we get $x^2 = x+2$,which im

Vedclass · Accepted Answer

$\frac{9}{8}$

Vedclass · Answer

(A) Given curves are $x^2=4y$ and $x=4y-2$. From the second equation,$4y = x+2$. Substituting this into the first equation,we get $x^2 = x+2$,which implies $x^2 - x - 2 = 0$. Solving for $x$,we get $(x-2)(x+1) = 0$,so $x=2$ and $x=-1$. When $x=2$,$y=1$. When $x=-1$,$y=1/4$. The points of intersection are $(2,1)$ and $(-1, 1/4)$. The required area is given by the integral of the upper curve minus the lower curve from $x=-1$ to $x=2$: Area $= \int_{-1}^{2} [\frac{x+2}{4} - \frac{x^2}{4}] dx$ $= \frac{1}{4} [\frac{x^2}{2} + 2x - \frac{x^3}{3}]_{-1}^{2}$ $= \frac{1}{4} [(\frac{4}{2} + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3})]$ $= \frac{1}{4} [(6 - \frac{8}{3}) - (\frac{1}{2} - \frac{5}{3})]$ $= \frac{1}{4} [\frac{10}{3} - (-\frac{7}{6})] = \frac{1}{4} [\frac{20+7}{6}] = \frac{27}{24} = \frac{9}{8} 	ext{ sq. units.}$

The area (in sq. units) of the region bounded by the curve $x^2=4y$ and the straight line $x=4y-2$ is

Similar Questions

The ratio in which the $x$-axis divides the area of the region bounded by the curves $y = x^2 - 4x$ and $y = 2x - x^2$ is:

The area of the region bounded by $y^2=x$ and $y=|x|$ is

The area enclosed between the curves $y = \sin x$,$y = \cos x$ and the $x$-axis for $0 \le x \le \frac{\pi}{2}$ is:

The area of the region $A = \{(x, y): 4x^2 + y^2 \le 8 \text{ and } y^2 \le 4x\}$ is:

If the area lying in the first quadrant and bounded by the circle $x^2+y^2-4x=0$, the parabola $y^2=x$ and the $X$-axis is $A$, then $6A-9\sqrt{3}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module