The area of the region bounded by the curve y | vedclass

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

Question

(C) To find the area bounded by the curve $y^2=9x$ and the line $y=3x$,we first find the points of intersection.Substituting $y=3x$ into $y^2=9x$,we g

Vedclass · Accepted Answer

$\frac{1}{2}$ sq.units

Vedclass · Answer

(C) To find the area bounded by the curve $y^2=9x$ and the line $y=3x$,we first find the points of intersection.Substituting $y=3x$ into $y^2=9x$,we get $(3x)^2=9x$,which implies $9x^2=9x$,so $x^2-x=0$,giving $x(x-1)=0$. Thus,$x=0$ and $x=1$.For $x=0$,$y=0$. For $x=1$,$y=3$.The area is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=1$.$	ext{Area} = \int_0^1 (\sqrt{9x} - 3x) dx$$= \int_0^1 (3\sqrt{x} - 3x) dx$$= 3 \left[ \frac{2}{3}x^{3/2} - \frac{x^2}{2} ight]_0^1$$= 3 \left( \frac{2}{3}(1)^{3/2} - \frac{(1)^2}{2} ight) - 0$$= 3 \left( \frac{2}{3} - \frac{1}{2} ight)$$= 3 \left( \frac{4-3}{6} ight) = 3 	imes \frac{1}{6} = \frac{1}{2} 	ext{ sq.units}$.

The area of the region bounded by the curve $y^2=9x$ and the line $y=3x$ is

Similar Questions

If $(a, 0); a > 0$ is the point where the curve $y = \sin 2x - \sqrt{3} \sin x$ cuts the $x$-axis first,and $A$ is the area bounded by this part of the curve,the origin,and the positive $x$-axis,then:

The area of the region bounded by the curve $y = \cos x$ between $x = 0$ and $x = \frac{3\pi}{2}$ is . . . . . . square units.

If a curve $y = a\sqrt{x} + bx$ passes through the point $(1, 2)$ and the area bounded by the curve,the line $x = 4$,and the $X$-axis is $8$ sq. units,then:

The area of the figure bounded by the parabolas $x=-2y^{2}$ and $x=1-3y^{2}$ is

The area (in sq unit) of the region bounded by the curves $2x = y^2 - 1$ and $x = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module