The area (in sq. units) bounded by the curves $x^2=9y$,$(x-6)^2=9y$ and the $X$-axis is

  • A
    $0$
  • B
    $1$
  • C
    $2$
  • D
    $4$

Explore More

Similar Questions

One of the points of intersection of the curves $y=1+3x-2x^2$ and $y=\frac{1}{x}$ is $\left(\frac{1}{2}, 2\right)$. Let the area of the region enclosed by these curves be $\frac{1}{24}(\ell \sqrt{5}+m)-n \log_{e}(1+\sqrt{5})$,where $\ell, m, n \in N$. Then $\ell+m+n$ is equal to

If the area enclosed between the curves $y^2 = 4kx$ and $y = kx$ for $k > 0$ is $\frac{2}{3}$ sq. units, then $k =$ ?

The area bounded by the curves $y = \sqrt{x}$,$2y + 3 = x$ and the $x$-axis in the $1^{st}$ quadrant is

The area between the parabolas $y^2 = 4ax$ and $x^2 = 8ay$ is

Difficult
View Solution

If $A$ is the area bounded by the curves $y = |\cos x|$ and $y = 5 - \frac{4}{\pi} |x - \pi|$ in $x \in [0, 2\pi]$,then the value of $\left( \frac{A}{2} + 2 \right)$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo