Application of ...

$$\int_{0}^{\pi }max\left \{ \sin x, \cos x \right \}dx$$ is equal to

The ratio of the area's bounded by the curves $$ \displaystyle y^{2}=12x $$ and $$ \displaystyle x^{2}=12y $$ is divided by the line $$ \displaystyle x=3 $$ is

The function $$ \displaystyle f\left ( x \right )=\max \left \{ x^{2},\left ( 1-x \right )^{2},2x\left ( 1-x \right ) \forall 0\leq x\leq 1\right \} $$ then area of the region bounded by the curve $$ \displaystyle y=f\left ( x \right ) $$, x-axis and $$ \displaystyle x=0,x=1 $$ is equals,

The area lying in the first quadrant inside the circle $${ x }^{ 2 }+{ y }^{ 2 }=12$$ and bounded by the parabolas $${ y }^{ 2 }=4x,{ x }^{ 2 }=4y$$ is:

The area bounded by $$ \displaystyle x=a\cos ^{3}\theta,y=a\sin ^{3}\theta $$ is:

Compute the area of the curvilinear triangle bounded by the y-axis & the curve, $$\displaystyle\ y=\tan x$$ & $$ \displaystyle\ y=(2/3) \cos x$$

Let $$\displaystyle f\left ( x \right )=min \left \{ 1, 1-\cos x, 2\sin x \right \}$$ then $$\displaystyle \int_{0}^{\pi}f\left ( x \right )dx$$ is

The area of the plane region bounded by the curves $$x+2y^{2}=0$$ and $$x+3y^{2}=1$$ is

Find the area bounded by the curves $$\displaystyle x = y^{2}$$ and $$\displaystyle x = 3-2y^{2}$$

The area bounded by the curves $$y=\log x$$, $$y=\log \left | x \right |$$, $$y=\left | \log x \right |$$ and $$y=\left | \log \left | x \right | \right |$$