Application of ...

The area(in sq. units) of the smaller portion enclosed between the curves, $$x^2+y^2=4$$ and $$y^2=3x$$, is.

For $$a > 0$$, let the curves $$C_1 : y^2 = ax$$ and $$C_2 : x^2 = ay$$ intersect at origin $$O$$ and a point $$P$$. Let the line $$x = b (0 < b < a)$$ intersect the chord $$OP$$ and the x-axis at points $$Q$$ and $$R$$, rspectively. If the line $$x=b$$ bisects the area bounded by the curves, $$C_1$$ and $$C_2$$, and the area of $$\Delta OQR = \dfrac{1}{2}$$, then '$$a$$' satisfies the equation:

Given $$f(x)=\begin{cases} x,0\le x<\dfrac { 1 }{ 2 }  \\ \dfrac { 1 }{ 2 } ,x=\dfrac { 1 }{ 2 }  \\ 1-x,\dfrac { 1 }{ 2 } <x\le 1 \end{cases}$$ and $$g(x)=\left(x-\dfrac{1}{2}\right)^{2},x\epsilon R$$, Then the area (in sq.units) of the region bounded by the curves $$y=f(x)$$ and $$y=g(x)$$ between the lines $$2x=1$$ and $$2x=\sqrt{3}$$, is:

If the area enclosed between the curves $$y=kx^2$$ and $$x=ky^2$$, $$(k > 0)$$, is $$1$$ square unit. Then $$k$$ is?

If the area of the region bounded by the curves, $$y=x^2, y=\displaystyle\frac{1}{x}$$ and the lines $$y=0$$ and $$x=t(t > 1)$$ is $$1$$ sq. unit, then t is equal to?

The area (in sq. units) of the region $$\{(x, y) \in R^2 |4x^2 \le y \le 8x + 12\}$$ is :

The area enclosed between the curves $$\mathrm{y}=\mathrm{a}\mathrm{x}^{2}$$ and $$\mathrm{x}=\mathrm{a}\mathrm{y}^{2} (\mathrm{a}>0)$$ is 1 sq. unit, then the value of a is

The area of the region $$\left \{(x, y) : xy \leq 8, 1 \leq y\leq x^{2}\right \}$$ is

The area of the region between the curves $$\mathrm{y}=\sqrt{\dfrac{1+\sin \mathrm{x}}{\cos \mathrm{x}}}$$ and $$\mathrm{y}=\sqrt{\dfrac{1-\sin \mathrm{x}}{\cos \mathrm{x}}}$$ bounded by the lines $$\mathrm{x}=0$$ and $$\displaystyle \mathrm{x}=\frac{\pi}{4}$$ is

Let the functions $$f:R\rightarrow R$$ and $$𝑔:R\rightarrow R$$ be defined by $$f(x)=e^{ x-1 }-e^{ -|x-1| }$$ and $$g(x)=\dfrac{1}{2}(e^{x-1}+e^{1-x})$$. Then the area of the region in the first quadrant bounded by the curves $$𝑦=𝑓(𝑥), 𝑦=𝑔(𝑥) $$ and $$x=0$$ is