Which of the following is the possible value/values of c for which the area of the figure bounded by the curves $y=\sin 2x$, the straight lines $x=\pi /6,x=c$ and the abscissa axis is equal to 1/2?

The area enclosed by $f\left(x\right)=12+ax\pm x^2$ coordinates axes and the ordinates at $x=3\left(f\left(3\right)>0\right)$ is 45 sq. units. If m and n are the x-axis intercepts of the graph of y=f(x), then the value of (m+n+a) si ___.

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

Computing area with parametrically represented boundaries : If the boundary of a figure is represented by parametric equation, i.e., $x=x\left(t\right),y=\left(t\right),$ then the area of the figure is evaluated by one of the three formulas :
$S=-\stackrel{\beta }{\underset{\alpha }{\int }}y\left(t\right)x^{\prime }\left(t\right)dt,$
$S=\stackrel{\beta }{\underset{\alpha }{\int }}x\left(t\right)y^{\prime }\left(t\right)dt,$
$S=\frac{1}{2}\stackrel{\beta }{\underset{\alpha }{\int }}\left(x{y}^{\prime }-y{x}^{\prime }\right)dt,$
Where $\alpha \text{and}\beta$ are the values of the parameter t corresponding respectively to the beginning and the end of the traversal of the curve corresponding to increasing t.
If the curve given by parametric equation $x=t-t^3,y=1-t^4$ forms a loop for all values of $t\in \left[-1,1\right]$ then the area of the loop is

$\text{Two curves }{C}_1\equiv {\left[f\left(y\right)\right]}^{2/3}+{\left[f\left(x\right)\right]}^{1/3}=0\text{and}{C}_2\equiv {\left[f\left(y\right)\right]}^{2/3}+{\left[f\left(x\right)\right]}^{2/3}=12,\text{ satisfying the relation }\left(x-y\right)f\left(x+y\right)-\left(x+y\right)f\left(x-y\right)=4xy\left(x^2-y^2\right)$
The area bounded by $C_1\text{and}x+y+2=0$ is

If the area enclosed by curve $y=f\left(x\right)\text{and}y=x^2+2$ between the abscissa $x=2\text{and}x=\alpha ,\alpha >2,is\left({\alpha }^3-4{\alpha }^2+8\right)$ sq. units then find function f(x). It is known that curve y=f(x) lies below the parabola $y=x^2+2.$

Area bounded by the min. $\left\{\left|x\right|,\left|y\right|\right\}=1$ and the max. $\left\{\left|x\right|,\left|y\right|\right\}=2$ is

The area of the figure bounded by the parabola ${\left(y-2\right)}^2=x-1,$ the tangent to it at the point with the ordinate x=3, and the x-axis is