Given curve \(y^{2}=4 x\) is a parabola having vertex (0, 0) and open right side and the given line is \(x+y=3 \text { or } \frac{x}{3}+\frac{y}{3}=1\) .
which intersects both coordinate axes at (3, 0) and (0, 3).

It is clear from the figure that, the required region is OACBO.
Now, the point of intersection of line x + y = 3 and curve is given by
\( y^{2} =4(3-y) \)
\(\Rightarrow y^{2}+4 y-12 =0 \)
\(\Rightarrow y^{2}+6 y-2 y-12 =0 \)
\(\Rightarrow y(y+6)-2(y+6) =0 \) 
\(\Rightarrow (y-2)(y+6)=0\) 
\(\Rightarrow y=2,-6\) 
When y = - 6, then x - 6 = 3 \(\Rightarrow\) x = 9.
So, the points of intersection are A(1, 2) and D( 9, - 6).
Now, required area = Area of shaded region OABO + Area of shaded region ABCA
\(=\int_{0}^{2} x \text { (parabola) } d y+\int_{2}^{3} x \text { (line) } d y\) 
\(=\int_{0}^{2}\left(\frac{y^{2}}{4}\right) d y+\int_{2}^{3}(3-y) d y=\frac{1}{4}\left[\frac{y^{3}}{3}\right]_{0}^{2}+\left[3 y-\frac{y^{2}}{2}\right]_{2}^{3}\) 
\(=\frac{1}{12}\left[(2)^{3}-(0)^{3}\right]+\left[9-\frac{9}{2}-\left(6-\frac{4}{2}\right)\right]\) 
\(=\frac{1}{12}(8)+\left(3-\frac{5}{2}\right)=\frac{8}{12}+\frac{1}{2}=\frac{8+6}{12}=\frac{14}{12}=\frac{7}{6} \text { sq units }\) 
Hence, the required area is \(\frac{7}{6}\) sq units.

Area of region, \(A=\int_{1}^{2} y d x=\int_{1}^{2} \frac{10}{x^{2}} d x\) 
\(=10 \int_{1}^{2} x^{-2} d x=10\left[\frac{x^{-1}}{-1}\right]_{1}^{2}\) 
\(=-10\left[\frac{1}{x}\right]_{1}^{2}=-10\left[\frac{1}{2}-\frac{1}{1}\right]\) 
\(=-10\left(-\frac{1}{2}\right)=5 \text { sq units }\) 
Also, given area of region B = area of region C
\(\Rightarrow \int_{2}^{p} \frac{10}{x^{2}} d x=\int_{p}^{5} \frac{10}{x^{2}} d x\) 
\(\Rightarrow 10\left[\frac{x^{-2+1}}{-2+1}\right]_{2}^{p}=10\left[\frac{x^{-2+1}}{-2+1}\right]_{p}^{5}\) 
\(\Rightarrow \frac{1}{-1}\left[\frac{1}{x}\right]_{2}^{p}=\frac{1}{-1}\left[\frac{1}{x}\right]_{p}^{5}\) 
\(\Rightarrow -\left[\frac{1}{p}-\frac{1}{2}\right]=-\left[\frac{1}{5}-\frac{1}{p}\right]\) 
\(\Rightarrow \frac{1}{p}+\frac{1}{p}=\frac{1}{5}+\frac{1}{2} \)
\(\Rightarrow \frac{2}{p}=\frac{2+5}{10} \Rightarrow \frac{2}{p}=\frac{7}{10} \)
\(\therefore \ p=\frac{20}{7} \) 
Hence, the value of p is \(\frac{20}{6}\) 

Getting the points of intersection as (4, 0), (0, 3).
\(\therefore\) Required area 
\(=\int _{ 0 }^{ 4 }{ \frac { 3 }{ 4 } } \sqrt { 16-{ x }^{ 2 } } dx-\frac { 1 }{ 4 } \int _{ 0 }^{ 4 }{ \left( 12-3x \right) dx } \)
\(=\left[ \frac { 3 }{ 4 } \left[ \frac { x }{ 2 } \sqrt { 16-{ x }^{ 2 } } +8{ sin }^{ -1 }\frac { x }{ 4 } \right] -\frac { 1 }{ 4 } \left( 12x-\frac { { 3x }^{ 2 } }{ 2 } \right) \right] \)
\(=\left( \frac { 3 }{ 4 } .8.\frac { \pi }{ 2 } -6 \right) =\left( 3\pi -6 \right) \ sq.units\)

Given, x² \(\le \) y....(i)
and y \(\le \) |x| ...(ii)
Clearly curve (i) is parabola directed upward with vertex at (0, 0) and symmetrical about y-axis.
Also,\(y=\left| x \right| =\begin{cases} x\quad ,\quad if\quad x\ge 0 \\ -x,\quad if\quad x<0 \end{cases}\)
The lines y = x and y = -x both passes through origin and have slope of +1 & -1 respectively.
\(\Rightarrow\)Required Area = 2 Standard Area on a side 
\(=-\left[ \left( \frac { 1 }{ 2 } -1 \right) -\left( \frac { 16 }{ 2 } -4 \right) \right] \)
\(=2\left[ \frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] _{ 0 }^{ 1 }\)
\(=2\left[ \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right] =2\times \frac { 1 }{ 6 } \)
\(=\frac { 1 }{ 3 } sq.units\)

Let the line AB, BC and CA have equations 2x + y = 4, 3x - 2y = 6 and x - 3y + 5 = 0 resp. B(2, 0), C(4, 3) and A(1, 2)

Area \(=\int _{ 1 }^{ 4 }{ \frac { 1 }{ 3 } } \left( x+5 \right) dx-\int _{ 1 }^{ 2 }{ \left( 4-2x \right) } dx-\int _{ 2 }^{ 4 }{ \frac { 1 }{ 2 } } \left( 3x-6 \right) dx\)
\(=\frac { 1 }{ 3 } \left[ \frac { { \left( x+5 \right) }^{ 2 } }{ 2 } \right] _{ 1 }^{ 4 }+2\left[ \frac { { \left( 2-x \right) }^{ 2 } }{ 2 } \right] _{ 1 }^{ 2 }-\frac { 3 }{ 2 } \left[ \frac { { \left( x-2 \right) }^{ 2 } }{ 2 } \right] _{ 2 }^{ 4 }\)
\(=\left( \frac { 81 }{ 6 } -\frac { 36 }{ 6 } \right) +\left( 0-1 \right) -\frac { 3 }{ 4 } .4\)
\(=\frac { 15 }{ 2 } -1-3=\frac { 7 }{ 2 } sq.units.\)

Eqns. of PQ, QR and PR are:
PQ : y = \(\frac{5}{2}\)(x-2)
QR : y = 9-x
PR : y = \(\frac{3}{4}\) (x-2)
Req. Area = \(=\int _{ 2 }^{ 4 }{ \frac { 5 }{ 2 } \left( x-2 \right) dx+\int _{ 4 }^{ 6 }{ \left( 9-x \right) dx-\int _{ 2 }^{ 6 }{ \frac { 3 }{ 4 } \left( x-2 \right) dx } } } \)
​​​\(=\left[ \frac { 5 }{ 4 } { \left( x-2 \right) }^{ 2 } \right] _{ 2 }^{ 4 }-\frac { 1 }{ 2 } \left[ \left( 9-x \right) ^{ 2 } \right] _{ 4 }^{ 6 }-\frac { 3 }{ 8 } \left[ \left( x-2 \right) ^{ 2 } \right] _{ 2 }\)
= 5 + 8 - 6 = 7 sq.units.

\(\therefore\)  Point of intersection are (2, 1) and  (-1. 1/4).

Required Area of region (\(\triangle\) OABO)
\(=\int _{ -1 }^{ 2 }{ \frac { x+2 }{ 4 } } dx-\int _{ -1 }^{ 2 }{ \frac { { x }^{ 2 } }{ 4 } } dx\)
\(=\frac { 1 }{ 4 } \left[ \frac { { \left( x+2 \right) }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] _{ -1 }^{ 2 }\)
\(=\frac { 1 }{ 4 } \left[ \frac { 16 }{ 3 } -\frac { 5 }{ 6 } \right] =\frac { 27 }{ 24 } sq.units\)
\(=\frac { 9 }{ 8 } sq.units\)

Area (EABCDE) = Area (ABFE) + Area (CBFD)
​​​\(=\int _{ -3 }^{ -1 }{ \left( \left| x+1 \right| +1 \right) dx+\int _{ -1 }^{ 3 }{ \left( \left| x+1 \right| +1 \right) dx } } \)
\(=\int _{ -3 }^{ -1 }{ \left( -x \right) dx } +\int _{ -1 }^{ 3 }{ \left( x+2 \right) dx } \)
\(=\left[ -\frac { { x }^{ 2 } }{ 2 } \right] _{ -3 }^{ -1 }+\left[ \frac { { \left( x+2 \right) }^{ 2 } }{ 2 } \right] _{ -1 }^{ 3 }\)
\(=-\frac { 1 }{ 2 } \left( 1-9 \right) +\frac { 1 }{ 2 } \left( 25-1 \right) =16\ sq.unit\)

Given 4y = x²
\(\Rightarrow 4\frac { dy }{ dx } =2x\)
\(\Rightarrow \frac { dy }{ dx } =\frac { x }{ 2 } \)
\(\Rightarrow \left( \frac { dy }{ dx } \right) _{ x=2 }=1\)
The equation of tangent is y = x - 1.

The required Area = Shaded Area of graph
\(\Rightarrow -\left[ \int _{ 2 }^{ 3 }{ \left\{ \left( x-1 \right) -\frac { x }{ 2 } \right\} dx+\int _{ 3 }^{ 6 }{ \left[ \frac { \left( x+3 \right) }{ 3 } -\frac { x }{ 2 } \right] dx } } \right] \)
\(\Rightarrow -\left[ \int _{ 2 }^{ 3 }{ \left( x-1 \right) dx+\frac { 1 }{ 3 } \int _{ 3 }^{ 6 }{ \left( x+3 \right) dx-\frac { 1 }{ 2 } \int _{ 2 }^{ 6 }{ x } dx } } \right] \)
\(\Rightarrow -\left[ \left[ \frac { { x }^{ 2 } }{ 2 } -x \right] _{ 2 }^{ 3 }+\frac { 1 }{ 3 } \left[ \frac { { x }^{ 2 } }{ 2 } +3x \right] _{ 3 }^{ 6 }-\frac { 1 }{ 4 } \left[ { x }^{ 2 } \right] _{ 2 }^{ 6 } \right] \)
\(\Rightarrow \left[ \frac { 9 }{ 2 } -3-2+2 \right] -\frac { 1 }{ 3 } \left[ 18+18-\frac { 9 }{ 2 } -9 \right] +\frac { 1 }{ 4 } \left[ 36-4 \right] \) 
= 1 sq.unit

x² + y² = 32; y = x, point of intersection is y = 4.

Required Area = \(\int _{ 0 }^{ 4 }{ y\quad dy+\int _{ 4 }^{ 4\sqrt { 2 } }{ \sqrt { 32-{ y }^{ 2 }dy } } } \)
\(=\left[ \frac { { y }^{ 2 } }{ 2 } \right] _{ 0 }^{ 4 }+\left[ \frac { y }{ 2 } \sqrt { 32-{ y }^{ 2 } } +16{ sin }^{ -1 }\frac { y }{ 4\sqrt { 2 } } \right] _{ -4 }^{ 4\sqrt { 2 } }\)
\(\Rightarrow =8+\left( 0+16.\frac { \pi }{ 2 } \right) -\left( 8+16.\frac { \pi }{ 4 } \right) =\ 4\pi \)