\(\pi\)sq.units.

\(\left| x \right| +\left| y \right| =1\Rightarrow x+y=1,x-y=1,-x+y=1,-x-y=1\)
i.e., x + y = 1, x - y = 1, x - y = -1, x + y = -1
These are shown as in the following figure:

Therefore, Required area = 4 (area OAB)
\(=4\overset { 1 }{ \underset { 0 }{ \int { } } } (1-x)dx\)
\(=4\left[ x-\frac { { x }^{ 2 } }{ 2 } \right] _{ 0 }^{ 1 }\)
\(=4\left[ \left( 1-\frac { 1 }{ 2 } \right) -\left( 0-0 \right) \right] \)
\(=4\left( \frac { 1 }{ 2 } \right) =2sq.units\)

The given parabola is x² = y
The given line is y = x + 2
From (1) and (2), 
\({ x }^{ 2 }=x+2\Rightarrow { x }^{ 2 }-x-2=0\Rightarrow (x-2)(x+1)=0\Rightarrow x=2,-1\)
When x = 2, y = (2)² = 4.
When x = -1, y = (-1)² = 1
Thus (1) and (2) intersect at (2,4) and (-1,1).
Therefore, Reqd.area = Shaded area

\(=\overset { 2 }{ \underset { -1 }{ \int { } } } (x+2)dx-\overset { 2 }{ \underset { -1 }{ \int { } } } { x }^{ 2 }dx\)
\(=\left[ \frac { { x }^{ 2 } }{ 2 } +2x \right] _{ -1 }^{ 2 }-\left[ \frac { { x }^{ 3 } }{ 3 } \right] _{ -1 }^{ 2 }\)
\(=\left[ \left( \frac { 4 }{ 2 } +4 \right) -\left( \frac { 1 }{ 2 } -2 \right) \right] -\frac { 1 }{ 3 } [8-(-1)]\)
\(=\left( 6+\frac { 3 }{ 2 } \right) -3=3+\frac { 3 }{ 2 } =\frac { 9 }{ 2 } sq.units.\)

The given parabola is x² = y
This is an upward parabola with vertex (0, 0)
y = IxI represents the st. lines:
y = x and y = -x
y = x meets (1) at O (0, 0) and A (1, 1)
y = -x meets (1) at O (0, 0) and A (-1, 1)
Therefore, Required area = 2 (Shaded area in first quadrant)
\(=2\left( \overset { 1 }{ \underset { 0 }{ \int { } } } xdx-\overset { 1 }{ \underset { 0 }{ \int { } } } { x }^{ 2 }dx \right) \)
\(=2\left( \left[ \frac { { x }^{ 2 } }{ 2 } \right] _{ 0 }^{ 1 }-\left[ \frac { { x }^{ 3 } }{ 3 } \right] _{ 0 }^{ 1 } \right) \)
\(=2\left( \left( \frac { 1 }{ 2 } -0 \right) -\left( \frac { 1 }{ 3 } -0 \right) \right) =2\left( \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right) \)
\(=2\left( \frac { 1 }{ 6 } \right) =\frac { 1 }{ 3 } sq.unit.\)

We have : x = y² and x = 4
Now ar (OAB) \(=2\overset { 4 }{ \underset { 0 }{ \int { } } } \sqrt { x } dx\)

\(=2\left[ \frac { { x }^{ 3/2 } }{ 3/2 } \right] _{ 0 }^{ 4 }=\frac { 4 }{ 3 } \left[ { 4 }^{ 3/2 }-0 \right] \)
\(=\frac { 4 }{ 3 } [8-0]=\frac { 32 }{ 3 } \)
\(and\ ar(OCD)=2\overset { a }{ \underset { 0 }{ \int { } } } \sqrt { x } dx\)
\( =2\left[ \frac { { x }^{ 3/2 } }{ 3/2 } \right] _{ 0 }^{ a }=\frac { 4 }{ 3 } [{ a }^{ 3/2 }-0]\)
\(=\frac { 4 }{ 3 } { a }^{ 3/2 }\)
\(By\ the\ question,\frac { 4 }{ 3 } { a }^{ 3/2 }=\frac { 1 }{ 2 } \left( \frac { 32 }{ 3 } \right) \)
\(\Rightarrow \frac { 4 }{ 3 } { a }^{ 3/2 }=\frac { 16 }{ 3 } \Rightarrow { a }^{ 3/2 }=4\)
\(\Rightarrow a={ (4) }^{ 2/3 }\)
\(Hence, a={ 4 }^{ 2/3 }\)

The given line is \(x=\sqrt { 3 } y\) and the given circle is x² + y² = 4
From (1) \(y=\frac { x }{ \sqrt { 3 } } \)
Putting in (2), \({ x }^{ 2 }+\frac { { x }^{ 2 } }{ 3 } =4\Rightarrow \frac { { 4x }^{ 2 } }{ 3 } =4\Rightarrow { x }^{ 2 }=3\Rightarrow x=\pm \sqrt { 3 } \)
When \(x=\pm \sqrt { 3 } ,y=\pm 1\)
Thus P is \((\sqrt { 3 } ,1)\)
Now Reqd, area = ar (OPL) + ar (PLA)
\(\overset { \sqrt { 3 } }{ \underset { 0 }{ \int { } } } \frac { x }{ \sqrt { 3 } } dx+\overset { 2 }{ \underset { \sqrt { 3 } }{ \int { } } } \sqrt { 4-{ x }^{ 2 } } dx\ [Taking\ vertical\ strips]\)
\(=\frac { 1 }{ \sqrt { 3 } } \overset { \sqrt { 3 } }{ \underset { 0 }{ \int { } } } xdx+\overset { 2 }{ \underset { \sqrt { 3 } }{ \int { } } } \sqrt { { 2 }^{ 2 }-{ x }^{ 2 } } dx\)
\(=\frac { 1 }{ \sqrt { 3 } } \left[ \frac { { x }^{ 2 } }{ 2 } \right] _{ 0 }^{ \sqrt { 3 } }+\left[ \frac { x\sqrt { 4-{ x }^{ 2 } } }{ 2 } +\frac { 4 }{ 2 } { sin }^{ -1 }\frac { x }{ 2 } \right] _{ \sqrt { 3 } }^{ 2 }\)
\(=\frac { 1 }{ 2\sqrt { 3 } } [3-0]+\left[ \left\{ { 0+2sin }^{ -1 }(1) \right\} -\left\{ \frac { \sqrt { 3 } (1) }{ 2 } +2{ sin }^{ -1 }\frac { \sqrt { 3 } }{ 2 } \right\} \right] \)
\(=\frac { \sqrt { 3 } }{ 2 } +2\left( \frac { \pi }{ 2 } \right) -\frac { \sqrt { 3 } }{ 2 } -2\left( \frac { \pi }{ 3 } \right) \)
\(=\pi -\frac { 2\pi }{ 3 } =\frac { \pi }{ 3 } sq.units\)

The area of the region bounded by the curve, y² = x, the lines, x = 1 and x = 4, and the x-axis is the area ABCD.

\(\text { Area of } \mathrm{ABCD} =\int^{1} y d x \)
\(=\int^{4} \sqrt{x} d x \)
\(=\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{4} \)
\(=\frac{2}{3}\left[(4)^{\frac{3}{2}}-(1)^{\frac{3}{2}}\right] \)
\(=\frac{2}{3}[8-1] \)
\(=\frac{14}{3} \text { units } \)

\(\frac { 15 }{ 2 } \) sq units.

On the graph it represents the area bounded by the curve \(y=\left| x-2 \right| \), x-axis and between the ordinates at x = -1 and x = 3.

\(\text { Curve is } y=\frac{1}{x}, x \text { -axis and between } x=1, x=4\)
\(\text { Area } =\int_{1}^{4} \frac{1}{x} d x=[\log |x|]_{1}^{4} \)
\(=\log 4-\log 1 \)
\(=\log 4 \text { sq units. } \)

Let given equation of lines are
AB : 3x - y - 3 = 0   ...(i)
BC: 2x + y - 12 = 0    ...(ii)
CA: x - 2y - 1 = 0  ..(iii)
Solving equation (i) and (ii), we get
x = 3, y = 6 ⇒ B(3, 6)
Solving equation (ii) and (iii), we get
x = 5, Y = 2 ⇒ C (5, 2)
Solving equation (i) and (iii), we get
x = 1, y = 0 ⇒ A (1, 0)

Required area of ΔABC = Area of ΔABP + Area of trapezium BCQP- Area of ΔACQ
=   \(\int _{ AB }^{ }{ y\quad dx } +\int _{ BC }^{ }{ y\quad dx } -\int _{ AC }^{ }{ y\quad dx } \)  
=  \(\int _{ 1 }^{ 3 }{ (3x-3) } dx+\int _{ 3 }^{ 5 }{ (12-2x) } dx-\int _{ 1 }^{ 5 }{ \frac { 1 }{ 2 } } (x-1)dx\)
\(=3{ \left[ \frac { { x }^{ 2 } }{ 2 } -x \right] }_{ 1 }^{ 3 }+{ \left[ 12x-{ x }^{ 2 } \right] }_{ 3 }^{ 5 }-\frac { 1 }{ 2 } { \left[ \frac { { x }^{ 2 } }{ 2 } -x \right] }_{ 1 }^{ 5 }\)
= \(3\left[ \left( \frac { 9 }{ 2 } -3 \right) -\left( \frac { 1 }{ 2 } -1 \right) \right] +[(60-25)-(36-9)]-\frac { 1 }{ 2 } \left[ \left( \frac { 25 }{ 2 } -5 \right) -\left( \frac { 1 }{ 2 } -1 \right) \right] \)
= \(3[2]+[8]-\frac { 1 }{ 2 } [8]\)
= 6 + 8 - 4 = 10 sq. units

Given region {(x,y) : (x² + y² \(\le \) 1 \(\le \) (x + y)}
\(\Rightarrow\) The given region is bounded inside the circle x² + y² = 1 and above the line x + y = 1 as shown in the figure.

\(\therefore\) Required area of the shaded portion
\(=\left| \int _{ circle }^{ }{ ydx } \right| -​​​​\left| \int _{ line }^{ }{ ydx } \right| \)
\(\left| \int _{ 0 }^{ 1 }{ \sqrt { 1-{ x }^{ 2 } } dx } \right| -\left| \int _{ 0 }^{ 1 }{ (1-x)dx } \right| \)
\(=\left| { \left[ \frac { x }{ 2 } \sqrt { 1-{ x }^{ 2 } } +\frac { 1 }{ 2 } { sin }^{ -1 }x \right] }_{ 0 }^{ 1 } \right| -{ \left| \left[ { x-\frac { { x }^{ 2 } }{ 2 } } \right] _{ 0 }^{ 1 } \right| }\)
\(=\left| 0+\frac { 1 }{ 2 } { sin }^{ -1 }(1)-0-0 \right| -\left| 1-\frac { 1 }{ 2 } -0+0 \right| \)
\(=\frac { 1 }{ 2 } \left( \frac { \pi }{ 2 } \right) -\frac { 1 }{ 2 } =\frac { \pi }{ 4 } -\frac { 1 }{ 2 } Sq.units\)

Point of intersection of Circle and line is (3, 3).
\(\therefore \ Area=\int _{ 0 }^{ 3 }{ x\quad dx } +\int _{ 3 }^{ 3\sqrt { 2 } }{ \sqrt { 18-{ x }^{ 2 } } } dx\)
\(=\left[ \frac { { x }^{ 2 } }{ 2 } \right] _{ 0 }^{ 3 }+\left[ \frac { x }{ 2 } \sqrt { 18-{ x }^{ 2 } } +9{ sin }^{ -1 }\frac { x }{ 3\sqrt { 2 } } \right] _{ 3 }^{ 3\sqrt { 2 } }\)
\(=\frac { 9 }{ 2 } +\frac { 9\pi }{ 2 } -\frac { 9 }{ 2 } -\frac { 9\pi }{ 4 } \)
\(=\frac { 9\pi }{ 4 } sq.units.\)