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.

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

\({ A }_{ 1 }=\int _{ x=0 }^{ x=5 }{ ydx } =\int _{ 0 }^{ 5 }{ { x }^{ 2 }dx } \)
\(=\frac { 1 }{ 3 } \left| { x }^{ 3 } \right| _{ 0 }^{ 5 }=\frac { 125 }{ 3 } units^{ 2 }\)
\({ A }_{ 2 }=\int _{ y=0 }^{ x=5 }{ xdy } =\int _{ 0 }^{ 5 }{ { y }^{ 2 }dy } \quad \begin{bmatrix} \because & y=\sqrt { x } \\ or & x={ y }^{ 2 } \end{bmatrix}\)
\({ A }_{ 2 }=\left[ \frac { y^{ 3 } }{ 3 } \right] _{ 0 }^{ 5 }=\frac { 125 }{ 3 } units^{ 2 }\)
Thus, A = A,
Value: Team work is better than individual work.

(i) As usual, area of the field - 6 sq. units.
(ii) No.of trees = 4 \(\times\)26 = 104.
(iii) Plants provide us oxygen and play major role in rains, so plantation is necessary for all human beings.

Let us first sketch the region whose area is to be found out. The required area is the area included between the curves:
x ² = y and y = IxI.
The graph of x² = y is a parabola with vertex (0,0) and axis as y - axis.
The graph of y - IxI is the union of lines y = x, x \(\ge \) 0 and y = x, x < 0.
Solving x² = y and y = x, we get the points of intersection as O (0,0) and A (1,1).

Solving, x² = y and y = -x, we get the points of intersection as O (0,0) and B (-1,1).
Therefore, Required area = area OAL + area OBM = 2 area OAL
\(=2\left[ \overset { 1 }{ \underset { 0 }{ \int { } } } xdx-\overset { 1 }{ \underset { 0 }{ \int { } } } { x }^{ 2 }dx \right] =2\left[ \left| \frac { { x }^{ 2 } }{ 2 } \right| _{ 0 }^{ 1 }-\left| \frac { { x }^{ 3 } }{ 3 } \right| _{ 0 }^{ 1 } \right] \)
\(=2\left[ \left( \frac { 1 }{ 2 } -0 \right) -\left( \frac { 1 }{ 3 } -0 \right) \right] =2\left[ \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right] \)
\(=2\left( \frac { 1 }{ 6 } \right) =\frac { 1 }{ 3 } sq.units\)

The given ellipse is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
The ellipse is symmetrical about x - axis.
[because On changing y to -y, (1) remains unchanged]
The ellipse is symmetrical about y - axis.
[because On changing x to -x, (1) remains unchanged]
Thus area of the ellipse = (Shaded area) = (area OAB)

\(=\overset { a }{ \underset { 0 }{ \int { } } } ydx\)
[Taking vertical strips]
\(=\overset { a }{ \underset { 0 }{ \int { } } } \frac { b }{ a } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } dx\)
\(=\left[ \because \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\Rightarrow \frac { { y }^{ 2 } }{ { b }^{ 2 } } =1-\frac { { x }^{ 2 } }{ { a }^{ 2 } } \\ \Rightarrow { y }^{ 2 }=\frac { { b }^{ 2 } }{ { a }^{ 2 } } ({ a }^{ 2 }-{ x }^{ 2 })\Rightarrow y=\frac { b }{ a } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } (\because y>0) \right] \)
\(=\frac { b }{ a } \overset { a }{ \underset { 0 }{ \int { } } } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } dx=\frac { b }{ a } \left[ \frac { \sqrt { { a }^{ 2 }-{ x }^{ 2 } } }{ 2 } +\frac { { a }^{ 2 } }{ 2 } { sin }^{ -1 }\frac { x }{ a } \right] _{ 0 }^{ a }\)
\(=\frac { b }{ a } \left[ \left\{ \frac { a }{ 2 } (0)+\frac { { a }^{ 2 } }{ 2 } { sin }^{ -1 }(1) \right\} -\left\{ 0+0 \right\} \right] =\frac { b }{ a } \left[ \frac { { a }^{ 2 } }{ 2 } .\frac { \pi }{ 2 } \right] =\frac { \pi ab }{ 4 } \)

y² = 4x is right - handed parabola.

Therefore, Required area, ABCD = \(\overset { 4 }{ \underset { 1 }{ \int { } } } ydx\) [Taking vertical strips]
\(\overset { 4 }{ \underset { 1 }{ \int { } } } 2\sqrt { x } dx\\ \)
[\({ y }^{ 2 }=4x\Rightarrow y=\pm 2\sqrt { x } .\) But region ABCD lies in 1^st quadrant, Therefore y is +ve]
\(=2\overset { 4 }{ \underset { 1 }{ \int { } } } { x }^{ 1/2 }dx=2\left[ \frac { { x }^{ 3/2 } }{ 3/2 } \right] _{ 1 }^{ 4 }=\frac { 4 }{ 3 } [{ 4 }^{ 3/2 }-1]\)
\(=\frac { 4 }{ 3 } [8-1]=\frac { 28 }{ 3 } =9\frac { 1 }{ 3 } sq.units.\)

Part (B) is the correct answer.
Reason: The given curves are y = cos x and \(y=sinx;0\le x\le \frac { \pi }{ 2 } \)
The curves intersect, where
\(cosx=sinx\Rightarrow tanx=1\Rightarrow x=\frac { \pi }{ 4 } .\)
And \(y=cos\frac { \pi }{ 4 } =sin\frac { \pi }{ 4 } =\frac { 1 }{ \sqrt { 2 } } \)

Therefore, Reqd, area = ar (OPBO) = ar (OPA) + ar (APBA)
\(=\overset { \frac { 1 }{ \sqrt { 2 } } }{ \underset { 0 }{ \int { } } } { sin }^{ -1 }ydy+\overset { 1 }{ \underset { \frac { 1 }{ \sqrt { 2 } } }{ \int { } } } { cos }^{ -1 }ydy\)
\(=\sqrt { 2 } -1\)

Part (C) is the correct answer.
Reason: The given curve is y = x IxI.
When x > 0, the curve is y = x²
When x < 0, the curve is y = - x²

\(\therefore \ Reqd.area=\left| \overset { 0 }{ \underset { -1 }{ \int { } } } { -x }^{ 2 }dx \right| +\overset { 1 }{ \underset { 0 }{ \int { } } } { x }^{ 2 }dx\)
\(=\left| \left[ -\frac { { x }^{ 3 } }{ 3 } \right] _{ -1 }^{ 0 } \right| +\left[ \frac { { x }^{ 3 } }{ 3 } \right] _{ 0 }^{ 1 }\)
\(=\left| -0-\frac { 1 }{ 3 } \right| +\left[ \frac { 1 }{ 3 } -0 \right] =\frac { 1 }{ 3 } +\frac { 1 }{ 3 } =\frac { 2 }{ 3 } \)

The given ellipse is \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1\)
Since (1) is symmetrical about both axes,
Therefore, area of the ellipse = 4 (Shaped area).

\(=4(area\ OAB)\)
\(But\ area\ OAB=\overset { 4 }{ \underset { 0 }{ \int { } } } ydx\quad [Taking\ vertical\ strips]\)
\(=\overset { 4 }{ \underset { 0 }{ \int { } } } \frac { 3 }{ 4 } \sqrt { 16-{ x }^{ 2 } } dx\)
\([\because \frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1\Rightarrow \frac { { y }^{ 2 } }{ 9 } =1-\frac { { x }^{ 2 } }{ 16 } \Rightarrow y=\frac { 3 }{ 4 } \sqrt { 16-{ x }^{ 2 } } (\because y>0)]\)
\(=\frac { 3 }{ 4 } \left[ \frac { x\sqrt { 16-{ x }^{ 2 } } }{ 2 } +\frac { 16 }{ 2 } { sin }^{ -1 }\frac { x }{ 4 } \right] _{ 0 }^{ 4 }\)
\(=\frac { 3 }{ 4 } [[2(0)+8{ sin }^{ -1 }(1)]-[0-0]]\)
\(=\frac { 3 }{ 4 } \left[ 8\frac { \pi }{ 2 } \right] =3\pi \)
\(\therefore From\ (2),area\ of\ the\ ellipse =4(3\pi )=12\pi sq.units.\)

\(\left( 4\pi +\frac { 32 }{ 3 } \right) sq\quad units.\)

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 curves are y = 2x....(i)
and x - y = 4...(ii)
Obviously, curve (i) is right handed parabola having vertex at (0, 0) and axis along +ve direction of x-axis while curve (ii) is a straight line.

For intersection point of curve (i) and (ii) (x-4)² = 2x
\(\Rightarrow\) x²-8x + 16 = 2x
\(\Rightarrow\) x²-10x +16 = 0
\(\Rightarrow\) x²-8x-2x+16 = 0
\(\Rightarrow\) x(x-8)-2(x-8) = 0
\(\Rightarrow\) (x-8) (x-2) = 0
\(\Rightarrow\) x = 2, 8
\(\Rightarrow\) y = -2, 4
Intersection points are (2, -2), (8, 4)
Therefore, required area =  Area of shaded region
\(=\int _{ -2 }^{ 4 }{ \left( y+4 \right) dy-\int _{ -2 }^{ 4 }{ \frac { { y }^{ 2 } }{ 2 } } dy } \)
\(=\int _{ -2 }^{ 4 }{ \left( y+4 \right) dy-\int _{ -2 }^{ 4 }{ \frac { { y }^{ 2 } }{ 2 } } dy } \)
\(=\frac { 1 }{ 2 } .\left[ 64-4 \right] -\frac { 1 }{ 6 } \left[ 64+8 \right] \)
\(=30-\frac { 72 }{ 6 } =18\ sq.units\)

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.