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
\(=2\left[ \overset { 1 }{ \underset { 0 }{ \int { } } } xdx-\overset { 1 }{ \underset { 0 }{ \int { } } } { x }^{ 2 }dx \right] =2\left[ \left| \frac { { x }^{ 2 } }{ 2 } \right| _{ 0 }^{ 1 }-\left| \frac { { x }^{ 3 } }{ 3 } \right| _{ 0 }^{ 1 } \right] \)
\(=2\left[ \left( \frac { 1 }{ 2 } -0 \right) -\left( \frac { 1 }{ 3 } -0 \right) \right] =2\left[ \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right] \)
\(=2\left( \frac { 1 }{ 6 } \right) =\frac { 1 }{ 3 } sq.units\)

Let the sides AB, BC and CA of \(\Delta \)ABC be:
3x - 2y + 1 = 0
2x + 3y - 21 = 0 and 
x - 5y + 9 = 0
Solving (3) and (1), we get A as (1,2).
Solving (1) and (2), we get B as (3,5)
Solving (2) and (3), we get C as (6,3).
Now ar (\(\Delta \)ABC) = ar (ALMB) + ar (BMNC) - ar (ALNC)
\(=\overset { 3 }{ \underset { 1 }{ \int { } } } \frac { 3x+1 }{ 2 } dx+\overset { 6 }{ \underset { 3 }{ \int { } } } \frac { 21-2x }{ 3 } dx-\overset { 6 }{ \underset { 1 }{ \int { } } } \frac { x+9 }{ 5 } dx\)
\(=\frac { 1 }{ 2 } \left[ \frac { 3 }{ 2 } { x }^{ 2 }+x \right] _{ 1 }^{ 3 }+\frac { 1 }{ 3 } \left[ 21x-{ x }^{ 2 } \right] _{ 3 }^{ 6 }-\frac { 1 }{ 5 } \left[ \frac { { x }^{ 2 } }{ 2 } +9x \right] _{ 1 }^{ 6 }\)

\(=\frac { 1 }{ 2 } \left[ \left( \frac { 27 }{ 2 } +3 \right) -\left( \frac { 3 }{ 2 } +1 \right) \right] +\frac { 1 }{ 3 } \left[ \left( 126-36 \right) -\left( 63-9 \right) \right] -\frac { 1 }{ 5 } \left[ \left( 18+54 \right) -\left( \frac { 1 }{ 2 } +9 \right) \right] \)
\(=\frac { 1 }{ 2 } \left[ 12+2 \right] +\frac { 1 }{ 3 } \left[ 90-54 \right] -\frac { 1 }{ 5 } \left[ 72-\frac { 19 }{ 2 } \right] \)
\(=7+12-\frac { 125 }{ 10 } =19-\frac { 125 }{ 10 } =\frac { 190-125 }{ 10 } \)
\(=\frac { 65 }{ 10 } =\frac { 13 }{ 2 } =6.5\quad sq.units.\)

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]
\(=2\overset { 4 }{ \underset { 1 }{ \int { } } } { x }^{ 1/2 }dx=2\left[ \frac { { x }^{ 3/2 } }{ 3/2 } \right] _{ 1 }^{ 4 }=\frac { 4 }{ 3 } [{ 4 }^{ 3/2 }-1]\)
\(=\frac { 4 }{ 3 } [8-1]=\frac { 28 }{ 3 } =9\frac { 1 }{ 3 } sq.units.\)

The given parabola is y² = 2x and the given st. line is x - y = 4
Solving (1) and (2):
From (2), x = 4 + y
Putting in (1), y² = 8 + 2y \(\Rightarrow\) y² - 2y - 8 = 0 \(\Rightarrow\) (y - 4) (y + 2) = 0 \(\Rightarrow\) y = 4, - 2
When y = 4, then from (3), x = 4 + 4 = 8
When y = -2, then from (3), x = 4 - 2 = 2
Thus the line (2) cuts parabola (2) in the points A (2, -2) and B(8, 4).

The region is shown as shaded in the above figure.
Therefore, Reqd.area
\(=\overset { 4 }{ \underset { -2 }{ \int { } } } \left( 4+y\frac { { y }^{ 2 } }{ 2 } \right) dy=\left[ 4y+\frac { { y }^{ 2 } }{ 2 } -\frac { { y }^{ 3 } }{ 6 } \right] _{ -2 }^{ 4 }\)
\(=\left( 4(4)+\frac { 16 }{ 2 } -\frac { 64 }{ 6 } \right) -\left( -8+\frac { 4 }{ 2 } +\frac { 8 }{ 6 } \right) \)
\(=\left( 16+8\frac { 64 }{ 6 } \right) -\left( -8+2+\frac { 8 }{ 6 } \right) \)
\(=\left( 24-\frac { 64 }{ 6 } \right) +\left( 6-\frac { 8 }{ 6 } \right) =30-\frac { 72 }{ 6 } =30-12=18sq.units\)

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

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

Part (B) is the correct answer.
Reason: The parabola y² = 4x
intersects the line y = 2x
at. O(0,0) and A (1,2)

\(\therefore \ Reqd.area=\overset { 1 }{ \underset { 0 }{ \int { } } } \sqrt { 4x } dx-\overset { 1 }{ \underset { 0 }{ \int { } } } (2x)dx\)
\(=2\left[ \frac { { x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right] _{ 0 }^{ 1 }-2\left[ \frac { { x }^{ 2 } }{ 2 } \right] _{ 0 }^{ 1 }\)
\(=\frac { 4 }{ 3 } [1-0]-\frac { 2 }{ 2 } [1-0]=\frac { 4 }{ 3 } -1\)
\(=\frac { 1 }{ 3 } sq.unit.\)

The given curve is x² = 4y
which is an upward parabola with vertex (0,0).
The given line is x = 4y - 2
Solving (1) and (2):

\( { (4y-2) }^{ 2 }=4y\)
\(\Rightarrow { 16y }^{ 2 }-16y+4=4y\)
\(\Rightarrow { 16y }^{ 2 }-20y+4=0\Rightarrow { 4y }^{ 2 }-5y+1=0\)
\(\Rightarrow (4y-1)(y-1)=0\Rightarrow y=\frac { 1 }{ 4 } ,1\)
\(When \ y=\frac { 1 }{ 4 } ,thenx=4\left( \frac { 1 }{ 4 } \right) -2\)
\(=1-2=-1\)
\(When \ y=1,thenx=4(1)-2=4-2=2\)
\(Thus(2)meets(1)at\)
\(A\left( -1,\frac { 1 }{ 4 } \right) andB(2,1).\)
\(\therefore Reqd.area=ar(ALOMBDA)-ar(LMBOAL)\)
\(=\overset { 2 }{ \underset { -1 }{ \int { } } } \frac { x+2 }{ 4 } dx-\overset { 2 }{ \underset { -1 }{ \int { } } } \frac { { x }^{ 2 } }{ 4 } dx\)
\(=\frac { 1 }{ 4 } \left[ \frac { { x }^{ 2 } }{ 2 } +2x \right] ^{ 2 }_{ -1 }-\frac { 1 }{ 4 } \left[ \frac { { x }^{ 3 } }{ 3 } \right] _{ -1 }^{ 2 }\)

\(=\frac { 1 }{ 4 } \left[ \left( 2+4 \right) -\left( \frac { 1 }{ 2 } -2 \right) \right] -\frac { 1 }{ 12 } \left[ 8-\left( -1 \right) \right] \)
\(=\frac { 1 }{ 4 } \left[ 6+\frac { 3 }{ 2 } \right] -\frac { 1 }{ 12 } [9]=\frac { 15 }{ 8 } -\frac { 3 }{ 4 } \)
\(=\frac { 15-6 }{ 8 } =\frac { 9 }{ 8 } sq.units.\)

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)
\(\overset { \sqrt { 3 } }{ \underset { 0 }{ \int { } } } \frac { x }{ \sqrt { 3 } } dx+\overset { 2 }{ \underset { \sqrt { 3 } }{ \int { } } } \sqrt { 4-{ x }^{ 2 } } dx\ [Taking\ vertical\ strips]\)
\(=\frac { 1 }{ \sqrt { 3 } } \overset { \sqrt { 3 } }{ \underset { 0 }{ \int { } } } xdx+\overset { 2 }{ \underset { \sqrt { 3 } }{ \int { } } } \sqrt { { 2 }^{ 2 }-{ x }^{ 2 } } dx\)
\(=\frac { 1 }{ \sqrt { 3 } } \left[ \frac { { x }^{ 2 } }{ 2 } \right] _{ 0 }^{ \sqrt { 3 } }+\left[ \frac { x\sqrt { 4-{ x }^{ 2 } } }{ 2 } +\frac { 4 }{ 2 } { sin }^{ -1 }\frac { x }{ 2 } \right] _{ \sqrt { 3 } }^{ 2 }\)
\(=\frac { 1 }{ 2\sqrt { 3 } } [3-0]+\left[ \left\{ { 0+2sin }^{ -1 }(1) \right\} -\left\{ \frac { \sqrt { 3 } (1) }{ 2 } +2{ sin }^{ -1 }\frac { \sqrt { 3 } }{ 2 } \right\} \right] \)
\(=\frac { \sqrt { 3 } }{ 2 } +2\left( \frac { \pi }{ 2 } \right) -\frac { \sqrt { 3 } }{ 2 } -2\left( \frac { \pi }{ 3 } \right) \)
\(=\pi -\frac { 2\pi }{ 3 } =\frac { \pi }{ 3 } sq.units\)

Curves are 4y² = 9x and 3x² = 16y
Eliminating y from two equations, we get
\(4\left(\frac{3 x^{2}}{16}\right)^{2}=9 x\)
\(\Rightarrow x^{4}= \frac{9 \times 16^{2}}{4 \times 9} x=64 x \Rightarrow x=4 \text { or } x=0 \)
\(\therefore \text { area } =\int_{0}^{4}\left(y_{1}-y_{2}\right) d x=\int_{0}^{4}\left(\sqrt{\frac{9 x}{4}}-\frac{3 x^{2}}{16}\right) d x \)
\(=4 \text { sq units } \)

\(y_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 ; y_{2}=\frac{x}{a}+\frac{y}{b}=1\)

\(\text { Area } =\int_{0}^{a}\left(y_{1}-y_{2}\right) d x \)
\(=\int_{0}^{a}\left\{\frac{b}{a} \sqrt{a^{2}-x^{2}}-\frac{b}{a}(a-x)\right\} d x\)
\(=\frac{b}{a}\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}-a x+\frac{x^{2}}{2}\right]_{0}^{a} \)
\(=\frac{b}{a}\left[\left(0+\frac{a^{2}}{2} \cdot \sin ^{-1} 1-a^{2}+\frac{a^{2}}{2}\right)-0\right] \)
\(=\frac{b}{a}\left[\frac{a^{2}}{2} \cdot \frac{\pi}{2}-\frac{a^{2}}{2}\right]=\frac{b}{a} \cdot \frac{a^{2}}{2}\left(\frac{\pi}{2}-1\right) \)
\(\frac { ab }{ 4 } \left( \pi -2 \right) \) 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 
\(=-\left[ \left( \frac { 1 }{ 2 } -1 \right) -\left( \frac { 16 }{ 2 } -4 \right) \right] \)
\(=2\left[ \frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] _{ 0 }^{ 1 }\)
\(=2\left[ \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right] =2\times \frac { 1 }{ 6 } \)
\(=\frac { 1 }{ 3 } sq.units\)

The given curves are
y = x² (parabola)
y = x (line)
These intersect at
O(0, 0) and A(1, 1).
 
The area bounded by the curves = Shaded area
 \(=\int _{ 0 }^{ 1 }{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) } dx\)
\(=\int _{ 0 }^{ 1 }{ \left( x-{ x }^{ 2 } \right) } dx\)
\(=\left[ \frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] _{ 0 }^{ 1 }\)
\(=\frac { 1 }{ 2 } -\frac { 1 }{ 3 } =\frac { 1 }{ 6 } sq.units\)