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

The given curves viz. parabolas are :
y² = 4ax and x² = 4ay
These two intersects at O (0,0) and B (4a,4a). [Solve!]
Therefore, Reqd.area = Shaded area OABCO
= area OABL - area OCBL
= (area under parabola y² = 4ax) - (area under parabola x² = 4ay)

\(=\overset { 4a }{ \underset { 0 }{ \int { } } } \sqrt { 4ax } dx-\overset { 4a }{ \underset { 0 }{ \int { } } } \frac { { x }^{ 2 } }{ 4a } dx\)
\(=2\sqrt { a } \overset { 4a }{ \underset { 0 }{ \int { } } } { x }^{ 1/2 }dx-\frac { 1 }{ 4a } \overset { 4a }{ \underset { 0 }{ \int { } } } { x }^{ 2 }dx\)
\(=2\sqrt { a } \left[ \frac { { x }^{ 3/2 } }{ 3/2 } \right] _{ 0 }^{ 4a }-\frac { 1 }{ 4a } \left[ \frac { { x }^{ 3 } }{ 3 } \right] _{ 0 }^{ 4a }\)
\(=\frac { 4\sqrt { a } }{ 3 } \left[ { (4a) }^{ 3/2 }-0 \right] -\frac { 1 }{ 2a } \left[ { 64a }^{ 3 }-0 \right] \)
\(=\frac { 32 }{ 3 } { a }^{ 2 }-\frac { 16 }{ 3 } { a }^{ 2 }=\frac { { 16a }^{ 2 } }{ 3 } sq.units.\)

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)
\(=\overset { 3 }{ \underset { 1 }{ \int { } } } \frac { 3x+1 }{ 2 } dx+\overset { 6 }{ \underset { 3 }{ \int { } } } \frac { 21-2x }{ 3 } dx-\overset { 6 }{ \underset { 1 }{ \int { } } } \frac { x+9 }{ 5 } dx\)
\(=\frac { 1 }{ 2 } \left[ \frac { 3 }{ 2 } { x }^{ 2 }+x \right] _{ 1 }^{ 3 }+\frac { 1 }{ 3 } \left[ 21x-{ x }^{ 2 } \right] _{ 3 }^{ 6 }-\frac { 1 }{ 5 } \left[ \frac { { x }^{ 2 } }{ 2 } +9x \right] _{ 1 }^{ 6 }\)

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

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
\(=\overset { 4 }{ \underset { -2 }{ \int { } } } \left( 4+y\frac { { y }^{ 2 } }{ 2 } \right) dy=\left[ 4y+\frac { { y }^{ 2 } }{ 2 } -\frac { { y }^{ 3 } }{ 6 } \right] _{ -2 }^{ 4 }\)
\(=\left( 4(4)+\frac { 16 }{ 2 } -\frac { 64 }{ 6 } \right) -\left( -8+\frac { 4 }{ 2 } +\frac { 8 }{ 6 } \right) \)
\(=\left( 16+8\frac { 64 }{ 6 } \right) -\left( -8+2+\frac { 8 }{ 6 } \right) \)
\(=\left( 24-\frac { 64 }{ 6 } \right) +\left( 6-\frac { 8 }{ 6 } \right) =30-\frac { 72 }{ 6 } =30-12=18sq.units\)

The given curve is ay² = x³
The region is shown as shaded in the figure:

Therefore, Required area, ALMB
\(=\overset { 2a }{ \underset { a }{ \int { } } } xdy=\overset { 2a }{ \underset { a }{ \int { } } } { a }^{ 1/3 }{ y }^{ 2/3 }dy\)
\(={ a }^{ 1/3 }\left[ \frac { { y }^{ 5/3 } }{ 5/3 } \right] _{ a }^{ 2a }=\frac { 3 }{ 5 } { a }^{ 1/3 }\left[ { (2a) }^{ 5/3 }-{ a }^{ 5/3 } \right] \)
\(=\frac { 3 }{ 5 } { a }^{ 1/3 }{ a }^{ 5/3 }\left[ { 2.2 }^{ 2/3 }-1 \right] =\frac { 3 }{ 5 } { a }^{ 2 }({ 2.2 }^{ 2/3 }-1)sq.units\)

The given ellipse is :
\(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\) and the st. line is \(\frac { x }{ a } +\frac { y }{ b } =1\)
Reqd. area is the shaded area, as shown in the figure.

\(\therefore \ Shaded\ area=9\overset { a }{ \underset { 0 }{ \int { } } } \frac { b }{ a } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } dx-\overset { a }{ \underset { 0 }{ \int { } } } \frac { b }{ a } (a-x)dx\)
\(=\frac { b }{ a } \left[ \frac { x }{ 2 } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } +\frac { { a }^{ 2 } }{ 2 } { sin }^{ -1 }\frac { x }{ a } \right] _{ 0 }^{ a }-\frac { b }{ a } \left[ ax-\frac { { x }^{ 2 } }{ 2 } \right] _{ 0 }^{ a }\)
\(=\frac { b }{ a } \left[ \left( 0+\frac { { a }^{ 2 } }{ 2 } { sin }^{ -1 }(1) \right) -(0+0) \right] -\frac { b }{ a } \left[ \left( { a }^{ 2 }-\frac { { a }^{ 2 } }{ 2 } \right) -(0-0) \right] \)
\(=\frac { b }{ a } \left[ \frac { { a }^{ 2 } }{ 2 } \left( \frac { \pi }{ 2 } \right) \right] -\frac { b }{ a } \left[ \frac { { a }^{ 2 } }{ 2 } \right] \)
\(=\frac { \pi ab }{ 4 } -\frac { ab }{ 2 } =\frac { ab }{ 4 } (\pi -2)sq.units.\)

The given parabola is y =x²
i.e., x² = \(\frac{1}{4}y\)
This is an upward parabola

\(\therefore \ Reqd.areaABCD=\overset { 4 }{ \underset { 1 }{ \int { } } } xdy\quad [Taking\ horizontal\ strips]\)
\(=\overset { 1 }{ \underset { 1 }{ \int { } } } \frac { 1 }{ 2 } \sqrt { y } dy\)
\([\because { x }^{ 2 }=\frac { 1 }{ 4 } y\Rightarrow x=\pm \frac { 1 }{ 2 } \sqrt { y }\)
\(But\ region\ ABCD\ lies\ in\ 1st\)
\( \therefore x\ is\ +ve]\)
\(=\frac { 1 }{ 2 } \overset { 4 }{ \underset { 1 }{ \int { } } } { y }^{ \frac { 1 }{ 2 } }dy=\frac { 1 }{ 2 } \left[ \frac { { y }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right] _{ 1 }^{ 4 }\)
\(=\frac { 1 }{ 3 } \left[ { 4 }^{ \frac { 3 }{ 2 } }-{ 1 }^{ \frac { 3 }{ 2 } } \right] =\frac { 1 }{ 3 } [8-1]\)
\(=\frac { 7 }{ 3 } sq.units\)

Part (B) is the correct answer.
Reason: Reqd.area = ar (quad. OAB) - ar (\(\Delta \)OAB)

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

The given lines are :
y = 2x + 1
y = 3x + 1 and x = 4
Solving (1) and (2): we get x = 0, y = 1.
Thus lines (1) and (2) intersect at A (0,1).
Similarly lines (2) and (3) intersect at B (4,13) and lines (3) and (1) intersect at C (4,9).
Therefore, Reqd. area = ar(ABC)

\(=ar(OABL)-ar(OACL)\)
\(=\overset { 4 }{ \underset { 0 }{ \int { } } } (3x+1)dx-\overset { 4 }{ \underset { 0 }{ \int { } } } (2x+1)dx\)
\(=\left[ \frac { { 3x }^{ 2 } }{ 2 } +x \right] _{ 0 }^{ 4 }-\left[ { x }^{ 2 }+x \right] _{ 0 }^{ 4 }\)
\(=\left[ \left( \frac { 3 }{ 2 } (16)+4 \right) -(0-0) \right] -[(16+4)-(0+0)]\)
\(=28-20=8sq.units\)

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 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).

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

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.

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

\(\left( \frac { \pi }{ 4 } -\frac { 2 }{ 3 } \right) { a }^{ 2 }\) sq units.

Given, x² ≤ y..(i)
and y ≤ |x|...(ii)
Clearly, curve (i) is parabola directed upward with vertex at (0, 0) and symmetrical about y-axis.
Also \(y=|x|=\begin{cases} x\quad if\quad x\ge 0 \\ -x\quad ,if\quad x<\quad 0 \end{cases}\)
The lines y = x and y = - x both passes through origin and have slope of + 1& - 1 respectively.

⇒ Required Area = 2x Standard Area on a side
= \(-\left[ \left( \frac { 1 }{ 2 } -1 \right) -\left( \frac { 16 }{ 2 } -4 \right) \right] \)
= \(2{ \left[ \frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 2 } \right] }_{ 0 }^{ 1 }\)
= \(2\left[ \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right] =2\times \frac { 1 }{ 6 } \)
= \(\frac { 1 }{ 3 } \) sq.units

Eq.of line AB : y-3 \(=\frac { 7-3 }{ 4-2 } \left( x-2 \right) \)
\(\Rightarrow\) y = 2x-1
Eq. of line BC : y-2 \(=\frac { 7-2 }{ 4-6 } \left( x-6 \right) \)
\(\Rightarrow y=\frac { 5 }{ 2 } x-13\)
Eq. of line AC: y-3 \(=\frac { 3-2 }{ 2-6 } \left( x-2 \right) \)
\(\Rightarrow y=-\frac { x }{ 4 }+\frac { 7 }{ 2 }\)
Area of \(\triangle \)ABC = Area of \(\triangle \)ABD + Area of \(\triangle \)BDC
\(=\int _{ 2 }^{ 4 }{ \left\{ \left( 2x-1 \right) -\left( -\frac { x }{ 4 } +\frac { 7 }{ 2 } \right) \right\} dx } \)
\(=\left[ \frac { 9 }{ 4 } \frac { { x }^{ 2 } }{ 2 } -\frac { 9 }{ 2 } x \right] _{ 2 }^{ 4 }+\left[ \frac { +11 }{ 4 } \frac { { x }^{ 2 } }{ 2 } +\frac { 33 }{ 2 } x \right] _{ -4 }^{ 6 }\)
\(=\frac { 9 }{ 8 } \left( 16-4 \right) -\frac { 9 }{ 2 } \left( 4-2 \right) +\frac { 11 }{ 8 } \left( 36-16 \right) +\frac { 33 }{ 2 } \left( 6-4 \right) \)
\(=\frac { 9 }{ 8 } \times 12-9+\frac { 11 }{ 8 } \times 20+\frac { 33 }{ 2 } \times 2\)
\(=\frac { 27 }{ 2 } -9+\frac { 55 }{ 2 } +33\)
\(=24+\frac { 82 }{ 2 } =24+41=65\ sq.units.\)

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