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 us first sketch the region whose area is to be found out. The required area is the area included between the curves:
x² + y² = 1 and x + y = 1
x² + y² = 1 is a circle having centre at (0, 0) and radius 1 unit.
x + y = 1 is a st. line passing through the points (1, 0) and (0, 1).
Required area is shown shaded as in the sd joining figure.

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

We have : 4x ² + y² \(\le \) 4 and 2x + y \(\ge \) 2
Draw the ellipse \(\frac { { x }^{ 2 } }{ 1 } +\frac { { y }^{ 2 } }{ 4 } =1\), which is an upward ellipse and the line \(\frac { x }{ 1 } +\frac { y }{ 2 } =1\).
The reqd.region is shown shaded in the figure.
Therefore, Reqd.area = \(\overset { 1 }{ \underset { 0 }{ \int { } } } \sqrt { { 4-4x }^{ 2 } } dx-\overset { 1 }{ \underset { 0 }{ \int { } } } (2-2x)dx\)

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

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

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

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

Let us first sketch the region whose area is to be found out. The required area is the area included between the curves x² = y and y = IxI.
The graph of x² = y is parabola with vertex (0,0) and axis as y - axis.
The graph of y = I x I 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
\(=\left[ \overset { 1 }{ \underset { 0 }{ \int { } } } xdx-\overset { 1 }{ \underset { 0 }{ \int { } } } { x }^{ 2 }dx \right] \)
\(=\left[ \left| \frac { { x }^{ 2 } }{ 2 } \right| _{ 0 }^{ 1 }-\left| \frac { { x }^{ 3 } }{ 3 } \right| _{ 0 }^{ 1 } \right] \)
\(=2\left[ \left( \frac { 1 }{ 2 } -0 \right) -\left( \frac { 1 }{ 3 } -0 \right) \right] =2\left[ \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right] \)
\(=2\left( \frac { 1 }{ 6 } \right) =\frac { 1 }{ 3 } sq.unit.\)

(i) Required area = \(\overset { 2 }{ \underset { 1 }{ \int { } } } ydx\)  [Taking vertical strips]

\(=\overset { 2 }{ \underset { 1 }{ \int { } } } { x }^{ 2 }dx=\left[ \frac { { x }^{ 3 } }{ 3 } \right] _{ 1 }^{ 2 }\)
\(=\frac { 8 }{ 3 } -\frac { 1 }{ 3 } =\frac { 7 }{ 3 } sq.units.\)
(ii) Required. area
\(=\overset { 5 }{ \underset { 1 }{ \int { } } } ydx=\overset { 5 }{ \underset { 1 }{ \int { } } } { x }^{ 4 }dx\)
\(=\left[ \frac { { x }^{ 5 } }{ 5 } \right] _{ 1 }^{ 5 }=\frac { { 5 }^{ 5 } }{ 5 } -\frac { 1 }{ 5 } =625-\frac { 1 }{ 5 } \)
\(=\frac { 3125-1 }{ 5 } =\frac { 3124 }{ 5 } \)
\(=624.8sq.units\)

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):

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

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

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

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, x² = 4y, y = 2, and y = 4, and the y-axis is the area ABCD.


\(\text { Area of } \mathrm{ABCD} =\int_{2}^{4} x d y \)
\(=\int_{2}^{4} 2 \sqrt{y} d y \)
\(=2 \int_{2}^{4} \sqrt{y} d y \)
\(=2\left[\frac{y^{\frac{3}{2}}}{3}\right.\)
\(=\left(\frac{4}{2}\right]_{2}^{4} \)
\(=\frac{4}{3}\left[(4)^{\frac{3}{2}}-(2)^{\frac{3}{2}}\right] \)
\(=\frac{4}{3}[8-2 \sqrt{2}] \)
\(=\left(\frac{32-8 \sqrt{2}}{3}\right) \text { units } \)