Given 2a = 20 cm \(\Rightarrow\) a = 10 cm;
2b = 10 cm \(\Rightarrow\)a = 5 cm
\(\therefore\) Equation of the ellipse is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\Rightarrow \frac { { x }^{ 2 } }{ 100 } +\frac { { y }^{ 2 } }{ 25 } =1\Rightarrow \frac { { y }^{ 2 } }{ 25 } =1-\frac { { x }^{ 2 } }{ 100 } =\frac { 100-{ x }^{ 2 } }{ 100 } \)
\(\Rightarrow { y }^{ 2 }=\frac { 25 }{ 100 } (100-{ x }^{ 2 })\)

\(\therefore\) Required volume \(=2\pi \int _{ 0 }^{ 10 }{ { y }^{ 2 }dx } \)
\(=2\pi \int _{ 0 }^{ 10 }{ \frac { 25 }{ 100 } (100-{ x }^{ 2 })dx=\frac { 50\pi }{ 100 } \int _{ 0 }^{ 10 }{ (100-{ x }^{ 2 })dx } } \)
\(=\frac { \pi }{ 2 } { \left[ 100x-\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 10 }=\frac { \pi }{ 2 } \left[ 100-\frac { 1000 }{ 3 } \right] \)
\(V=\frac { \pi }{ 2 } \left( \frac { 3000-1000 }{ 3 } \right) =\frac { \pi }{ 2 } \left( \frac { 2000 }{ 3 } \right) \)
\(=\frac { 1000\pi }{ 3 } \)

Volume of the right circular conical frustum is obtained by revolving the line y = x between x = a and x = b around the x - axis

\(\therefore\) Height of the frustum h = b - a
\(\therefore\)Volume \(=\pi \int _{ a }^{ b }{ { x }^{ 2 }dx } =\pi { \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ a }^{ b }\)
\(=\frac { \pi }{ 3 } [{ b }^{ 3 }-{ a }^{ 3 }]\)
\(=\frac { \pi }{ 3 } (b-a)({ b }^{ 2 }+ab+{ a }^{ 2 })\)
Now, substitute h = b - a, r = a and R = b we get Volume of the conical frustum
\(\frac { \pi }{ 3 } [h({ R }^{ 2 }+rR+{ r }^{ 2 })]\)
Given h = 2 m, r = 1 m, R = 2 m we get
Required volume \(=\frac { \pi }{ 3 } [2(4+2+1)]\)
\(=\frac { \pi }{ 3 } (14)\)
\(=\frac { 14\pi }{ 3 } \)

Find the intersccting point of y = x and y = x²
x² = x
x(x-1) = 0 
x =0, x = 1
If x = 0, y = 0 and if x = 1, y = 1
Points of intersection are (0, 0) and (1, 1)
Equation of the given line is y = x and the parabola is y = x²

Volume  V \(=\pi \int _{ 0 }^{ 1 }{ ({ x }^{ 2 }-{ x }^{ 4 }) } dx\)
\([\because y=x\Rightarrow { y }^{ 2 }={ x }^{ 2 },y={ x }^{ 2 }\Rightarrow { y }^{ 2 }={ x }^{ 4 }]\)
\(=\pi { \left[ \frac { { x }^{ 3 } }{ 3 } -\frac { { x }^{ 5 } }{ 5 } \right] }_{ 0 }^{ 1 }=\pi \left[ \frac { 1 }{ 3 } -\frac { 1 }{ 5 } \right] -0\)
\(=\pi \left[ \frac { 5-3 }{ 15 } \right] =\frac { 2\pi }{ 15 } \)cu.units

Equation of the given curve is y + 1 = x²
\(\therefore\) The vertex of this open upward parabola is (0, -1)

Required volume \(=\int _{ -1 }^{ 3 }{ { x }^{ 2 }dy } \)
\(=\pi \int _{ -1 }^{ 3 }{ (1+y)dy } \)
\(=\pi { \left[ y+\frac { { y }^{ 2 } }{ 2 } \right] }_{ -1 }^{ 3 }=\pi \left[ \left( 3+\frac { 9 }{ 2 } \right) -\left( -1+\frac { 1 }{ 2 } \right) \right] \)
\(=\pi \left[ \left( \frac { 6+9 }{ 2 } \right) -\left( -\frac { 1 }{ 2 } \right) =\pi \left( \frac { 15 }{ 2 } +\frac { 1 }{ 2 } \right) =\left( \frac { 16 }{ 2 } \right) \right] \)
V = 8\(\pi\)

Equation of the given curve is y = e^-2x
Required Volume = \(\pi \int _{ 0 }^{ 1 }{ { { (e }^{ -2x }) }^{ 2 }dx } \)
\(=\pi \int _{ 0 }^{ 1 }{ { e }^{ -4x } } dx=\pi { \left[ \frac { { e }^{ -4x } }{ -4 } \right] }_{ 0 }^{ 1 }\)
\(=\frac { -\pi }{ 4 } \left[ { e }^{ -4 }-{ e }^{ -0 } \right] =-\frac { \pi }{ 4 } \left( { e }^{ -4 }-1 \right) \)
\(V=\frac { \pi }{ 4 } (1-{ e }^{ -4 })\) cubic units

If the region in the first quadrant bounded by the circle x² + y² = r², the x-axis, the lines x = r − h and x = r is revolved about the x-axis, then the solid generated is a spherical cap of height h cut of from a sphere of radius r. Hence, the required volume is given by

\(V=\pi \int _{ r-h }^{ r }{ { y }^{ 2 }dx=\pi \int _{ r-h }^{ r }{ \left( { r }^{ 2 }-{ x }^{ 2 } \right) dx } =\pi } { \left( { r }^{ 2 }x-\frac { { x }^{ 3 } }{ 3 } \right) }_{ r-h }^{ r }\)
\(=\pi \left( { r }^{ 2 }(r-(r-h)-\frac { \left( { r }^{ 3 }-{ (r-h) }^{ 3 } \right) }{ 3 } \right) =\pi \left( { r }^{ 2 }h-\frac { \left( { r }^{ 3 }-\left( { r }^{ 3 }-3{ r }^{ 2 }h+3{ rh }^{ 2 }-{ h }^{ 3 } \right) \right) }{ 3 } \right) \)
\(\\ =\pi \left( \frac { { 3rh }^{ 2 }-{ h }^{ 3 } }{ 3 } \right) =\frac { 1 }{ 3 } \pi { h }^{ 2 }(3r-h)\)

Equation of the given circle is x² + y² = 16 ...(1)
and the parabola is y² = 6x. ...(2)
Substituting (2) in (1) we get,
x² + 6x - 16 = 0 \(\Rightarrow\) (x + 8) (x - 2) = 0
\(\Rightarrow\) (x-4) (x-2) = 0 \(\Rightarrow\) x = 2, 4
\(\Rightarrow\) x = -8, 2

\(\therefore\) Required area = 2
\(\\ \\ \\ \\ \\ \\ \\ \int _{ 0 }^{ 2 }{ Area\ below\ the\ parabola } +\int _{ 2 }^{ 4 }{ Area\ below\ the\ circle } \)
\(=2\left[ \int _{ 0 }^{ 2 }{ \sqrt { 6x } } dx+\int _{ 2 }^{ 4 }{ \sqrt { 16-{ x }^{ 2 } } dx } \right] \)
\(=2\left[ { \left( \frac { \sqrt { 6 } .{ x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right) }_{ 0 }^{ 2 }+{ \left( \frac { x }{ 2 } \sqrt { 16-{ x }^{ 2 } } +\frac { 16 }{ 2 } { sin }^{ -1 }\left( \frac { x }{ 4 } \right) \right) }_{ 0 }^{ 4 } \right] \)
\(=2\left[ \frac { 2 }{ 3 } \sqrt { 6. } 2\sqrt { 2 } +8{ sin }^{ -1 }\left( \frac { 1 }{ 2 } \right) \right] \)
\(=2\left[ \frac { 4\sqrt { 12 } }{ 3 } +8\left( \frac { \pi }{ 2 } \right) -\sqrt { 12 } -8\left( \frac { \pi }{ 6 } \right) \right] \)
\(=2\left[ \frac { 4\sqrt { 12 } }{ 3 } +4\pi -2\sqrt { 3 } -\frac { 4\pi }{ 3 } \right] \)
\(=2\left[ \frac { 8\sqrt { 3 } -6\sqrt { 3 } }{ 3 } +\frac { 12\pi -4\pi }{ 3 } \right] \)
\(=\left[ \frac { 2\sqrt { 3 } }{ 3 } +\frac { 8\pi }{ 3 } \right] =2\times \frac { 2 }{ 3 } \left[ \sqrt { 3 } +4\pi \right] \)
\(=\frac { 4 }{ 3 } \left[ 4\pi +\sqrt { 3 } \right] \)sq.units

Given equation of the parabola is (y-1) = (x-2)²
\(\Rightarrow\) y = (x - 2)² + 1
It vertex is (2, 1) which is the minimum point P. Let Q(x, y) be a point on the parabola given slope of PQ = 2
\(\Rightarrow \frac { y-1 }{ x-2 } =2\ \left[ \because slope=\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \right] \)
\(\Rightarrow\) y-1 = 2(x-2) \(\Rightarrow\) y-1 = 2x-4 \(\Rightarrow\) y = 2x-4+1
\(\Rightarrow\) y = 2x + 3
From (1) and (2), (x-2)²+1 = 2x-3
\(\Rightarrow\) x²-4x + 4 + 1 = 2x - 3 \(\Rightarrow\) x²- 6x + 8 = 0
\(\Rightarrow\) (x-4) (x-2) = 0 \(\Rightarrow\) x = 2, 4

\(\therefore\) Required area \(=\int _{ 2 }^{ 4 }{ ({ y }_{ 1 }-{ y }_{ 2 }) } dx\)
\(=\int _{ 2 }^{ 4 }{ (2x-3)-{ (x-2) }^{ 2 }-1dx } \)
\(=\int _{ 2 }^{ 4 }{ (2x-3-{ x }^{ 2 }+4x-4-1)dx } \)
\(=\int _{ 2 }^{ 4 }{ (-{ x }^{ 2 }+6x-8)dx } \)
\({ \left[ \frac { -{ x }^{ 3 } }{ 3 } +3{ x }^{ 2 }-8x \right] }_{ 2 }^{ 4 }=\left( \frac { -64 }{ 3 } +48-32 \right) -\left( \frac { -8 }{ 3 } +12-16 \right) \)
\(=\left( \frac { -64 }{ 3 } +16 \right) -\left( -\frac { 8 }{ 3 } -4 \right) \)
\(=\left( \frac { -64+48 }{ 3 } \right) -\left( \frac { -8-12 }{ 3 } \right) \)
\(=\frac { 16 }{ 3 } \) sq.units

Equation of the given curves are y² = 4x and x² = 4y

\(\therefore \) Required area \(=\int _{ 0 }^{ 4 }{ \left( \sqrt { 4x } -\frac { { x }^{ 2 } }{ 4 } \right) dx } \)
\(=\int _{ 0 }^{ 4 }{ \left( 2\sqrt { x } -\frac { { x }^{ 2 } }{ 4 } \right) dx } \)
\(={ \left[ \frac { { 2x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right] }_{ 0 }^{ 4 }{ \left[ \frac { 4 }{ 3 } x\sqrt { x } -\frac { { x }^{ 3 } }{ 12 } \right] }_{ 0 }^{ 4 }\)
\(=\frac { 4 }{ 3 } (4)(2)-\frac { 64 }{ 12 } \)
\(=\frac { 32 }{ 3 } -\frac { 32 }{ 6 } =\frac { 64-32 }{ 6 } =\frac { 32 }{ 6 } \)
\(=\frac { 16 }{ 3 } \) sq.units
Yes the area can be divided into 3 equal parts and the area to the divided among his, wife daughter and son is  \(=\frac { 16 }{ 3 } \)sq.units

Given equation of the parabola is y² = x and the
\(\therefore\) y² = y + 2 \(\Rightarrow\) y² - y - 2 = 0
\(\Rightarrow\) (y-2) (y+1) = 0
\(\Rightarrow\) y = -1, 2
y = x - 2
x 0 2 y -2 0 \(\therefore \) Required area = \(\int _{ -1 }^{ 2 }{ ({ x }_{ 1 }-{ x }_{ 2 }) } dy\)

\(=\int _{ -1 }^{ 2 }{ (y+2)-{ y }^{ 2 }dy } ={ \left[ \frac { { y }^{ 2 } }{ 2 } +2y-\frac { { y }^{ 3 } }{ 3 } \right] }_{ -1 }^{ 2 }\)
\(=\left( 2+4-\frac { 8 }{ 3 } \right) -\left( \frac { 1 }{ 2 } -2+\frac { 1 }{ 3 } \right) \)
\(=\left( \frac { 5+12-8 }{ 3 } \right) -\left( \frac { 3-12+2 }{ 6 } \right) \)
\(=\left( \frac { 10 }{ 3 } \right) -\left( \frac { -7 }{ 6 } \right) =\frac { 10 }{ 3 } +\frac { 7 }{ 6 } =\frac { 20+7 }{ 6 } =\frac { 27 }{ 6 } \)
Area \(=\frac{9}{2}\) sq.units

Given equation of the curves are y = tan x, y = cot x.
The intersection of y = tan x and y = cot x are
tan x = cot x \(\Rightarrow\) x = \(\frac{\pi}{2}\)
\(\therefore\) Required area \(=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ (tan\ x-cot\ x)dx } \)
\(={ [-log\ sin\ x+log\ sec\ x] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(=log{ \left[ \frac { sec\ x }{ sin\ x } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }==log{ \left[ \frac { 1 }{ sin\ x\ cos\ x } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(=-log{ (sin\quad x\ cos\ x) }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(=-log\left( sin\frac { \pi }{ 4 } .cos\frac { \pi }{ 4 } \right) +log(sin0\quad cos0)\)
\(=-log\left( \frac { 1 }{ \sqrt { 2 } } .\frac { 1 }{ \sqrt { 2 } } \right) +0\)
\(=-log\left( \frac { 1 }{ 2 } \right) =-(log1-log2)=log2\)

Equation of the given curves are y = sin x  ..(1)
Y = cos x ...(2)
from (1) and (2), sin x = cos x
y = sin x

x	0	\(\pi\)/2
y	0	1

y = cos x

x	0	\(\pi\)/2
y	1	0

\(\Rightarrow x=\frac { \pi }{ 4 } \)

\(\therefore \) Required area = \(2\int _{ \frac { \pi }{ 4 } }^{ \frac { 3\pi }{ 4 } }{ (sinx-cosx)dx } \)
[\(\because\) the area is symmetrical about X - axis]
\(=2{ \left[ -cosx-sinx \right] }_{ \frac { \pi }{ 4 } }^{ \frac { 3\pi }{ 4 } }\)
\(=-2\left[ \left( cos\frac { 3\pi }{ 4 } +sin\frac { 3\pi }{ 4 } \right) -\left( cos\frac { \pi }{ 4 } +sin\frac { \pi }{ 4 } \right) \right] \)
= -2\(\left[ \left( -\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } \right) -\left( \frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } } \right) \right] \)
[cos 135^o = cos(180^o- 45) = -cos 45^o sin135^o = sin(180^o- 45) = -sin 45^o]
\(=-2\left[ \frac { -2 }{ \sqrt { 2 } } \right] =\frac { 4 }{ \sqrt { 2 } } \times \frac { \sqrt { 2 } }{ \sqrt { 2 } } =\frac { 4\sqrt { 2 } }{ 2 } =2\sqrt { 2 } \) 

Given equation of the parabola is y = x²-2x  ...(1)
and the line is y = 2x + 5   ...(2)
From (1) and (2),
x²-2x = 2x+5
\(\Rightarrow\) x²-4x-5 = 0
\(\Rightarrow\)(x-5)(x+1) = 0
\(\Rightarrow\)x = 5, -1
For parabola

x	0	2
y	0	0

For the line

x	0	-5/2
y	5	0

\(\therefore\) Required area \(=\int _{ -1 }^{ 5 }{ ({ y }_{ 1 }-{ y }_{ 2 })dx } \)
\(=\int _{ -1 }^{ 5 }{ (2x+5)-({ x }^{ 2 }-2x) } \)
\(=\int _{ -1 }^{ 5 }{ (2x+5-{ x }^{ 2 }+2x) } dx\)
\(=\int _{ -1 }^{ 5 }{ (4x5-{ x }^{ 2 })dx } \)
\(={ \left( \frac { { 4x }^{ 2 } }{ 2 } +5x-\frac { { x }^{ 3 } }{ 3 } \right) }_{ -1 }^{ 5 }\)
\(={ \left( { 2x }^{ 2 }+5x-\frac { { x }^{ 3 } }{ 3 } \right) }_{ -1 }^{ 5 }\)
\(=\left( 50+25-\frac { 125 }{ 3 } \right) -\left( 2-5+\frac { 1 }{ 3 } \right) \)
\(=\left( \frac { 100 }{ 3 } \right) -\left( \frac { -8 }{ 3 } \right) =\frac { 100 }{ 3 } +\frac { 8 }{ 3 } =\frac { 108 }{ 3 } \)
= 36 sq.units

Equation of the given curve is 2+x-x²+y = 0

x	0	2	-1
y	-2	0	0

\(\Rightarrow\)y = x²-x-2

\(\therefore\) Required area= \(\int _{ -3 }^{ -1 }{ ydx } +\int _{ -1 }^{ 2 }{ -y } dx+\int _{ 2 }^{ 3 }{ ydx } \)
\(\\ =\int _{ -3 }^{ -1 }{ \left( { x }^{ 2 }-x-2 \right) } dx\int _{ -1 }^{ 2 }{ (2+x-{ x }^{ 2 })dx } +\int _{ 2 }^{ 3 }{ ({ x }^{ 2 }-x-2) } dx\)
\(={ \left( \frac { { x }^{ 3 } }{ 3 } -\frac { { x }^{ 2 } }{ 2 } -2x \right) }_{ -3 }^{ -1 }+{ \left( 2x+\frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right) }_{ -1 }^{ 2 }+{ \left( \frac { { x }^{ 3 } }{ 3 } -\frac { { x }^{ 2 } }{ 2 } -2x \right) }_{ 2 }^{ 3 }\)
\(=\left( -\frac { 1 }{ 3 } -\frac { 1 }{ 2 } +2 \right) -\left( -9-\frac { 9 }{ 2 } +6 \right) +\left( 4+2-\frac { 8 }{ 3 } \right) -\left( -2+\frac { 1 }{ 2 } +\frac { 1 }{ 3 } \right) +\left( 9-\frac { 9 }{ 2 } -6 \right) -\left( \frac { 8 }{ 3 } -2-4 \right) \)
\(\\ =\left( \frac { -2-3+6 }{ 6 } \right) -\left( \frac { -6-9 }{ 2 } \right) +\left( \frac { 18-8 }{ 3 } \right) -\left( \frac { -12+3+2 }{ 6 } \right) +\left( \frac { 6-9 }{ 2 } \right) -\left( \frac { 8-24 }{ 3 } \right) \)
\(=\frac { 1 }{ 6 } +\frac { 15 }{ 2 } +\frac { 10 }{ 3 } +\frac { 7 }{ 6 } -\frac { 3 }{ 2 } +\frac { 16 }{ 3 } \)
\(=\frac { 1+45+20+7-9+32 }{ 6 } =\frac { 90 }{ 6 } =15 \)

The lines 5x − 2y = 15, x + y + 4 = 0 intersect at (1, −5). The line 5x − 2y =15 meets the x-axis at (3, 0). The line x + y + 4 = 0 meets the x-axis at (−4, 0). The required area is shaded. It lies below the x-axis. It can be computed either by considering vertical strips or horizontal strips.
When we do by vertical strips, the region has to be divided into two sub-regions by the line x = 1. Then, we get

\(A=\left| \int _{ -4 }^{ 1 }{ ydx } \right| +\left| \int _{ 1 }^{ 3 }{ ydx } \right| \)
\(=\left| \int _{ -4 }^{ 1 }{ (-4-x)dx } \right| +\left| \int _{ 1 }^{ 3 }{ \left( \frac { 5x-15 }{ 2 } \right) dx } \right| \)
\(=\left| { \left( -4x-\frac { { x }^{ 2 } }{ 2 } \right) }_{ -4 }^{ 1 } \right| +\left| { \left( \frac { { 5x }^{ 2 } }{ 4 } -\frac { 15x }{ 2 } \right) }_{ 1 }^{ 3 } \right| \)
\(=\left| \left( -\frac { 9 }{ 2 } \right) -\left( 8 \right) \right| +\left| \left( -\frac { 45 }{ 4 } \right) -\left( -\frac { 25 }{ 4 } \right) \right| \)
\(=\frac { 25 }{ 2 } +5\)
\(=\frac { 35 }{ 2 } \)
When we do by horizontal strips, there is no need to subdivide the region. In this case, the area is bounded on the right by the line 5x − 2y = 15 and on the left by x + y + 4 = 0. So, we get
\(A=\int _{ -5 }^{ 0 }{ [{ x }_{ R }-{ x }_{ L }]dy } =\int _{ -5 }^{ 0 }{ \left[ \frac { 15+2y }{ 5 } -(-4-y) \right] dy } \)
\(=\int _{ -5 }^{ 0 }{ \left[ 7+\frac { 7y }{ 5 } \right] dy={ \left[ 7y+\frac { { 7y }^{ 2 } }{ 10 } \right] }_{ -5 }^{ 0 } } \)
\(\\ =0-\left[ -35+\frac { 35 }{ 2 } \right] =\frac { 35 }{ 2 } \)

First, we find the points of intersection of x + y = 3 and y² − 4x:
x + y = 3 ⇒ y = 3− x.
\(\therefore\) y² = 4x ⇒ (3 -x)² = 4x
⇒ x² −10x + 9 = 0
⇒ x = 1, x = 9 .
\(\therefore\) x = 1 in x + y = 3 \(\Rightarrow\) y = 2, and x = 9 in x + y = 3 ⇒ y = −6 .
\(\therefore\) (1, 2) and (9,−6) are the points of intersection.
The line x + y = 3 meets the y -axis at (0, 3).
The required area is sketched

Viewing in the direction of y -axis, on the right bounding curve is given by
\(x=\begin{cases} \frac { { y }^{ 2 } }{ 4 } ,0\le y\le 2 \\ 3-y,2\le y\le 3 \end{cases}\)
\(\therefore A=\int _{ 0 }^{ 2 }{ xdy+\int _{ 2 }^{ 3 }{ xdy } =\int _{ 0 }^{ 2 }{ \frac { { y }^{ 2 } }{ 4 } dy+\int _{ 2 }^{ 3 }{ (3-y) } dy } } \)
\(={ \left( \frac { { y }^{ 3 } }{ 12 } \right) }_{ 0 }^{ 2 }+{ \left( 3y-\frac { { y }^{ 3 } }{ 2 } \right) }_{ 2 }^{ 3 }=\left( \frac { 8 }{ 12 } -0 \right) +\left( 9-\frac { 9 }{ 2 } \right) -\left( 6-\frac { 4 }{ 2 } \right) =\frac { 7 }{ 6 } \)

The smaller segment is sketched. Here 0

\(A=2\int _{ h }^{ a }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 } } } dx=2{ \left[ \frac { x\sqrt { { a }^{ 2 }-{ x }^{ 2 } } }{ 2 } +\frac { { a }^{ 2 } }{ 2 } { sin }^{ -1 }\left( \frac { x }{ a } \right) \right] }_{ h }^{ a }\)
\(=2\left[ 0+\frac { { a }^{ 2 } }{ 2 } { sin }^{ -1 }(1) \right] -2\left[ \frac { h\sqrt { { a }^{ 2 }-{ h }^{ 2 } } }{ 2 } +\frac { { a }^{ 2 } }{ 2 } { sin }^{ -1 }\left( \frac { h }{ a } \right) \right] \)
\(={ a }^{ 2 }\left( \frac { \pi }{ 2 } \right) -h\sqrt { { a }^{ 2 }-{ h }^{ 2 } } -{ a }^{ 2 }{ sin }^{ -1 }\left( \frac { h }{ a } \right) \)
\(=a^{2}\left[\frac{\pi}{2}-\sin ^{-1}\left(\frac{h}{a}\right)\right]-h \sqrt{a^{2}-h^{2}}\)
\(={ a }^{ 2 }{ cos }^{ -1 }\left( \frac { h }{ a } \right) -h\sqrt { { a }^{ 2 }-{ h }^{ 2 } } \)

The region is sketched. The upper boundary of the region is y = sin x for \(\\ \\ \frac { \pi }{ 4 } \le x\le \frac { 5\pi }{ 4 } \) and the lower boundary of the region is y = cos x for \(\frac { \pi }{ 4 } \le x\le \frac { 5\pi }{ 4 } \). So the required area A is given by

\(A=\int _{ \frac { \pi }{ 4 } }^{ \frac { 5\pi }{ 4 } }{ ({ y }_{ U }-{ y }_{ L })dx= } \int _{ \frac { \pi }{ 4 } }^{ \frac { 5\pi }{ 4 } }{ (sinx-cosx)dx={ [-cosx=sinx] }_{ \frac { \pi }{ 4 } }^{ \frac { 5\pi }{ 4 } } } \)
\(=\left( -sin\frac { 5\pi }{ 4 } -cos\frac { 5\pi }{ 4 } \right) -\left( -sin\frac { \pi }{ 4 } -cos\frac { \pi }{ 4 } \right) \)
\(=\left( -\left( -\frac { 1 }{ \sqrt { 2 } } \right) -\left( -\frac { 1 }{ \sqrt { 2 } } \right) -\left( -\left( \frac { 1 }{ \sqrt { 2 } } \right) -\left( \frac { 1 }{ \sqrt { 2 } } \right) \right) \right) \)
\(=\frac { 2 }{ \sqrt { 2 } } +\frac { 2 }{ \sqrt { 2 } } =2\sqrt { 2 } \)

Both the curves are symmetrical about y -axis
The curve y = |x| is \(y=\begin{cases} x\quad if\quad x\ge 0 \\ -x\quad if\quad x\le 0 \end{cases}\)
It intersects the parabola x² = y at (1, 1) and (−1, 1). The area of the region bounded by the curves is sketched. It lies in the first quadrant as well as in the second quadrant. By symmetry, the required area is twice the area in the first quadrant.

In the first quadrant, the upper curve is y = x, 0 \(\le x\le\)1 and the lower curve is y = x² \(\le x\le\),0 1. Hence, the required area is given by
\(A=2\int _{ 0 }^{ 1 }{ \left[ { y }_{ U }-{ y }_{ L } \right] } dx=2\int _{ 0 }^{ 1 }{ \left[ x-{ x }^{ 2 } \right] dx } \)
\(=2{ \left[ \frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 1 }\)
\(=2\left( \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right) =\frac { 1 }{ 3 } \)

The required area is sketched. One portion of the region lies above the x−axis between x = 0 and x = \(\pi\), and the other portion lies below x−axis between x = \(\pi\) and x = 2\(\pi\). So, the required area is given by 

The equation of the latus-rectum is x = a. It intersects the parabola at the points L(a, 2a) and L₁ (a, −2). The required area is sketched. By symmetry, the required area A is twice the area bounded by the portion of the parabola

y = 2\(\sqrt a \sqrt x\), x -axis, x = 0 and x = a.
Hence, by taking vertical strips, we get
\(A=2\int _{ 0 }^{ a }{ ydx=2\int _{ 0 }^{ a }{ 2\sqrt { a } \sqrt { x } dx=4 } \sqrt { a } } { \left[ \frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } } \right] }_{ 0 }^{ a }\)
\(=4\sqrt { a } \times \frac { 2 }{ 3 } { a }^{ \frac { 3 }{ 2 } }=\frac { 8{ a }^{ 2 } }{ 3 } \)

The ellipse is symmetric about both major and minor axes. It is sketched. So, viewing in the positive direction of y -axis, the required area A is four times the area of the region bounded by the portion of the ellipse in the first quadrant 
\(\left( y=\frac { b }{ a } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } ,0

x -axis, x = 0 and x = a .
Hence, by taking vertical strips, we get \(A=4\int _{ 0 }^{ a }{ ydx=4\int _{ 0 }^{ a }{ \frac { b }{ a } } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } dx } \)
\(=\frac { 4b }{ a } { \left[ \frac { x\sqrt { { a }^{ 2 }-{ y }^{ 2 } } }{ 2 } +\frac { { a }^{ 2 } }{ 2 } { sin }^{ -1 }\left( \frac { x }{ a } \right) \right] }_{ 0 }^{ a }=\frac { 4b }{ a } \times \frac { \pi { a }^{ 2 } }{ 4 } =\pi ab\)

\(Let\quad I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { dx }{ 4x{ sin }^{ 2 }x+5{ cos }^{ 2 }x } } \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sec }^{ 2 }x }{ 4{ tan }^{ 2 }x+5 } } dx\)
(Dividing both numerator and denominator by cos² x).
Let u = tan x 
Then du = sec² x dx
When x = 0, u = tan 0 = 0
When x = \(\frac{\pi}{2}\), u = tan\(\frac{\pi}{2}\) = \(\infty\)
\(\therefore I=\int _{ 0 }^{ \infty }{ \frac { du }{ 4{ u }^{ 2 }+5 } } \) (This is an improper integral)
\(\frac { 1 }{ 4 } \int _{ 0 }^{ \infty }{ \frac { du }{ \left[ { u }^{ 2 }+\left( \frac { \sqrt { 5 } }{ 2 } \right) \right] } =\frac { 1 }{ 4 } \times \frac { 2 }{ \sqrt { 5 } } { \left[ { tan }^{ -1 }\left( \frac { u }{ \frac { \sqrt { 5 } }{ 2 } } \right) \right] }_{ 0 }^{ \infty } } =\frac { 1 }{ 2\sqrt { 5 } } \left( { tan }^{ -1 }\infty -{ tan }^{ -1 }0 \right) =\frac { 1 }{ 2\sqrt { 5 } } \left( \frac { \pi }{ 2 } \right) =\frac { \pi }{ 4\sqrt { 5 } } \)

\(Let\ I=\int _{ 0 }^{ \pi }{ x\left[ { sin }^{ 2 }(sinx)+{ cos }^{ 2 }(cosx) \right] } dx\quad ...(1)\)
\(\left[ \because \int _{ 0 }^{ a }{ f(x)dx=\int _{ 0 }^{ a }{ f(a-x)dx } } \right] \)
\(I=\int _{ 0 }^{ \pi }{ (\pi -x) } [{ sin }^{ 2 }(sin(\pi -x))+{ cos }^{ 2 }(cos(\pi -x)]dx\)
\(=\int _{ 0 }^{ \pi }{ (\pi -x)[{ sin }^{ 2 }(sinx)+{ cos }^{ 2 }(-cosx)]dx } \)
\(=\int _{ 0 }^{ \pi }{ (\pi -x)[{ sin }^{ 2 }(sin\quad x)+{ cos }^{ 2 })(cosx)]dx } \)
\([\because cos(-x)=cos\quad x]\)
\(=\int _{ 0 }^{ \pi }{ \pi [{ sin }^{ 2 }(sinx)+{ cos }^{ 2 }(cosx)] } dx\)
\(-\int _{ 0 }^{ \pi }{ x[{ sin }^{ 2 }(sinx)+{ cos }^{ 2 }(cosx)] } dx\)
\(=\int _{ 0 }^{ \pi }{ \pi [{ sin }^{ 2 }(sinx)+{ cos }^{ 2 }(cosx)]dx-I } [from\quad (1)]\)
\(2I=\pi \int _{ 0 }^{ \pi }{ [{ sin }^{ 2 }(sinx)+{ cos }^{ 2 }(cosx)] } dx\)
\(I=\pi \int _{ 0 }^{ \pi /2 }{ \left[ { sin }^{ 2 }(sin\quad x)+{ cos }^{ 2 }(cos\quad x) \right] } dx\quad ...(2)\)
\(I=\pi \int _{ 0 }^{ \pi /2 }{ \left[ { sin }^{ 2 }(sin\frac { \pi }{ 2 } -x)+{ cos }^{ 2 }(cos\frac { \pi }{ 2 } -x) \right] dx } \)
\(I=\pi \int _{ 0 }^{ \pi /2 }{ [{ sin }^{ 2 }(cosx)+{ cos }^{ 2 }(sin\quad x)]dx] } ....(3)\)
Adding (2) and (3) we get,
\(2I=\pi \int _{ 0 }^{ \pi /2 }{ \left[ { sin }^{ 2 }(sinx)+{ cos }^{ 2 }(cosx) \right] dx } +{ sin }^{ 2 }(cosx)+{ cos }^{ 2 }(sin\quad x)]dx\)
\(I=\pi \int _{ 0 }^{ \pi /2 }{ (1+1)dx } \ [\because { sin }^{ 2 }\theta +{ cos }^{ 2 }\theta =1]\)
\(=\pi (2){ [x] }_{ 0 }^{ \frac { \pi }{ 2 } }=2\pi \left[ \frac { \pi }{ 2 } -0 \right] ={ \pi }^{ 2 }\ \)
\( \therefore I=\frac { { \pi }^{ 2 } }{ 2 } \)

Let \(I=\int _{ 0 }^{ \pi }{ \frac { xsinx }{ 1+sinx } dx }...(1)\)
By the property, \(\left[ \because \int _{ 0 }^{ a }{ f(x)dx } =\int _{ 0 }^{ a }{ f(a-x)dx } \right] \)
we get \(I=\int _{ 0 }^{ \pi }{ \frac { (\pi -x)sin(\pi -x) }{ 1+sin(\pi -x) } } dx\)
\(=\int _{ 0 }^{ \pi }{ \frac { (\pi -x)sinx }{ 1+sinx } dx } ...(2)\)
\((1)+(2)\rightarrow \)
\(2I=\int _{ 0 }^{ \pi }{ \frac { xsinx }{ 1+sinx } dx+\int _{ 0 }^{ \pi }{ \frac { (\pi -x)sinx }{ 1+sinx } dx } } \)
\(=\int _{ 0 }^{ \pi }{ \frac { xsinx+(\pi -x)sinx }{ 1+sinx } dx } \)
\(=\int _{ 0 }^{ \pi }{ \frac { xsinx+(\pi -x)sinx }{ 1+sinx } dx } \)
\(=\int _{ 0 }^{ \pi }{ \frac { \pi sinx }{ 1+sinx } dx=\pi \int _{ 0 }^{ \pi }{ \frac { sinx(1-sinx) }{ (1+sinx)(1-sinx) } dx } } \)
\(=\pi \int _{ 0 }^{ \pi }{ \frac { sin\quad x-{ sin }^{ 2 }x }{ 1-{ sin }^{ 2 }x } } =\pi \int _{ 0 }^{ \pi }{ \frac { sinx-{ sin }^{ 2 } }{ { cos }^{ 2 }x } dx } \)
\(=\pi \int _{ 0 }^{ \pi }{ \frac { sinx }{ { cos }^{ 2 }x } dx-\pi \int _{ 0 }^{ \pi }{ \frac { { sin }^{ 2 } }{ { cos }^{ 2 } } dx } } \)
\(=\pi \int _{ 0 }^{ \pi }{ tanxsec\quad x\quad dx-\pi \int _{ 0 }^{ \pi }{ { tan }^{ 2 }xdx } } \)
\(=\pi { \left[ secx \right] }_{ 0 }^{ \pi }-\pi ({ sec }^{ 2 }x-1)dx\)
\(={ \pi \left[ sec-tan\quad x+x \right] }_{ 0 }^{ \pi }\)
\(=\pi [(sec\pi -tan\pi +\pi )-(sec0-tan0+0)]\)
\(=\pi [(-1-0+\pi )-(1-0+0)]\)
\(2I=\pi [\pi -2]\ \)
\(\therefore I=\frac { \pi }{ 2 } [\pi -2]\)

Let x = tan \(\theta \Rightarrow\)dx = sec²\(\theta d \theta\)

x	0	1
\(\theta\)	0	\(\frac{\pi}{4}\)

\(\therefore I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \frac { log(1+tan\theta ) }{ { sec }^{ 2 }\theta } } { sec }^{ 2 }\theta d\theta \)
\(=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log(1+tan\theta )d\theta } \quad ...(1)\)
Using property,
\(I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log(1+tan(\frac { \pi }{ 4 } -\theta ))d\theta } \)
\(I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log\left( \frac { 1+tan\theta +1-tan\theta }{ 1+tan\theta } \right) } d\theta \)
\(I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log\left( \frac { 2 }{ 1+tan\theta } \right) } d\theta \quad ..(2)\)
(1)+(2)
\(2I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log(1+tan\theta )d\theta } +\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log\left( \frac { 2 }{ 1+tan\theta } \right) } d\theta \)

\(=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log2d\theta =log2{ [\theta ] }_{ 0 }^{ \frac { \pi }{ 4 } } } \)
\(2I=log2(\frac { \pi }{ 4 } -0)=\frac { \pi }{ 4 } log2\)
\(\therefore I=\frac { \pi }{ 8 } log2\)

Let I = \(\int ^{\pi}_{-\pi} \frac{cos ^2 x}{1+ a^x} dx\)-- (1) 
Using \(\int ^{b}_{a}\) f(x) dx =\(\int ^{b}_{a}\) f(a+b -x)dx  we get,
I = \(\int ^{\pi}_{-\pi} \frac{cos ^2 (\pi -\pi - x)}{1+ a^{\pi -\pi - x}} dx\)
= \(\int ^{\pi}_{-\pi} \frac{cos ^2 (-x)}{1+ a^{-x}} dx\)
= \(\int ^{\pi}_{-\pi} a^x (\frac{cos ^2 x }{a^x+ 1} )dx\) --- (2)
Adding (1) and (2) we get
2I =  \(\int ^{\pi}_{-\pi} \frac{cos ^2 x }{a^x+ 1}(a^x+ 1) dx\) =  \(\int ^{\pi}_{-\pi} cos ^2 x dx\)
= 2 \(x =\int ^{\pi}_{-\pi} cos ^2 x dx\)  (since cos² x is an even function)
Hence, I =  \(\int ^{\pi}_{0} (\frac{1 + cos2x }{2} )dx\) 
=  \(\frac {1}{2} [ x + \frac {sin 2x}{2}]^{x}_{0}\) 
= \(\frac {1}{2} [\pi]\)
= \(\frac {\pi}{2}\)

I =\(\int ^{1}_{0} (tan ^{-1} x + tan ^{-1}(1-x))\) dx
=\(\int ^{1}_{0} (tan ^{-1} x dx + \int ^{1}_{0} tan ^{-1}(1-x)) dx\)
= \(\int ^{1}_{0} (tan ^{-1} x dx + \int ^{1}_{0} tan ^{-1}(1-(1-x)) dx\),  Since\(\int ^{a}_{0} f (x) dx = \int ^{a}_{0} f(a-x) dx \)
=\(\int ^{1}_{0} tan ^{-1} x dx + \int ^{1}_{0} tan ^{-1} x dx\)
= 2\(\int ^{1}_{0} tan^{-1} x dx\)
= \([2 \int udv]^1_0\) ,  Where u = tan^-1 x and dν = dx
= 2\([uν - \int udv]^1_0,\) applying integration by parts
= 2\((x tan^{-1} x - \int x \frac{dx}{1+x^2})^{1}_{0}\)
= 2\((x tan^{-1} x - \frac {1}{2} log (1+x^2))^{1}_{0}\)
= \(\frac{\pi}{2} \)- log 2

Let us put I = \(\int ^\frac{\pi}{4}_{0}\) log(1 + tan x) dx
Applying the property \(\int ^{a}_{0}\) f(x) dx =\(\int ^{a}_{0}\)f(a-x) dx in equation (1), we get
I = \(\int ^\frac{\pi}{4}_{0}\) log \([1+tan (\frac{\pi}{4}-x)]\)dx = \(\int ^\frac{\pi}{4}_{0}\) log\([1 + \frac{tan \frac {\pi}{4}- tan x}{1 + tan {\frac {\pi}{4} tan x}}]\) dx
=\(\int ^\frac{\pi}{4}_{0}\) log \([1 + \frac {1 - tan x}{1 + tan x}]\)dx = \(\int ^\frac{\pi}{4}_{0}\) log\([\frac {1+tan x +1 - tan x}{1 + tan x}]\) dx
=\(\int ^\frac{\pi}{4}_{0}\) log \([\frac{2}{1+tanx}]\) dx = \(\int ^\frac{\pi}{4}_{0}\) [log 2 - log (1+tan x)] dx
= log 2  \(\int ^\frac{\pi}{4}_{0}\) dx - \(\int ^\frac{\pi}{4}_{0}\)log (1+tan x)] dx
= \(\frac{\pi}{4}\)log 2 - I
So, we get 2I = \(\frac{\pi}{4}\)log 2.
Hence, we get I = \(\frac{\pi}{8}\)log 2.

Let I =\(\int ^{\pi}_{0} \frac{x}{1+sin x}\)
= \(\int ^{\pi}_{0} x \frac{x}{1+sin x}\)dx
Let f (x) = \(\frac{1}{1+ sin x}\) 
Then f(π-x) = \(\frac{1}{1+ sin (\pi - x)}\) = \(\frac{1}{1+ sin x}\) = f(x)
∴  \(\int ^{\pi}_{0} \frac{x}{1+sin x}\) dx = \(\frac{\pi}{2}\)  \(\int ^{\pi}_{0} \frac{1}{1+sin x}\) dx,      \((\because \int _{ 0 }^{ a }{ xf(x)dx=\frac { a }{ 2 } \int _{ 0 }^{ a }{ f(x)dx } if } f(a-x)=f(x))\)
= \(\pi\) \(\int ^{ \frac{\pi}{2}}_0 \frac{1}{1+sin x}\) dx,       (∴  \(\int ^{\pi}_0\) g(sin x)dx = 2  \(\int ^{ \frac{\pi}{2}}_0\) g(sin x)dx)
= \(\pi\)  \(\int ^{ \frac{\pi}{2}}_0 \frac{1}{1+sin (\frac{\pi}{2} -x)}\) dx  (∴ \(\int ^{a}_{0} \) f(x) dx =  \(\int ^{a}_{0} \) f(a-x)dx)
= \(\pi\)  \(\int ^{ \frac{\pi}{2}}_0 \frac{1}{1+cos x}\) dx = \(\pi\) \(\int ^{ \frac{\pi}{2}}_0 \frac{1}{2cos^2 \frac{x}{2}}\) dx = \(\frac{\pi}{2}\) \(\int ^{ \frac{\pi}{2}}_0 sec ^2 \frac{x}{2}\) dx
= \(\pi\)  \([tan \frac{x}{2}]^\frac{\pi}{2}_0\) =  \(\pi\) \([tan \frac{\pi}{4} - tan 0]\) = \(\pi\)

I = \(\int ^\frac{\pi}{4}_{0} \frac{1}{sin x+cos x}\) dx = \(\int ^\frac{\pi}{4}_{0} \frac{1} {\sqrt 2( \frac {1}{\sqrt 2}sin x+ \frac {1}{\sqrt 2}cos x)}\) dx
= \(\frac{1}{\sqrt 2}\int ^\frac{\pi}{4}_{0} \frac{1} {( cos \frac {\pi}{4}cos x+ sin \frac {\pi}{4}sin x)}\) dx = \(\frac{1}{\sqrt 2}\int ^\frac{\pi}{4}_{0} \frac{1} {cos (\frac{\pi}{4}- x)}\) dx
= \(\frac{1}{\sqrt 2}\int ^\frac{\pi}{4}_{0} \frac{1} {cos x}\)dx since\(\int ^{a}_{0}\) f(x) dx =\(\int ^{a}_{0}\) f (a-x)dx
\(=\frac { 1 }{ \sqrt { 2 } } \int _{ 0 }^{ \frac { \pi }{ 4 } } secxdx=\frac { 1 }{ \sqrt { 2 } } { \left[ log(secx+tanx) \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
=  \(\frac{1}{\sqrt 2} [log ( \sqrt {2} + 1) - log (1+0)]\)
= \(\frac{1}{\sqrt 2} [log ( \sqrt {2} + 1)\).

Put u = tan \(\frac{x}{2}\) 
Then, sin x = \(\frac{2 tan \frac {x}{2}}{1 + tan^2 \frac{x}{2}}\) = \(\frac{2u}{1+u^2}\), du = \(\frac{1}{2}\) sec² \(\frac{x}{2}\) dx  \(\Rightarrow \) dx = \(\frac {2du}{1+u^2}\)
When x = 0,u = tan 0 = 0
When x = \(\frac{\pi}{2},u=tan\frac{\pi}{4}=1\)
∴ I =\(\int ^\frac{\pi}{2}_0\) \(\frac{dx}{4+5 sin x}\) =\(\int ^{1}_0\) \(\frac{\frac {2du}{1+u^2}}{4+5 (\frac{2u}{1+u^2})}\) =\(\int ^{1}_0\) \(\frac {du}{2u^2+5u +2}\)=\(\frac{1}{2}\)\(\int ^{1}_{0}\) \(\frac {du}{u^2+\frac{5}{2}u +1}\)
\(\frac { 1 }{ 2 } \int _{ 0 }^{ 1 } \frac { du }{ (u+\frac { 5 }{ 4 } )^2-(\frac { 3 }{ 4 } )^{ 2 } } { \left[ \frac { 1 }{ 2 } \times \frac { 1 }{ 2\times \left( \frac { 3 }{ 4 } \right) } log\left( \frac { (u+\frac { 5 }{ 4 } )-\frac { 3 }{ 4 } }{ (u+\frac { 5 }{ 4 } )+\frac { 3 }{ 4 } } \right) \right] }_{ 0 }^{ 1 } = \frac{1}{3} [log (\frac{u+\frac{1}{2}}{u+2})] =\frac { 1 }{ 3 } log2\)

By definition, we have |x + 3| =  \(\begin{cases} \begin{matrix} x+3 & x\ge -3 \\ -x-3 & x<-3 \end{matrix}\end{cases}\)
for the graph of y =| x + 3 | in −4 \(\le\) x \(\le\) 4.
∴ \(\int ^4_{-4}\)|x+3| dx =  \(\int ^{-3}_{-4}\) |x+3| dx + \(\int ^{4}_{-3}\) |x+3| dx = \(\int ^{-3}_{-4}\)|-x-3| dx+ \(\int ^{4}_{-3}\)(x+3)dx
= \([- \frac {x^2}{2}-3x]^{-3}_{-4}\) +  \([ \frac {x^2}{2}+3x]^{4}_{-3}\)
= \(( -\frac {9}{2}+9)\) - \(( -\frac {16}{2}+12)\) + \(( \frac {16}{2}+12)\) - \(( \frac {9}{2}-9)\) = \(( \frac {9}{2})\) - 4+20 + \(( \frac {9}{2})\) = 25

Here a = 0, b = 0.5, n = 5, f(x) = x²
So, the width of each subinterval is \(h=\Delta x=\frac { b-a }{ n } =\frac { 0.5-0 }{ 5 } =0.1\)
The partition of the interval is given by the points
x₀ = 0,
x₁ = x₀ + h = 0 + 0.1 = 0.1
x₂ = x₁ + h = 0.1+ 0.1 = 0.2
x₃ = x₂ + h = 0.2 + 0.1 = 0.3
x₄ = x₃ + h = 0.3+ 0.1 = 0.4
x₅ = x₄ + h = 0.4 + 0.1 = 0.5
(i) The left-end rule for Riemann sum with equal width \(\Delta\)x is
S = [f(x₀) + f(x₁) +....+f(x_n-1)]\(\Delta x\)
\(\therefore\) S = [f(0) + f(0.1) + f(0.2) + f(0.3) + f(0.4)](0.1)
= [0.00 + 0.01+ 0.04 + 0.09 + 0.16](0.1) = 0.03
\(\therefore \int _{ 0 }^{ 0.5 }{ x^{ 2 } }\) dx is approximately 0.03.
(ii) The right-end rule for Riemann sum with equal width \(\Delta\)x is
S = [f(x₁)+f(x₂)+...+f(x_n)]\(\Delta\)x
\(\therefore\) S = [f(0.1) + f(0.2) + f(0.3) + f(0.4) + f(0.5)](0.1)
= [0.01 + 0.04 + 0.09 + 0.16 + 0.25](0.1) = 0.055
\(\therefore \int _{ 0 }^{ 0.5 }{ x^{ 2 } }\) dx is approximately 0.055.
(iii) The mid-point rule for Riemann sum with equal width \(\Delta\)x is
\(S=\left[ f\left( \frac { { x }_{ 0 }+{ x }_{ 1 } }{ 2 } \right) +f\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } \right) +..+f\left( \frac { { x }_{ n-1 }+{ x }_{ n } }{ 2 } \right) \right] \Delta x\)
\(\therefore\) S = [f[f(0.05) + f(0.15) + f(0.25) + f(0.35) + f(0.45)](0.1)
= [0.0025 + 0.0225 + 0.0625 + 0.1225 + 0.2025](0.1)
= 0.04125
\(\therefore \int _{ 0 }^{ 0.5 }{ { x }^{ 2 } } \) dx is approximately 0.04125

Equation of the given curve is y = 2x²
Volume \(=\pi \int _{ a }^{ b }{ { y }^{ 2 }dx } \)
\(=\pi \int _{ 0 }^{ 1 }{ { (2{ x }^{ 2 }) }^{ 2 } } dx=\pi \int _{ 0 }^{ 1 }{ { 4x }^{ 4 }dx } \)
\(=4\pi { \left[ \frac { { x }^{ 5 } }{ 5 } \right] }_{ 0 }^{ 1 }=\frac { 4\pi }{ 5 } (1-0)\)
\(v=\frac { 4\pi }{ 5 } \) cubic units

Given equation of line is 2x - y + 1 = 0

x	0	-1/2
y	1	0

\(\Rightarrow 2x=y-1\Rightarrow x=\frac { y-1 }{ 2 } \)
[\(\because\) the area is left of the x-axis]

\(\therefore Area\ =\int _{ -1 }^{ 1 }{ -xdy+\int _{ 1 }^{ 3 }{ xdy } } \)
\(=-\frac { 1 }{ 2 } \int _{ -1 }^{ 1 }{ (y-1)dy } +\frac { 1 }{ 2 } \int _{ 1 }^{ 3 }{ (y-1)dy } \)
\(={ -\frac { 1 }{ 2 } { \left[ \frac { { y }^{ 2 } }{ 2 } -y \right] }_{ -1 }^{ 1 }{ +\frac { 1 }{ 2 } }\left[ \frac { { y }^{ 2 } }{ 2 } -y \right] }_{ 1 }^{ 3 }\)
\(=\frac { 1 }{ 2 } \left[ { \left( y-\frac { { y }^{ 2 } }{ 2 } \right) }_{ -1 }^{ 1 }+{ \left( \frac { { y }^{ 2 } }{ 2 } -y \right) }_{ 1 }^{ 3 } \right] \)
\(=\frac { 1 }{ 2 } \left[ \left( 1-\frac { 1 }{ 2 } \right) -\left( 1-\frac { 1 }{ 2 } \right) +\left( \frac { 9 }{ 2 } -3 \right) -\left( \frac { 1 }{ 2 } -1 \right) \right] \)
\(=\frac { 1 }{ 2 } \left[ \frac { 1 }{ 2 } +\frac { 3 }{ 2 } +\frac { 3 }{ 2 } +\frac { 1 }{ 2 } \right] =\frac { 1 }{ 2 } \left[ \frac { 8 }{ 2 } \right] =\frac { 1 }{ 2 } (4)\)
= 2 sq.units

Solution the region is sketched. It lies below the x − axis. Hence, the required area is given by

\(A=\left| \int _{ -2 }^{ 3 }{ ydx } \right| =\left| \int _{ -2 }^{ 3 }{ \left( \frac { 7x-35 }{ 5 } \right) dx } \right| \)
\(=\frac { 1 }{ 5 } \left| { \left( 7\left( \frac { { x }^{ 2 } }{ 2 } \right) -35x \right) }_{ -2 }^{ 3 } \right| \)
\(=\frac { 1 }{ 5 } \left| \left( \left( \frac { 63 }{ 2 } \right) -105 \right) -(84) \right| =\frac { 63 }{ 2 } \)

\(Let\ I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { dx }{ 1+5{ cos }^{ 2 }x } } \)
Put U = tan x \(\Rightarrow\) du = sec 2x dx Dividing the numerator and denominator by cos²x we get,
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \frac { 1 }{ { cos }^{ 2 }x } }{ \frac { 1 }{ { cos }^{ 2 }x } +5 } } dx=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sec }^{ 2 }x }{ { sec }^{ 2 }x+5 } dx } \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sec }^{ 2 }x }{ 1+{ tan }^{ 2 }+x+5 } dx } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sec }^{ 2 }x\quad dx }{ { tan }^{ 2 }x+6 } } \)

x	0	\(\frac{\pi}{2}\)
u	0	\(\infty\)

\(=\int _{ 0 }^{ \infty }{ \frac { du }{ { u }^{ 2 }+6 } } \)
\(=\int _{ 0 }^{ \infty }{ \frac { du }{ { u }^{ 2 }+{ (\sqrt { 6 } ) }^{ 2 } } } \)
\(I=\frac { \pi }{ 2\sqrt { 6 } } \left[ \because \int _{ 0 }^{ \infty }{ \frac { dx }{ { a }^{ 2 }+{ x }^{ 2 } } =\frac { \pi }{ 2a } ,Here\quad a=\sqrt { 6 } } \right] \)

\(I=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { 1 }{ 1+\sqrt { tanx } } dx } \int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { 1 }{ 1+\frac { \sqrt { sin\quad x } }{ \sqrt { cos\quad x } } } dx } \)
\(=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { \sqrt { cos\quad x } }{ \sqrt { cos\quad x } +\sqrt { sin\quad x } } dx\quad \quad ..(1) } \)
By the property,\( \int _{ a }^{ b }{ f(x)dx=\int _{ a }^{ b }{ f(a+b-x)dx } } \)
we get \(I=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { \sqrt { cos(\frac { \pi }{ 8 } +\frac { 3\pi }{ 8 } -x) } }{ \sqrt { cos(\frac { \pi }{ 8 } +\frac { 3\pi }{ 8 } -x) } +\sqrt { sin(\frac { \pi }{ 8 } +\frac { 3\pi }{ 8 } -x) } } } dx\)
\(=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { \sqrt { cos(\frac { \pi }{ 2 } -x) } }{ \sqrt { cos(\frac { \pi }{ 2 } -x) } +\sqrt { sin(\frac { \pi }{ 2 } -x) } } } dx\)
\(=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { \sqrt { sin\quad x } }{ \sqrt { sin\quad x } +\sqrt { cos\quad x } } ...(2) } \)
\((1)+(2)\rightarrow \)
\(2I=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { \sqrt { cos\quad x } }{ \sqrt { cos\quad x } +\sqrt { sin\quad x } } dx+\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \frac { \sqrt { sin\quad x } }{ \sqrt { sin\quad x } +\sqrt { sin\quad x } } } } \)
\(=\int _{ \frac { \pi }{ 8 } }^{ 3\frac { \pi }{ 8 } }{ \frac { \sqrt { cos\quad x } +\sqrt { sin\quad x } }{ \sqrt { cos\quad x } +\sqrt { sin\quad x } } dx=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ dx={ [x] }_{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } } } } \)
\(2I=\frac { 3\pi }{ 8 } -\frac { \pi }{ 8 } =\frac { 2\pi }{ 8 } =\frac { \pi }{ 4 } \)
\(\therefore I=\frac { \pi }{ 8 } \)

Let us put I = \(\int ^{3}_{2} \frac{\sqrt {x}}{\sqrt {5-x}+\sqrt {x}}\) dx --- (1)
Applying the formula \(\int ^{a}_{b}\) f(x) dx =\(\int ^{a}_{b}\) f(a+b -x) dx, we get
I =  \(\int ^{3}_{2} \frac{\sqrt {(2+3-x)}}{\sqrt {5-(2+3-x)}+\sqrt { {(2+3-x)}}}\) dx = \(\int ^{3}_{2} \frac{\sqrt {(5-x)}}{\sqrt {x}+\sqrt { {(5-x}}}\) dx     -- (2)
Adding (1) and (2), we get 
2I =  \(\int ^{3}_{2} \frac{\sqrt {x} + \sqrt {5-x} }{\sqrt {x}+\sqrt { {5-x}}}\) dx =\(\int ^{3}_{2} \) dx = \([x]^{3}_{2}\) = 3 - 2 = 1
Hence, we get I = \(\frac{1}{2}\)

Let I = \(\int ^{a}_{0} \frac{f(x)}{f(x)+f(a-x)}\)    ------- (1)
Applying the formula \(\int ^{a}_{0}\) f(x) dx = \(\int ^{a}_{0}\)f(a-x)dx  in equation (1), we get
 I =  \(\int ^{a}_{0} \frac{f{(a-x)}}{f{(a-x)}+f(a-(a-x))}\) dx
= \(\int ^{a}_{0} \frac{f{(a-x)}}{f{(x)}+f(a-x)}\) dx----- (2)
Adding equations (1) and (2), we get
2I =  \(\int ^{a}_{0} \frac{f{(x)}}{f{(x)}+f(a-x)}\) dx + \(\int ^{a}_{0} \frac{f{(a-x)}}{f{(x)}+f(a-x)}\) dx
= \(\int ^{a}_{0} \frac{f(x) +f{(a-x)}}{f{(x)}+f(a-x)}\) dx
= \(\int ^{a}_{0} dx = a\)
Hence, we get I = \(\frac{a}{2}\)

Given equation of line is 3x - 2y + 6 = 0
2y = 3x + 6 \(\Rightarrow\) y = \(\frac{3x+6}{2}\)

x	0	-2
t	3	0

\(\therefore Area=\int _{ -3 }^{ -2 }{ -ydx+ydx } +\int _{ -2 }^{ 1 }{ ydx } \)
[\(\because\) the Area is below the x - axis]

\(=\frac { -1 }{ 2 } \int _{ -3 }^{ -2 }{ (3x+6)dx+\frac { 1 }{ 2 } \int _{ -2 }^{ 1 }{ (3x+6)dx } } \)
\(={ \left[ \frac { { 3x }^{ 2 } }{ 2 } +6x \right] }_{ -3 }^{ -2 }+\frac { 1 }{ 2 } { \left[ \frac { { 3x }^{ 2 } }{ 2 } +6x \right] }_{ -2 }^{ 1 }\)
\(=-\frac { 1 }{ 2 } \left[ \left( \frac { 12 }{ 2 } -12 \right) -\left( \frac { 27 }{ 2 } -18 \right) \right] +\frac { 1 }{ 2 } \left[ \left( \frac { 3 }{ 2 } +6 \right) -\left( \frac { 12 }{ 2 } -12 \right) \right] \)
\(=-\frac { 1 }{ 2 } \left[ (-6)-\left( \frac { 27-36 }{ 2 } \right) \right] +\frac { 1 }{ 2 } \left[ \left( \frac { 3+12 }{ 2 } \right) -(-6) \right] \)
\(=-\frac { 1 }{ 2 } \left[ -6+\frac { 9 }{ 2 } \right] +\frac { 1 }{ 2 } \left[ \frac { 15 }{ 2 } +6 \right] \)
\(\\ =-\frac { 1 }{ 2 } \left[ \frac { -3 }{ 2 } \right] +\frac { 1 }{ 2 } \left[ \frac { 27 }{ 2 } \right] =\frac { 3 }{ 4 } +\frac { 27 }{ 4 } =\frac { 30 }{ 4 } =\frac { 15 }{ 2 } \)
\(\therefore\)  A = 7.5 sq.units