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 \(=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ (tan\ x-cot\ x)dx } \)
\(={ [-log\ sin\ x+log\ sec\ x] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(=log{ \left[ \frac { sec\ x }{ sin\ x } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }==log{ \left[ \frac { 1 }{ sin\ x\ cos\ x } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(=-log{ (sin\quad x\ cos\ x) }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(=-log\left( sin\frac { \pi }{ 4 } .cos\frac { \pi }{ 4 } \right) +log(sin0\quad cos0)\)
\(=-log\left( \frac { 1 }{ \sqrt { 2 } } .\frac { 1 }{ \sqrt { 2 } } \right) +0\)
\(=-log\left( \frac { 1 }{ 2 } \right) =-(log1-log2)=log2\)

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

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
\(A=2\int _{ 0 }^{ 1 }{ \left[ { y }_{ U }-{ y }_{ L } \right] } dx=2\int _{ 0 }^{ 1 }{ \left[ x-{ x }^{ 2 } \right] dx } \)
\(=2{ \left[ \frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 1 }\)
\(=2\left( \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right) =\frac { 1 }{ 3 } \)

First, we get the points of intersection of the parabolas. For this, we solve y² x = 4 and x² y = 4 simultaneously Eliminating y between them, we get x⁴ = 64x and so x = 0 and x = 4. Then the points of intersection are (0, 0) and (4, 4). The required region is sketched.

Viewing in the direction of y -axis, the equation of the upper boundary is y = 2\(\sqrt x\) for 0\(\le x \le\) 4 and the equation of the lower boundary is \(y =\frac {x^2}{4}\)for \(0 \leq x \leq 4\). So, the required area \(\Delta\) is
\(A=\int_{0}^{4}\left(y_{U}-Y_{L}\right) d x=\int_{0}^{4}\left(2 \sqrt{x}-\frac{x^{2}}{4}\right) d x=\left[2\left(\frac{2 x^{3 / 2}}{3}\right)-\frac{x^{3}}{12}\right]_{0}^{4}=\left[2\left(\frac{2 \times 8}{3}\right)-\frac{64}{12}\right]-0=\frac{16}{3}\)

The given curve is \(y=\begin{cases} cosx,0\le x\le \frac { \pi }{ 2 } \\ -cosx,\frac { \pi }{ 2 } \le x\le \pi \end{cases}\)
It lies above the x − axis. The required area is sketched. So, the required area is given by

\(A=\int _{ 0 }^{ \pi }{ ydx=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ cosxdx } +\int _{ \frac { \pi }{ 2 } }^{ \pi }{ (-cosx)dx } ={ [sin\quad x] }_{ 0 }^{ \frac { \pi }{ 2 } }-{ [sin\quad x] }_{ \frac { \pi }{ 2 } }^{ \pi } } \)
= [1-0]-[0-1] = 2

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 \(A=4\int _{ 0 }^{ a }{ ydx=4\int _{ 0 }^{ a }{ \frac { b }{ a } } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } dx } \)
\(=\frac { 4b }{ a } { \left[ \frac { x\sqrt { { a }^{ 2 }-{ y }^{ 2 } } }{ 2 } +\frac { { a }^{ 2 } }{ 2 } { sin }^{ -1 }\left( \frac { x }{ a } \right) \right] }_{ 0 }^{ a }=\frac { 4b }{ a } \times \frac { \pi { a }^{ 2 } }{ 4 } =\pi ab\)