\(12\sqrt { 3 } \)

prove.

\(\frac { ab\left( \pi -2 \right) }{ 4 } \)

c=-1

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 \(=\frac { \pi }{ 3 } [2(4+2+1)]\)
\(=\frac { \pi }{ 3 } (14)\)
\(=\frac { 14\pi }{ 3 } \)

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
\(A=\int_{0}^{4}\left(y_{U}-Y_{L}\right) d x=\int_{0}^{4}\left(2 \sqrt{x}-\frac{x^{2}}{4}\right) d x=\left[2\left(\frac{2 x^{3 / 2}}{3}\right)-\frac{x^{3}}{12}\right]_{0}^{4}=\left[2\left(\frac{2 \times 8}{3}\right)-\frac{64}{12}\right]-0=\frac{16}{3}\)

Let I = \(\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ \frac { { sin }^{ -1 }x }{ { (1-{ x }^{ 2 }) }^{ \frac { 3 }{ 2 } } } dx } \)
Put u = sin^-1 x. Then, x = sin u and so, du =\(\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } dx\)
When x = 0, u = 0 
When \(x=\frac { 1 }{ \sqrt { 2 } } ,u=\frac { \pi }{ 4 } .\)
\(\therefore I=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \frac { u }{ { cos }^{ 2 }u } } du=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ u{ sec }^{ 2 }udu={ [utanu] }_{ 0 }^{ \frac { \pi }{ 4 } } } -\int _{ 0 }^{ \frac { \pi }{ 4 } }{ tan\quad udu={ [utanu] }_{ 0 }^{ \frac { \pi }{ 4 } }+{ \left[ logcosu \right] }_{ 0 }^{ \frac { \pi }{ 4 } } } \)
\(=\frac { \pi }{ 4 } +log\frac { 1 }{ \sqrt { 2 } } =\frac { \pi }{ 4 } -\frac { 1 }{ 2 } log2\)

Here f (x) = x³, a = 0 and b = 1. Hence, we get
\(\int _{ a }^{ b }{ f(x)dx } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ f } \left( \frac { r }{ n } \right) \Rightarrow \int _{ 0 }^{ 1 }{ x^3dx } =\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ n } \sum _{ r=1 }^{ n }{ \frac { r^3 }{ n^3 } } \)
\(=\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ { n }^{ 4 } } [{ 1 }^{ 3 }+{ 2 }^{ 3 }+...+{ n }^{ 3 }]=\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ { n }^{ 4 } } \frac { { n }^{ 2 }{ (n+1) }^{ 2 } }{ 4 } \)
\(=\underset { n\rightarrow \infty }{ lim } \frac { 1 }{ 4 } { \left( 1+\frac { 1 }{ n } \right) }^{ 2 }=\frac { 1 }{ 4 } \)

To find \(\int _{ 1 }^{ 1.5 }{ xdx } \) in  {1.1, 1.2, 1.3, 1.4, 1.5}.
Here a = 1,
b = 1.5, n = 5, f(x) = x
\(\therefore h=\Delta x=\frac { b-a }{ n } =\frac { 1.5-1 }{ 5 } =\frac { 0.5 }{ 5 } =0.1\)
The partition of the interval is given by 
x₀ = 1
x₁ = x₀ + h = 1 + 0.1 = 1.1
x₂ = x₁ + h = 1.1 + 0.1 = 1.2
x₃ = x₂ + h = 1.2 + 0.1 = 1.3
x₄ = x₃ + h = 1.3 + 0.1 = 1.4
x₅ = x₄ + h = 1.4 + 0.1 = 1.5
The left end rule for Riemann sum with equal width \(\Delta\)x is
\(\int _{ a }^{ b }{ xdx } =[f({ x }_{ 0 })+f({ x }_{ 1 })+...+f({ x }_{ (n-1 })]\Delta x\)
S = [f(1) + f(1.1) + f(1.2) + f(1.3) + f(1.4)](0.1)
= (1 + 1.1 + 1.2 + 1.3 + 1.4) (0.1)
= 6 \(\times\) 0.1
S = 0.6

\(4\sqrt { 3 } -4-\frac { \pi }{ 3 } \)

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { cot \ x } }{ \sqrt { cot \ x } +\sqrt { tan \ x } } dx } \) ....(1)
By the property, \(\int _{ 0 }^{ a }{ f(x)dx } =\int _{ 0 }^{ a }{ f(a-x) } dx\)
∴ I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { cot(\frac { \pi }{ 2 } -x) } }{ \sqrt { cot(\frac { \pi }{ 2 } -x) } +\sqrt { tan(\frac { \pi }{ 2 } -x) } } dx } \)
= \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { tanx } }{ \sqrt { tanx } +\sqrt { cotx } } dx } \) ...(2)
(1) + (2) ⇒ 2I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { cot \ x } +\sqrt { tan \ x } }{ \sqrt { cot \ x } +\sqrt { tan \ x } } dx } \)
=\(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ dx } ={ \left[ x \right] }_{ 0 }^{ \frac { \pi }{ 2 } }=\frac { \pi }{ 2 } -0=\frac { \pi }{ 2 } \)
∴ I = \(\frac { \pi }{ 4 } \)

Let \(f(x)={ sin }^{ 2 }x\)
\(f(-x)={ (sin(-x)) }^{ 2 }={ sin }^{ 2 }x=f(x)\)
\(\therefore \int _{ -\frac { \pi }{ 4 } }^{ \frac { \pi }{ 4 } }{ { sin }^{ 2 }xdx } =2\int _{ 0 }^{ \frac { \pi }{ 4 } }{ { sin }^{ 2 }xdx } \)
\(cos\ 2x=1-2{ sin }^{ 2 }x\)
\(2{ sin }^{ 2 }x=1-cos2\)
\({ sin }^{ 2 }x=\frac { 1-cos\quad 2x }{ 2 } \)
\(=2\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( \frac { 1-cos2x }{ 2 } \right) dx } \)
\(=\frac { 2 }{ 2 } { \left[ x-\frac { sin2x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(=\frac { \pi }{ 4 } -\frac { sin\frac { \pi }{ 4 } }{ 2 } -0+\frac { sin0 }{ 2 } \)
\(\\ =\frac { \pi }{ 4 } -\frac { 1 }{ 2 } =\frac { \pi -2 }{ 4 } \)

Let \(f(x)=xcos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \)
\(f(-x)=-x\quad cos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \)
\(=-x\quad cos\left( \frac { { e }^{ \frac { 1 }{ x } }-1 }{ { e }^{ \frac { 1 }{ x } }+1 } \right) \)
\(=-x\quad cos\left( \frac { 1-{ e }^{ x } }{ 1+{ e }^{ x } } \right) \)
\(=-x\quad cos\left( -\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \right) \)
\(=-xcos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) \)
\([\because cos(-\theta )=cos\theta ]\)
= -f(x)
\(\therefore\) f(x) is an odd function
\(\therefore \int _{ -5 }^{ 5 }{ xcos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) } dx=0\)

I = \(\int _{ 0 }^{ 1 }{ \frac { 1-{ x }^{ 2 } }{ { (1+{ x }^{ 2 }) }^{ 2 } } } dx\)
[Dividing the numerator and denominator by x]
\(I=\int _{ 0 }^{ 1 }{ \frac { \frac { 1 }{ { x }^{ 2 } } -\frac { { x }^{ 2 } }{ { x }^{ 2 } } }{ 0\left( \frac { 1 }{ x } +\frac { { x }^{ 2 } }{ { x }^{ 2 } } \right) } } dx=\int _{ 0 }^{ 1 }{ \frac { { x }^{ \frac { 1 }{ 2 } -1 } }{ { (\frac { 1 }{ x } +x) }^{ 2 } } } dx\)
\(put\ x+\frac { 1 }{ x } =t\Rightarrow (1-\frac { 1 }{ { x }^{ 2 } } )dx=dt\)
\(\Rightarrow \left( \frac { 1 }{ { x }^{ 2 } } -1 \right) dx=-dt\)

\(\therefore I=\int _{ \infty }^{ 2 }{ -\frac { dt }{ { t }^{ 2 } } } =\int _{ 2 }^{ \infty }{ \frac { dt }{ { t }^{ 2 } } } \)
\(=\int _{ 2 }^{ \infty }{ { t }^{ -2 }dt } ={ \left[ \frac { { t }^{ -1 } }{ -1 } \right] }_{ 2 }^{ \infty }={ \left[ -\frac { 1 }{ t } \right] }_{ 2 }^{ \infty }\)
\(=-\frac { 1 }{ \infty } +\frac { 1 }{ 2 } =0+\frac { 1 }{ 2 } \)
\(=\frac { 1 }{ 2 } \therefore \int _{ 0 }^{ 1 }{ \frac { 1-{ x }^{ 2 } }{ { (1+{ x }^{ 2 }) }^{ 2 } } dx=\frac { 1 }{ 2 } } \)

We write \(\int _{ 0 }^{ 2a }{ f(x)dx\int _{ 0 }^{ a }{ f(x)dx } } +\int _{ 0 }^{ 2a }{ f(x)dx } \) ....(1)
Consider \(\int _{ 0 }^{ 2a }{ f(x)dx } \)
Substituting x = a + u, we have dx = du ; when x = a, u = 0 and when x = 2a,u = a.
\(\therefore \int _{ 0 }^{ 2a }{ f(x)dx=\int _{ 0 }^{ a }{ f(a+u)du } =\int _{ 0 }^{ a }{ d(u)du } } \), since f(x) = f(a+x)
\(\\ \\ \\ =\int _{ 0 }^{ a }{ f(x)dx } \) ..(2)
Substituting (2) in (1), we get
\(\int _{ 0 }^{ 2a }{ f(x)dx } =2\int _{ 0 }^{ a }{ f(x)dx } \)

\(\int _{ 0 }^{ 3 }{ (3{ x }^{ 2 }-4x+5) } dx=\int _{ 0 }^{ 3 }{ 3{ x }^{ 2 } } dx-\int _{ 0 }^{ 3 }{ 4x } dx+\int _{ 0 }^{ 3 }{ 5 } dx\)
\(=3\int _{ 0 }^{ 3 }{ { x }^{ 2 } } dx-4\int _{ 0 }^{ 3 }{ c } dx+5\int _{ 0 }^{ 3 }{ dx } \)
\(=3{ \left[ \frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 3 }-4{ \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 3 }+5{ \left[ x \right] }_{ 0 }^{ 3 }\)
= (27 − 0) − 2(9 − 0) + 5(3− 0)
= 27 −18 +15 = 24.

\(\frac { 3\pi }{ 5 } \)

2

\(\frac { 1 }{ 2 } \)

\(\frac { 1 }{ 2 } \left[ 1-{ e }^{ \frac { \pi }{ 2 } } \right] \)

\(\frac { 1 }{ 10 } \left[ { e }^{ \frac { 3\pi }{ 2 } }-1 \right] \)

Given curve is y = \(\sqrt x\) + 1

Required area
\(\int _{ 0 }^{ 4 }{ ydx } =\int _{ 0 }^{ 4 }{ (\sqrt { x } +1)dx } \)
\({ \left[ \frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }+x \right] }_{ 0 }^{ 4 }=\frac { 2 }{ 3 } { (4) }^{ \frac { 3 }{ 2 } }+4\)
\(\frac { 2 }{ 3 } (4)\sqrt { 4 } +4=\frac { 16 }{ 3 } +4\)
\(\frac { 16+12 }{ 3 } =\frac { 28 }{ 3 } \) sq.units

\({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { n-1 }{ n } { I }_{ n-2 },n\ge 2 } \)
\(\therefore { I }_{ 7 }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { 6 }{ 7 } \times \frac { 4 }{ 5 } \times \frac { 2 }{ 3 } \times 1 } =\frac { 16 }{ 35 } \)

Let f (x) = x cos x
Then f (−x) = (−x) cos(−x) = −x cos x = − f (x).
So f (x) = x cos x is an odd function.
Hence, applying the property, for odd function f(x), \(\int _{ -a }^{ a }{ f(x)dx=0 } \)
\(\therefore\) we get \(\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ xcosx } dx=0\)

Take 2a = 2\(\pi\) and f(x) = g(cosx)
Then, f (2a−x) = f(2\(\pi\)-x) = g(cos(2\(\pi\)-x)) = g(cos x) = f(x)
\(\therefore \int _{ 0 }^{ 2a }{ f(x)dx=2 } \int _{ 0 }^{ a }{ f(x)dx } \)
\(\therefore \int _{ 0 }^{ 2\pi }{ g(cosx)dx=2\int _{ 0 }^{ \pi }{ g(cosx)dx } } \)