The region to be revolved is sketched.
Since revolution is made about the y-axis, the volume of the solid generated is given by

\(V=\pi \int _{ 0 }^{ 2 }{ { x }^{ 2 }dy=\pi \int _{ 0 }^{ 2 }{ { e }^{ y }dy } } \)
\(=\pi { \left[ { e }^{ y } \right] }_{ 0 }^{ 2 }=\pi ({ e }^{ 2 }-1)\)

We note that \(y=\frac { 3 }{ 4 } \sqrt { { x }^{ 2 }-16 } \Rightarrow \frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 9 } =1\). So, the given curve is a portion of the hyperbola \(\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 9 } =1\) between the lines y = 1 and y = 6 and it lies above the x-axis.

The region to be revolved is sketched.
Since revolution is made about y-axis, we write the equation of the portion of the hyperbola as \(x=\frac { 4 }{ 3 } \sqrt { 9+{ y }^{ 2 } } .\) 
So, the volume of the solid generated is given by 
\(V=\pi \int _{ 1 }^{ 6 }{ { x }^{ 2 }dy=\pi \int _{ 1 }^{ 6 }{ { \left( \frac { 4 }{ 3 } \sqrt { 9+{ y }^{ 2 } } \right) }^{ 2 } } } dy=\pi \left( \frac { 16 }{ 9 } \right) \int _{ 1 }^{ 16 }{ \left( 9+{ y }^{ 2 } \right) dy } \)
\(=\pi \left( \frac { 16 }{ 9 } \right) { \left( 9y+\frac { { y }^{ 3 } }{ 3 } \right) }_{ 1 }^{ 6 }=\pi \left( \frac { 16 }{ 9 } \right) \left[ (54+72)-(9+\frac { 1 }{ 3 } ) \right] =\frac { 5600 }{ 27 } \pi \)

Applying integration by parts, we get
\(\int _{ 0 }^{ \infty }{ { e }^{ -x }{ x }^{ n } } dx={ \left[ { x }^{ n }(-{ e }^{ -x }) \right] }_{ 0 }^{ \infty }-\int _{ 0 }^{ \infty }{ (-{ e }^{ -x }) } ({ nx }^{ n-1 })dx=n\int _{ 0 }^{ \infty }{ { e }^{ -x }{ x }^{ n-1 }dx } \)
\(Let\quad { I }_{ n }=\int _{ 0 }^{ \infty }{ { e }^{ -x }{ x }^{ n }dx.\ Then,\ { I }_{ n }={ nI }_{ n-1 } } \)
So, we get I_n= n(n−1) I_n−2.
Proceeding in this way, we get ultimately,
I_n = n(n-1)(n-2)...(2)(1)I₀.
But, \({ I }_{ 0 }=\int _{ 0 }^{ \infty }{ { e }^{ -x }{ x }^{ 0 }dx } ={ \left( -{ e }^{ -x } \right) }_{ 0 }^{ \infty }=0+1=1.\) So, we get n = n(n-1)(n-2)(2)(1) = n!
Hence, we get
\(\int _{ 0 }^{ \infty }{ { e }^{ -x }{ x }^{ n }dx=n! } \), where n is a nonnegative integer.

\(Let\quad I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { dx }{ 4x{ sin }^{ 2 }x+5{ cos }^{ 2 }x } } \)
\(=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sec }^{ 2 }x }{ 4{ tan }^{ 2 }x+5 } } dx\)
(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)
\(\frac { 1 }{ 4 } \int _{ 0 }^{ \infty }{ \frac { du }{ \left[ { u }^{ 2 }+\left( \frac { \sqrt { 5 } }{ 2 } \right) \right] } =\frac { 1 }{ 4 } \times \frac { 2 }{ \sqrt { 5 } } { \left[ { tan }^{ -1 }\left( \frac { u }{ \frac { \sqrt { 5 } }{ 2 } } \right) \right] }_{ 0 }^{ \infty } } =\frac { 1 }{ 2\sqrt { 5 } } \left( { tan }^{ -1 }\infty -{ tan }^{ -1 }0 \right) =\frac { 1 }{ 2\sqrt { 5 } } \left( \frac { \pi }{ 2 } \right) =\frac { \pi }{ 4\sqrt { 5 } } \)

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

\(\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { sin2xdx }{ sin^{ 4 }x+cos^{ 4 }x } =\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { sin2xdx }{ (sin^{ 2 }x+cos^{ 2 }x)-2sin^{ 2 }xcos^{ 2 }x } \)
\(\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { sin2xdx }{ (1-\frac { 1 }{ 2 } (2sinxcosx)^{ 2 } } =\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { 2sin2xdx }{ 2-sin^{ 2 }2x } =\int _{ 0 }^{ \frac { \pi }{ 4 } } \frac { 2sin2xdx }{ 1+cos^{ 2 }2x } \)
Put u = cos 2x, Then, du = −2sin 2x dx
When x = 0 , we have u = cos 0 = 1. When x = \(\frac{\pi}{4}\) we have u = cos\(\frac{\pi}{4}\) = 0
∴ I =  \(\int^{0}_{1} \frac {-du}{1+u^2}\) = \(\int^{0}_{1} \frac {du}{1+u^2}\) = [tan^-1 u\(]^1_0\) = \(\frac{\pi}{4}\)

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 =\(\int ^\frac{\pi}{2}_0\) \(\frac{dx}{4+5 sin x}\) =\(\int ^{1}_0\) \(\frac{\frac {2du}{1+u^2}}{4+5 (\frac{2u}{1+u^2})}\) =\(\int ^{1}_0\) \(\frac {du}{2u^2+5u +2}\)=\(\frac{1}{2}\)\(\int ^{1}_{0}\) \(\frac {du}{u^2+\frac{5}{2}u +1}\)
\(\frac { 1 }{ 2 } \int _{ 0 }^{ 1 } \frac { du }{ (u+\frac { 5 }{ 4 } )^2-(\frac { 3 }{ 4 } )^{ 2 } } { \left[ \frac { 1 }{ 2 } \times \frac { 1 }{ 2\times \left( \frac { 3 }{ 4 } \right) } log\left( \frac { (u+\frac { 5 }{ 4 } )-\frac { 3 }{ 4 } }{ (u+\frac { 5 }{ 4 } )+\frac { 3 }{ 4 } } \right) \right] }_{ 0 }^{ 1 } = \frac{1}{3} [log (\frac{u+\frac{1}{2}}{u+2})] =\frac { 1 }{ 3 } log2\)

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

The parabola x = y² +1 is y² x = −1. It is symmetrical about x-axis and has the vertex at (1, 0) and focus at \(\left( \frac { 5 }{ 4 } ,0 \right) \). The region for revolution is shaded. Hence, the required volume is given by
\(V=\pi \int _{ -1 }^{ 1 }{ { x }^{ 2 }dy } \)
\(=\pi \int _{ -1 }^{ 1 }{ { ({ y }^{ 2 }+1) }^{ 2 }dy } \)
\(=2\pi \int _{ 0 }^{ 1 }{ \left( { y }^{ 4 }+{ 1y }^{ 2 }+1 \right) dy } \), since the integrand is an even function
\(=2\pi { \left( \frac { { y }^{ 5 } }{ 5 } +2\frac { { y }^{ 3 } }{ 3 } +y \right) }_{ 0 }^{ 1 }=2\pi \left( \frac { 1 }{ 5 } +\frac { 2 }{ 3 } +1 \right) \pi \)

Using the formula \(n={ e }^{ { log }_{ e }n },\) we get
\(I=\int _{ 0 }^{ \infty }{ \frac { { x }^{ n } }{ { n }^{ x } } } dx=\int _{ 0 }^{ \infty }{ { n }^{ -x }{ x }^{ n }dx=\int _{ 0 }^{ \infty }{ { n }^{ -x }{ x }^{ n } } dx=\int _{ 0 }^{ \infty }{ { ({ e }^{ log\quad n }) }^{ -x }{ x }^{ n }dx=\int _{ 0 }^{ \infty }{ { e }^{ -xlog\quad n }{ x }^{ n }dx } } } \)
Using the substitution u = x log n, we get dx \(\frac { du }{ log\ n } \)
When x = 0,
we get u = 0
When x = \(\infty\), we get u = \(\infty\)
\(\therefore I=\int _{ 0 }^{ \infty }{ { e }^{ -u }{ \left( \frac { u }{ log\quad n } \right) }^{ n }\frac { du }{ log\quad n } } \)
\(=\frac { 1 }{ { (log\quad n) }^{ n+1 } } \int _{ 0 }^{ \infty }{ { e }^{ -u }{ u }^{ (n+1)-1 }du=\frac { Γ(n+1) }{ { (log\quad n) }^{ n+1 } } =\frac { n! }{ { (log) }^{ n+1 } } } \)

Let \(\sqrt{x}\) = u
Then x = u², and so dx = 2u du
When x = 0, u = 0 
When x = 9, u = 3
\(\therefore \int _{ 0 }^{ 9 }{ \frac { 1 }{ x+\sqrt { x } } dx=\int _{ 0 }^{ 3 }{ \frac { 1 }{ { u }^{ 2 }+u } (2u)du=2\int _{ 0 }^{ 3 }{ \frac { 1 }{ 1+u } du=2{ \left[ log|1+u \right] }_{ 0 }^{ 3 }=2[log4-0]=log16 } } } \)

\(\int _{ 0 }^{ 1 }{ \frac { 2x+7 }{ { 5x }^{ 2 }+9 } dx } =\int _{ 0 }^{ 1 }{ \frac { 2x }{ { 5x }^{ 2 }+9 } } +7\int _{ 0 }^{ 1 }{ \frac { dx }{ (5{ x }^{ 2 })+{ 3 }^{ 2 } } =\frac { 1 }{ 5 } } log{ [{ 5x }^{ 2 }+9] }_{ 0 }^{ 1 }+\frac { 7 }{ 5 } \int _{ 0 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }{ \left( \frac { 3 }{ \sqrt { 5 } } \right) }^{ 2 } } } \)
\(=\frac { 1 }{ 5 } [log14-log9]+\frac { 7 }{ 5 } \times \frac { \sqrt { 5 } }{ 3 } { \left[ { tan }^{ -1 }\frac { x }{ \left[ \frac { 3 }{ \sqrt { 5 } } \right] } \right] }_{ 0 }^{ 1 }=\frac { 1 }{ 5 } log\frac { 14 }{ 9 } +\frac { 7 }{ 3\sqrt { 5 } } { tan }^{ -1 }\frac { \sqrt { 5 } }{ 3 } \)

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

Let f(x) = e^-|-x| = e^-|x| = f(x)
So f (x) is an even function.
Hence, \(\int ^{log 2}_{-log 2} e ^{-|x|}\)dx = 2\(\int ^{log 2}_{0} e ^{-|x|}\)dx = 2\(\int ^{log 2}_{0} e ^{-x}\) dx
= 2(-e^-x)\(^{log2}_{0}\) = 2 (-e^-log2 + e⁰) = 2 \((-e ^{log \frac{1}{2}} + 1)\)
= 2\((-\frac {1}{2}+1)=1\).

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 \(\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ xcosx } dx=0\)