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

Let I = \(\int _{ 1 }^{ 2 }{ \frac { x }{ (x+1)(x+2) } dx } \)
\(I=\int _{ 1 }^{ 2 }{ \left[ \frac { -1 }{ (x+1) } +\frac { 2 }{ x+2 } \right] } dx\) (Using partial fractions)
\(={ [-log(x+1)+2log(x+2)] }_{ 1 }^{ 2 }\)
\(=log{ \left[ \frac { { (x+2) }^{ 2 } }{ x+1 } \right] }_{ 1 }^{ 2 }\)
\(=log\frac { 16 }{ 3 } -log\frac { 9 }{ 2 } \)
\(=log\frac { 32 }{ 27 } \)

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { cos\theta }{ (1+sin\theta )(2+sin\theta ) } } d\theta \)
Put u = 1 + sin\(\theta\)
Then, du = cos\(\theta\) d\(\theta\)
When \(\theta\) = 0, u = 1 
When \(\theta =\frac{\pi}{2}, u=2\)
\(\therefore I=\int _{ 1 }^{ 2 }{ \frac { du }{ u(1+u) } } =\int _{ 1 }^{ 2 }{ \frac { (1+u)-u }{ u(1+u) } du } =\int _{ 1 }^{ 2 }{ \left( \frac { 1 }{ u } -\frac { 1 }{ 1+u } \right) du=[logu-log(1+u)]_{ 1 }^{ 2 } } \)
\(=(log2-log3)-(log1-log2)=2log2-log3=log\frac { 4 }{ 3 } .\)

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

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 3 } }{ \frac { sec\ x\ tan\ x }{ 1+{ sec }^{ 2 }x } dx } \)
Put sec x = u. Then, sec x tan x dx = du.
When x = 0, u = sec0 = 1. When x = \(\frac{\pi}{3}, u=sec\frac{\pi}{2}=2\)
\(\therefore I=\int _{ 1 }^{ 2 }{ \frac { du }{ 1+{ u }^{ 2 } } ={ [{ tan }^{ -1 }u] }_{ 1 }^{ 2 }={ tan }^{ -1 }1={ tan }^{ -1 } } (2)-\frac { \pi }{ 4 } \)

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

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
\(\int _{ 0 }^{ 2a }{ f(x)dx } =2\int _{ 0 }^{ a }{ f(x)dx } \)

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

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

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
\(\therefore \int _{ -5 }^{ 5 }{ xcos\left( \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } \right) } dx=0\)

Let \(I=\int _{ 0 }^{ \pi }{ xlog\left( \frac { 3+cos\quad x }{ 3-cos\quad x } \right) \left( \frac { 3-cos\quad x }{ 3+cos\quad x } \right) dx } \quad ...(1)\)
\(I=\int _{ 0 }^{ 2\pi }{ (2\pi -x)log\left( \frac { 3+cos(2\pi -x) }{ 3-cos(2\pi -x) } \right) } dx\)
\(\left[ \because \int _{ 0 }^{ a }{ f(x)dx=\int _{ 0 }^{ a }{ (a-x)dx } } \right] \)
\(=\int _{ 0 }^{ 2\pi }{ 2\pi log\left( \frac { 3+cos\quad x }{ 3-cos\quad x } \right) dx } \)
\(-\int _{ 0 }^{ 2\pi }{ xlog\left( \frac { 3+cos\quad x }{ 3-cos\quad x } \right) dx } \)
\(\left[ \because cos(2\pi -x)=cos\quad x \right] \)
\(I=2\pi \int _{ 0 }^{ 2\pi }{ log\left( \frac { 3+cos\quad x }{ 3-cos\quad x } \right) dx-I } [using(1)]\)
\(\\ 2I=2\pi \int _{ 0 }^{ 2\pi }{ log\left( \frac { 3+cos\quad x }{ 3-cos\quad x } \right) dx } \quad \quad ...(2)\)
\(\left[ \because \int _{ 0 }^{ a }{ f(x)dx=2\int _{ 0 }^{ \frac { a }{ 2 } }{ f(x)dx\quad if(a-x)=f(x) } } \right] \)
I = 2 pi integral limit 0 to p_i
\(\left[ log3+\frac { x }{ 3-x } \right] dx\)
\(\Rightarrow I=2\pi \int _{ 0 }^{ \pi }{ log\left( \frac { 3+cos(\pi -x) }{ 3-cos(\pi -x) } \right) dx } \)
\(\Rightarrow I=2\pi \int _{ 0 }^{ \pi }{ log\left( \frac { 3-cos\quad x }{ 3+cos\quad x } \right) dx...(3) } \)
\([\because cos(\pi -x)=cosx]\)
Adding (2) and (3) we get,
\(2I=\int _{ 0 }^{ \pi }{ log\left( \frac { 3+cos\quad x }{ 3-cos\quad x } \right) \left( \frac { 3-cos\quad x }{ 3+cos\quad x } \right) dx } \)

\(\Rightarrow 2\pi \int _{ 0 }^{ 2\pi }{ 0dx=0 } \)
\(\Rightarrow 2I=0\Rightarrow I=0\)
\(\therefore \int _{ 0 }^{ 2\pi }{ xlog\left( \frac { 3+cos\ x }{ 3-cos\ x } \right) dx=0 } \)