Evaluate the following integrals:
`(i) int_(0)^(5)(x+1)dx` `(ii) int_(1)^(3)(2x+3)dx`

Explanation

`(i)` Let `f(x)=(x+1)`, `a=0`, `b=5` and `nh=(5-0)=5`. Then,
`int_(0)^(5)(x+1)dx=lim_(hto0)=lim_(hto0)h[f(0)+f(0+h)+f(0+2h)+….+f{0+(n-1)h}]`
`=lim_(hto0)h[1+(h+1)+(2h+1)+...+{(n-1)h+1}]`
`=lim_(hto0)h[n+{h+2h+3h+..+(n-1)h}]`
`lim_(hto0)h[n+{1+2+3+...+(n-1)}h]`
`=lim_(hto0)h[n+(n(n-1))/(2)h]=lim_(hto0)[nh+(nh(nh-h))/(2)]`
`=lim_(hto0)[5+(5(5-h))/(2)] [:'nh=5]`
`=(35)/(2)`.
`(ii)` Let `f(x)=(2x+3)`, `a=1`, `b=3` and `nh=(3-1)=2`.
Then, `int_(1)^(3)(2x+3)dx=lim_(hto0)h[f(1)+f(1+h)+....+f{1+(n-1)h}]`
`=lim_(hto0)h[5+(5+2h)+(5+4h)+...+{5+2(n-1)h}]`
[`:'f(1)=5,f(1+h)=5+2h`, etc.]
`=lim_(hto0)h[5n+{2h+4h+...+2(n-1)h}]`
`=lim_(hto0)h[5n+2{1+2+3+...+(n-1)}h]`
`=lim_(hto0)h[5n+2*(n(n-1))/(2)h]`
`=lim_(hto0)[5nh+nh(nh-h)]`
`=lim_(hto0)[10+2(2-h)]` `[:'nh=2]`
`=14`.