Evaluate `int_a^b(dx)/(sqrt(x)),w h e r ea , b > 0.`

Explanation

`I=int_(a)^(b)(dx)/(sqrt(x))` where `a,bgt0`
`=h lim_(nto oo) [1/(sqrt(a))+1/(sqrt(a+h))+1/(sqrt(a+2h))+…………..+1/(sqrt(a+(n-1)h))]`
We know that `sqrt(r)+sqrt(r-h) lt 2sqrt(r)ltsqrt(r+h)+sqrt(r)` (for sufficiently small `hgt0`). Thus
`1/(sqrt(r+h)+sqrt(r))lt1/(2sqrt(r)+sqrt(r))`
or `(sqrt(r+h)-sqrt(r))/hlt1/(2sqrt(r))lt(sqrt(r)-sqrt(r-h))/h`
Let put `r=a,a+h,a+2h,.........,a+(n-1)h`
`:.(sqrt(a+h)-sqrt(a))/hlt 1/(2sqrt(a)) lt (sqrt(a)-sqrt(a-h))/h`
`(sqrt(a+2h)-sqrt(a+h))/h lt 1/(2sqrt(a+h)) lt (sqrt(a+h)-sqrt(a))/h`
`(sqrt(a+3h)-sqrt(a+2h))/h lt 1/(2sqrt(a+2h)) lt (sqrt(a+2h)-sqrt(a+h))/h`
`(sqrt(a+nh)-sqrt(a+(n-1)h))/h lt 1/(2sqrt(a+(n-1)h))`
`lt(sqrt(a+(n-1)h)-sqrt(a+(n-2)h))/h`
Adding we get
`(sqrt(a+nh)-sqrt(a))/h lt sum_(r=0)^(n-1)1/(2sqrt(a+rh)) lt (sqrt(a+(n-1)h)-sqrt(a-h))/h`
or `2(sqrt(a+b-a)-sqrt(a))lth sum_(r=0)^(n-1)1/(sqrt(a+rh))`
`lt 2(sqrt(a+b-a-h)-sqrt(a-h))` (Put `nh=b-a`)
or `lim_(hto0) 2(sqrt(a+b-a)-sqrt(a))lt lim_(hto0)h sum_(r=0)^(n-1)1/(sqrt(a+rh))`
`lt lim_(hto0)2(sqrt(a+b-ah)-sqrt(a-h))`
or `2(sqrt(b)-sqrt(a)) lt lim_(hto0) h sum_(r=0)^(n-1)1/(sqrt(a+rh))lt 2(sqrt(b)-sqrt(a))`
or `(sqrt(b)-sqrt(a))lt int_(a)^(b)1/(sqrt(x)) dx lt 2(sqrt(b)-sqrt(a))`
or `int_(a)^(b)1/(sqrt(x)) dx=2(sqrt(b)-sqrt(a))`