Solve Reimann Sums and Integrals

We are already introduced to finding the value of definite integral by the anti-derivative approach
b
∫ f(x) dx = F(b) - F(a)
a
In this  approach we find  the anti-derivative F(x) of f(x)  and then subtract F(a) from F(b) to get the value of the definite integral
Reimann Sums is yet another method to compute the definite integral value from first principles
Let f(x)  be  a function defined on the closed interval [a,b] then the definite interval of f(x) from a to b is given by
      `int_a^bf(x)dx`          = `lim_(n->oo)(b-a)/n`     `sum_(r=1)^n f(a+r` Δx)
where Δx = `(b - a)/n`  ( provided the limit exits ) 

Ilustrative example to understand Reimann Sums as a method to determine the definite integral value

Examples for solve riemann sums and integrals
`int_1^2(2x+5)dx`
    Let  f(x) = 2x+5  and [a,b] = [1,2]
       Δx = `(b-a)/n`  = `(2-1)/n`  =`1/n`
       f(x) = 2x+5
f(a+rΔ) = f(1 +  r`(1)/n` )  =2 (1 +  r`(1)/n` )  + 5
By the Formula
      `int_a^bf(x)dx`          = `lim_(n->oo)(b-a)/n`     `sum_(r=1)^n f(a+r` Δx)
where Δx = `(b - a)/n`  ( provided the limit exits )
 `int_1^2(2x+5)dx`  = `lim_(n->oo)(1)/n`     `sum_(r=1)^n ``(2(1+ r/n)+5)`
= `lim_(n->oo)(1)/n`     `sum_(r=1)^n (7+ 2r/n)`
`=lim_(n->oo)(1)/n sum_(r=1)^n``(7) +lim_(n->oo)(1)/n (2/n)sum_(r=1)^n (r)`
=`lim_(x->oo)1/n(7n)+1/n(2/n)lim_(n->oo) (n)(n+1)/2'`

`='lim_(n->oo)(7) +lim_(n->oo) 1/n(2/n)(n)(n+1)/2'`

`= 7 + lim_(n->oo)(n+1)/n`
`= 7 + lim_(n->oo)(1+ 1/n)`
`= 7 + 1`
`= 8`
Ans is 8
Let us Validate the Answer by using the antiderivative Approach
`int_1^2(2x+5)dx= (2(x/2)^2 +5x)1^2`
`='(2^2-1^2) +5(2-1)'`
`= (4-1) + 5`
`= 3 +5`
`= 8`

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Conclusion for solve riemann sums and integrals

Thus Reimann Sums method divides the area under the curve into numerous rectangles and does the summation of the area of each rectangle to give the total  area under the curve which is essentially the value of the definite intergal