We are already introduced to finding the value of definite integral by the anti-derivative approach(adsbygoogle = window.adsbygoogle || []).push({});

\int_{a}^{b}f(x) dx = F(b) - F(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 )

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)` <br>

`=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 anti-derivative 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`

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Solve Reimann Sums and Integrals

**Physics Forums | Science Articles, Homework Help, Discussion**