# Solve Reimann Sums and Integrals

1. Apr 4, 2013

### SheldonAustin

We are already introduced to finding the value of definite integral by the anti-derivative approach

\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

2. Apr 4, 2013

### pwsnafu

It is the fundamental theorem of calculus that allows you to do this.

This "definition" is wrong! For example, take the interval [0,1] and consider the function f(x) = 1 when x is rational but f(x) = 0 when x is irrational. Then $0 + r \, \Delta x$ is always a rational hence the answer you'll get is 1. But the correct answer is "the function is not Riemann integral".

Take a look at the definition given on Wikipedia. It's very different to what you have written.