Solve Reimann Sums and Integrals

[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`

 

[pwsnafu]

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

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

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 )

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.

Edit: and it turns out Wikipedia talks about this as well. There you go.

 

[SheldonAustin]

Hello pwsnafu,

Thanks for replying, will rectify it soon.