Solve Reimann Sums and Integrals

In summary, a Riemann Sum is a method used to approximate the area under a curve by dividing it into smaller rectangles and adding up their areas. It is calculated by dividing the area into rectangles, calculating their areas, and adding them up. There are different types of Riemann Sums, but as the number of rectangles increases, the difference between them becomes smaller. Riemann Sums are important because they form the basis for calculating integrals, which are used in various fields to solve real-world problems involving rates of change and accumulation.
  • #1
SheldonAustin
2
0
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`
 
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  • #2
SheldonAustin said:
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.
 
Last edited:
  • #3
Hello pwsnafu,

Thanks for replying, will rectify it soon.
 

1. What is a Riemann Sum?

A Riemann Sum is a method used to approximate the area under a curve by dividing the area into smaller rectangles and adding up their areas. It is an important concept in calculus and is used to calculate integrals.

2. How do you calculate a Riemann Sum?

To calculate a Riemann Sum, you need to divide the area under the curve into smaller rectangles, calculate the area of each rectangle, and then add up the areas of all the rectangles. The more rectangles you use, the more accurate the approximation will be.

3. What is the difference between a left, right, and midpoint Riemann Sum?

A left Riemann Sum uses the left endpoint of each rectangle to calculate its area, a right Riemann Sum uses the right endpoint, and a midpoint Riemann Sum uses the midpoint. These methods can produce different approximations of the integral, but as the number of rectangles increases, the difference between them becomes smaller.

4. How does a Riemann Sum relate to integrals?

A Riemann Sum is the basis for calculating integrals. As the number of rectangles in a Riemann Sum approaches infinity, the approximation becomes more and more accurate, and it becomes equivalent to the value of the integral.

5. What is the importance of Riemann Sums and Integrals in real life?

Riemann Sums and Integrals are used in various fields such as physics, engineering, economics, and statistics. They are used to calculate important quantities such as area, volume, and displacement, and to solve real-world problems that involve rates of change and accumulation.

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