Between Sigma and Integral

Calculus Level 1

$\large A= \sum_{x=0}^n x \qquad, \qquad B= \int_0^n x \, dx$

If $n$ is an integer greater than 1, which of the above numbers yields a larger result?

A

They are both equal There is insufficient information

B

8 solutions

Athiyaman Nallathambi
Aug 26, 2015

Simplifying $A$ and $B$ we get, $A = \sum_{x=0}^{n}x = \frac{n\times(n+1)}{2} = \frac{n^2}{2}+\frac{n}{2}$ $B = \displaystyle\int_{0}^{n}xdx = \left[\frac{x^{2}}{2}\right]_{0}^{n} = \frac{n^2}{2}$ Clearly, $A > B$

Why isn't the sum of A equal to (0+n)*n/2?

Ruben Garcia Berasategui - 5 years, 9 months ago

We know that , $A = \sum_{x=0}^{n} x = 1+2+3+.....+n \rightarrow (i)$ If we reverse the sum $A$ , the sum remains the same, $A=n+(n-1)+(n-2)+.....+1 \rightarrow (ii)$ $(i)+(ii)$ , $2A = (n+1) + (n+1) +....... + (n+1) {\text{[ n terms ] }}$ $\Rightarrow 2A = n\times (n+1) \Rightarrow A = \dfrac {n\times (n+1)}{2}$

PS: You can check this if you still have doubts regarding this

Athiyaman Nallathambi - 5 years, 9 months ago

If you start counting from o number of terms is n+1 . It will be (0+n)*(n+1)/2

Venkata Joganpalli - 5 years, 9 months ago

Instead of n/2 it should be (n+1)/2 as there are n+1 terms(counting 0)

Het Thakkar - 5 years, 9 months ago

But summations only use integer values of n on the scale from zero to n. Integrals account for every single possible value of n. I don't see how the summation can possibly be greater than the integral and I am very confused due to this problem.

Jordan Reid - 5 years, 9 months ago

The question states that n is integer and greater than 1

Maeror Madidus - 5 years, 9 months ago

But it does not say x is an integer.

Venkata Joganpalli - 5 years, 9 months ago

@Venkata Joganpalli – If x is an integer greater than 1, which of the above numbers yields a larger result? - what does it mean then?

Maeror Madidus - 5 years, 9 months ago

@Maeror Madidus – A basic point which seems to miss the attention is the two quantities can not be compared. They are dimensionally different . In the summation the term equivalent to *dx * in the integral is not present.if one is area the other is only length.

Venkata Joganpalli - 5 years, 9 months ago

go through the real DEFINATION of integration ...(area) ...there is alwaya a small part left ....under the curve .. and summinatiion account for all that

shivanshu tiwari - 4 years, 11 months ago

The formulas for summations are only true when i=1 (the part they label x=0) ...... I am pretty sure.....

Tarah Gustafson - 5 years, 7 months ago

Good solution!!

Ankur Mukherjee - 5 years, 7 months ago

Thanks a lot :D

Athiyaman Nallathambi - 5 years, 7 months ago

Otto Bretscher
Aug 27, 2015

$A$ is an upper (Riemann) sum of the increasing function $f(x)=x$ between $x=0$ and $x=n$ .

and B is lower sum ?

Radinoiu Damian - 5 years, 9 months ago

No, the corresponding lower sum is $\sum_{x=0}^{n-1}x$ . The upper sum, $A$ , exceeds the integral $B$ , while the lower sum is less than $B$ .

Otto Bretscher - 5 years, 9 months ago

I get it now

Radinoiu Damian - 5 years, 9 months ago

@Radinoiu Damian – The lower Riemann integral would be $\frac{n(n-1)}{2}$

Curtis Clement - 5 years, 9 months ago

How can you know the sum inherently is an upper sum? I don't see how the upper bound n can change an upper sum into a lower sum and vice versa. This is how I see it: when x=0, it takes zero as the value until it gets to one, then one until it gets to two, and so on... If anything, it should be equivalent to a lower sum (using left endpoints).

http://pirate.shu.edu/~wachsmut/ira/integ/defs/ul_sum.html

For example, in this definition there's no emphasis placed on the bounds of a sum...

Anastasia Gladkina - 5 years, 9 months ago

Lance Fernando
Aug 28, 2015

Well, just logic will do. Since they're both of same numbers - we define the Sigma as the addition, whereas the Definite Integral refers to the subtraction.

In what way does integration refer to subtraction?

Niko Yochum - 5 years, 9 months ago

The upper and lower limits it suggests.

Grace Bismark - 5 years, 9 months ago

Alex Chandler
Sep 5, 2015

Think of the integral as an area under a straight line with slope 1. The sum can be interpreted as the area of a stair-step shape which always stays above the y=x line.

Hadia Qadir
Aug 28, 2015

A= n(n+1)/2 = n²/2+ n/2 B = n²/2.. .: A = B + n/2 A is greater.

Tamal Guha
Sep 7, 2015

Though the solution looks like A>B... But can we really relate these two quantities? Because A represents some length here, whereas, B stands for area... may be they have just magnitude difference... but that isn't really mean A>B...
(like 100m>10m^2!!! impossible to think!!!)

Parayus Mittal
Aug 27, 2015

A has got (n+1)terms = (n+1)(n+2)/2 B = n²/2

Notice from the sigma notation that the first term is for x=0.

Eric Escober - 5 years, 9 months ago

Shivanshu Tiwari
Jul 8, 2016

if you look at the basic defination of integration ...it states that it is the sum of the areas of the rectangle (of smallest width possible)....and when we take those areas the result (integration ) is always less than the true area (summesion)

Between Sigma and Integral

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