also riemann?

Question

i got great answers for the first question, but i didn't get what i was asking for.

the teacher solved it by applying the formula of the sum of riemann. he used the function f(x1) with x1=x2+ k/n

it is the change of variable that i don't understand. what is k/n?

Answer 1

Its simple,

For finding the sum you devide the interval of integraion in to certain partitions(these partitions are of equal length, se we decide some rule for it).

The rieman sum is

Rsum = sum(i=1 to n)f(y_i)*(x_i - x_(i-1))

Where, n is the total number of partitions (to some students this n is confusing, whether its number of elements or number of partitions??).

Now, each of the partition is of certain length (b-a)/n,

So for partitions a = x0 < x1 < x2 ... < xn = b, each partition would be of the form (x_i, x_i+(b-a)/n).

Some teachers also use some tricky notations with indices which is quite confusing. May be thats the case with you.

Anyways, you can also try some other examples to make your consept more clear.

All thebest again and thanks for good question

Answer 2

k is the index in the summation and n the number of points that you sum over so n/k is a constant and you will find that f(x1)=f'(x2) if you make the choice above.

Have a look at the example at the bottom of
http://mathworld.wolfram.com/RiemannIntegral.html

and see if that is any use.

Answer 3

2..... ..... ..... ..... 2
∫x^2dx = (1/3)x^3 | = 8/3 = 2.333333333
1..... ..... ..... ..... 1
Riemann intervals:
F(1) = 1*1.5^2 = 2.25
F(2) = 0.5*1.25^2 + 0.5*1.75^2 = 2.3125
F(4) = 0.25*(1.125^2 + 1.375^2 + 1.625^2 + 1.875^2)
F(4) = 2.32813
F(10) = 0.1(1.05^2 + 1.15^2 + 1.25^2 + 1.35^2 + 1.45^2 + 1.55^2 + 1.65^2 + 1.75^2 + 1.85^2 + 1.95^2)
F(10) = 2.3325
................. 20
F(20) = 0.05∑(1 + 0.25k)^2 = 2.333125
.................. k=1
..... ..... ..... n
F(n) = (R/n)∑(1+(R/2)(k/n))^2
..... ..... ..... k=1

where R = range of x
n = number of intervals
k = {1,2,3,4,...,n}

Answer 4

What was your earlier question.