Evaluate the definite integral from [0,4] 4x^2 dx, by using its definition as a limit of approximating sums.
First, I solve analytically so I know the answer I am trying to reach: 4/3 * x^3 over [0,4] = 4/3 * 4^3 = 256/3
Now, by approximating sums, I can get:
256 * lim(n -> infinity) of sum(j=0 to n-1) of j^2/n^3
I can use the computer to solve this and I reach the correct answer. In Mathematica Alpha notation: 256 * lim sum j^2/n^3, j=0..n as n -> infinity
How do I solve that limit by hand? Or is there a better way to solve the original problem?
You need to derive the formula for:
Sum from j = 0 to N of j^2
There are many different ways to do this, the more advanced math you know, the simpler it gets :).
An elementary method is to consider summing (j+1)^3 - j^3 instead of j^2. Obviously, if you sum a function of the form f(j+1) - f(j), all the terms except the first and last one will cancel:
Sum from j = 0 to N of [f(j+1) - f(j)] =
f(N+1) - f(0)
If we choose f(j) = j^3, then:
f(j+1) - f(j) =
(j+1)^3 - j^3 =
3 j^2 + 3 j + 1
So, if you know the summation of j from zero to N, you can find the summation of j^2. Of course, you can find the formula for the summation of j in the same way by taking f(j) = j^2.
You are awesome Iblis. That limit was the tricky part. I followed your proof and it makes perfect sense. Thanks
I meant that summation was the tricky part (not limit). thanks again.
To evaluate the limit of the sum without using a computer, you can use the formula for the sum of squares.
The formula for the sum of squares is given by:
1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6
In this case, we need to find the limit of the sum of j^2/n^3 as n approaches infinity.
Let's rewrite the sum in terms of n:
sum(j^2/n^3, j=0..n-1) = n^3 * sum(j^2/n^3, j=0..n-1)/n^3
Now, substitute the formula for the sum of squares:
sum(j^2/n^3, j=0..n-1) = (n-1)(n)(2n-1)/6
Substitute this expression back into the original numerator:
n^3 * (n-1)(n)(2n-1)/6 / n^3
Cancel out common factors:
(n-1)(n)(2n-1)/6
As n approaches infinity, we can ignore the terms involving n-1 and 2n-1, and the limit simplifies to:
lim(n->infinity) of (n(n)(2n))/6 = lim(n->infinity) of (2n^3)/6 = lim(n->infinity) of (n^3)/3
Now we have a simple limit involving only n, so we can evaluate it easily:
lim(n->infinity) of (n^3)/3 = infinity/3 = infinity
Therefore, the limit of the sum is infinity, and the value of the definite integral is also infinity.
This means that the original problem does not have a finite value for the definite integral.