Evaluate the definite integral from [0,4] 4x^2 dx, by using its definition as a limit of approximating sums.

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.