Calculate definite integral (4x^2) using definition as a limit of approximating sums

**VinceW** · May 31st 2011, 07:22 AM

Evaluate $\int_0^4 4x^2 \mathrm{d}x$ by using its definition as a limit of approximating sums.

I can easily calculate the target answer which is 256/3. But I'm having trouble solving it the way the problem asks:

Limit of approximating sums is:

$\lim\limits_{n \to \infty} \sum\limits_{j=1}^n f(c_j) (x_j - x_{j - 1}) = \lim\limits_{n \to \infty} \sum\limits_{j=1}^n f(\frac{4j}{n}) \frac{4}{n} = \lim\limits_{n \to \infty} \sum\limits_{j=1}^n \frac{4^3j^2}{n^2} \frac{4}{n} = 256 \lim\limits_{n \to \infty} \sum\limits_{j=1}^n \frac{j^2}{n^3}$

At that point I'm stuck. Any tips?

**Prove It** · May 31st 2011, 07:31 AM

$\displaystyle n$ is a constant in the sum, so write it as

$\displaystyle\begin{align*} \lim_{n \to \infty}\left(\frac{1}{n^3}\sum_{j = 1}^n{j^2}\right) &= \lim_{n \to \infty}\left\{\frac{1}{n^3}\left[\frac{n(n+1)(2n+1)}{6}\right]\right\}\\ &= \lim_{n \to \infty}\frac{(n+1)(2n+1)}{6n^2}\\ &= \lim_{n \to \infty}\frac{2n^2 + 3n + 1}{6n^2} \\ &= \lim_{n \to \infty}\left(\frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^2}\right)\\ &= \frac{1}{3}\end{align*}$

**VinceW** · May 31st 2011, 11:44 AM

duplicate

**VinceW** · May 31st 2011, 11:47 AM

Thanks! Solving the following summation was the really tricky part. I got it now thanks to your help.

$\sum\limits_{j=1}^n j^2 = \frac{n(n+1)(2n+1)}{6}$

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Calculate definite integral (4x^2) using definition as a limit of approximating sums

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