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Evaluate the integral as a Riemann sum $\int_{0}^{2} 4x^3dx$.

My book defines an definite integral as $$ \int_{a}^{b} f(x) dx = \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i) \Delta x $$
where ${x_i} = a+ i \Delta x$ and ${\Delta x} = \frac{b-a}{n}$.

Here is the answer key. My teacher decides to use the summation of $n^3$ integers to cancel out $i/n$.

$$\Delta x = \frac{2}{n}, x_i =\frac{2}{n}i$$

$$\begin{align} \int_{0}^{2} 4x^3dx &= \lim_{n\to\infty} \sum_{i=1}^{n}4\bigg(\frac{2i}{n}\bigg)^3 \frac{2}{n} \\ &= \lim_{n\to\infty} \frac{8}{n} \sum_{i=1}^{n}\frac{8i^3}{n^3} = \lim_{n\to\infty} \frac{64}{n^3} \sum_{i=1}^{n}i^3 \\ &= \lim_{n\to\infty} \sum_{i=1}^{n}\frac{64}{n^4} \bigg(\frac{n^2 + n}{n^2}\bigg)^2 = \lim_{n\to\infty} \sum_{i=1}^{n}16 \bigg(1 + \frac{1}{n} \bigg)^2 \\ &= 16(1)^2 = 16 \end{align} $$

Is there a shorter method to show that the integral approximates to around $16$? The following is all I could get:

$$ \int_{0}^{2} 4x^3dx = \lim_{n\to\infty} \sum_{i=1}^{n}4\bigg(\frac{2i}{n}\bigg)^3 \frac{2}{n} = \lim_{n\to\infty} \sum_{i=1}^{n} \frac{8i^3}{n^3} \frac{2}{n} = \lim_{n\to\infty} \sum_{i=1}^{n} \frac{16i^3}{n^4} \dots$$

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  • $\begingroup$ I don't think you can get rid of $\Sigma$ like that $\endgroup$ – J. W. Tanner Feb 4 at 19:51
  • $\begingroup$ You're right, I forgot to add them in. Fixed it $\endgroup$ – Evan Kim Feb 4 at 19:57
  • $\begingroup$ I still don't understand what exactly it is that you're asking. $\endgroup$ – Michael Rybkin Feb 4 at 20:04
  • $\begingroup$ There are still mistakes in what you wrote: what happened to $4$ after the second $=$ in the last line? $\endgroup$ – J. W. Tanner Feb 4 at 20:07
  • $\begingroup$ There are also mistakes in what you wrote for your teacher's solution $\endgroup$ – J. W. Tanner Feb 4 at 20:18
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Your last line should be $$ \int_{0}^{2} 4x^3dx = \lim_{n\to\infty} \sum_{i=1}^{n}4\bigg(\frac{2i}{n}\bigg)^3 \frac{2}{n} = \lim_{n\to\infty} \sum_{i=1}^{n} 4 \frac{8i^3}{n^3} \frac{2}{n} = \lim_{n\to\infty} \sum_{i=1}^{n} 4 \frac{16i^3}{n^4} = 64 \lim_{n\to\infty} \sum_{i=1}^{n} \frac {i^3}{n^4}$$

$$=64 \lim_{n\to\infty} \frac{n^2(n+1)^2}{4n^4} = \frac {64}{ 4 }= 16$$

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  • $\begingroup$ okay, thank you. That makes more sense $\endgroup$ – Evan Kim Feb 5 at 2:25

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