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:
$$ \int_{0}^{2} 4x^3dx = \lim_{n\to\infty} \sum_{i=1}^{n}4\bigg(\frac{2i}{n}\bigg)^3 \frac{2}{n} = \lim_{n\to\infty} \sum_{i=1}^{n} \frac{8i^3}{n^3} \frac{2}{n} = \lim_{n\to\infty} \sum_{i=1}^{n} \frac{16i^3}{n^4} \dots$$
Best Answer
Your last line should be $$ \int_{0}^{2} 4x^3dx = \lim_{n\to\infty} \sum_{i=1}^{n}4\bigg(\frac{2i}{n}\bigg)^3 \frac{2}{n} = \lim_{n\to\infty} \sum_{i=1}^{n} 4 \frac{8i^3}{n^3} \frac{2}{n} = \lim_{n\to\infty} \sum_{i=1}^{n} 4 \frac{16i^3}{n^4} = 64 \lim_{n\to\infty} \sum_{i=1}^{n} \frac {i^3}{n^4}$$
$$=64 \lim_{n\to\infty} \frac{n^2(n+1)^2}{4n^4} = \frac {64}{ 4 }= 16$$