# Evaluating an Integral as a Riemann sum

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:

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

• I don't think you can get rid of $\Sigma$ like that – J. W. Tanner Feb 4 at 19:51
• You're right, I forgot to add them in. Fixed it – Evan Kim Feb 4 at 19:57
• I still don't understand what exactly it is that you're asking. – Michael Rybkin Feb 4 at 20:04
• There are still mistakes in what you wrote: what happened to $4$ after the second $=$ in the last line? – J. W. Tanner Feb 4 at 20:07
• There are also mistakes in what you wrote for your teacher's solution – J. W. Tanner Feb 4 at 20:18

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