Express the limit of a Riemann sum as a definite integral

Question

$$\lim\limits_{n\rightarrow\infty}\sum\limits_{i=1}^n\left(\frac{12}{n}+\frac{8i}{n^2}\right)$$

I'm having trouble understanding how to find my limits of integration here. If I factor out $\frac{1}{n}$ I have $\Delta x=\frac{1}{n}$ and if I factor out $\frac{4}{n}$ I have $\Delta x=\frac{4}{n}$ and these result in different definite integrals.

What am I not understanding here?

Answer 1

You could write it as $$\int_0^1(12+8x)dx=\lim\limits_{n\to\infty}\sum\limits_{i=1}^n(12+8x)\Delta x$$ with $\Delta x=\dfrac1n$ and $x=\dfrac in$.

Answer 2

Recall Riemann sum: $$\lim_{n\to \infty} \frac{b-a}n \sum_{i=1}^n f\bigg(a+i\frac{b-a}n\bigg)=\int_a^bf(x)dx$$ We have: \begin{align} \lim_{n\to \infty} \sum_{i=1}^n \bigg(\frac{12}n+\frac{8i}{n^2}\bigg)&=\lim_{n\to \infty} \sum_{i=1}^n \frac{1}n \bigg(12+\frac{8i}n\bigg)\\ &=\lim_{n\to \infty} \frac{1}n \sum_{i=1}^n 12 +\frac{8i}n\\ &=\int_0^1 12 +8x dx\\ &=12x +\frac{8x^2}{2} \bigg\vert_0^1\\ &=16 \end{align} And : \begin{align} \lim_{n\to \infty} \sum_{i=1}^n \bigg(\frac{12}n+\frac{8i}{n^2}\bigg)&=\lim_{n\to \infty} \sum_{i=1}^n \frac{4}n \bigg(3+\frac{2i}n\bigg)\\ &=\lim_{n\to \infty} \frac{4}n \sum_{i=1}^n 3 +\frac{2i}n\\ &=\int_0^4 3+\frac{x}2 dx\\ &=3x+\frac{x^4}4 \bigg\vert_0^4\\ &=16 \end{align}