I try integrating $$\int_{-3}^{-1}\frac{4}{x} \, dx.$$
It's common that integral of power negative one can be expressed as natural logarithm i mean $\int \frac{1}{x} \, dx=\ln x$ but in this case both upper and lower integration limit are negative and this lead into undefined result.
so i try different approaching by using area interpretation of integrand and this give me a rectangle and triangle that have total area of $\frac{16}{3}$ and so i tried check it by riemann sum
$$\int_{-3}^{-1}f(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i) \, \Delta x$$
I took $\Delta x=\frac{2}{n}$ and $xi=\frac{2i}{n}-3$ but i can't procced any further after input $x_i$ into main equation and multiplying things and get :
$$\lim_{n\to\infty}\sum_{i=1}^n \frac{4n}{2i-3n} . \frac{2}{n} = \lim_{n\to\infty}\sum_{i=1}^n \frac{8}{2i-3n}$$
So how i can get past this type (quotient) of summation and got the integral result?
$$ \int \frac 4 x \, dx = 4\ln |x| + \text{constant}, $$ and unfortunately this means $$ 4\ln x + \begin{cases} \text{one constant} & \text{if } x>0, \\ \text{another constant} & \text{if } x<0. \end{cases} $$ Thus it is "constant" only on each interval separately.
At any rate, note the absolute value sign. Thus you have $$ \int_{-3}^{-1} \frac 4 x \, dx = 4\ln 1 - 4\ln 3 = -4\ln 3. $$