divide the specified interval into n=4 subintervals of equal length and then compute using the Area under a curve
with $\displaystyle x^*_k$as a left endpoint of each subinterval
f(x)=3x+1; [2,6]
$\displaystyle A \approx \sum_{k = 0}^{n} f(x_{k}^*) \Delta x$
Each interval is: $\displaystyle \Delta x = \frac{b-a}{n} = \frac{6-2}{4} = 1$
And since we're using left-end points, i.e. $\displaystyle x^*_k = x_{k-1}$, we have: $\displaystyle x_0 = 2, \ x_1 = 3, x_2 = 4, x_3 = 5$
So plugging it all in: $\displaystyle A \approx \sum_{k=1}^4 f(x^*_k) \Delta x = \sum_{k=1}^4 f(x_{k-1}) (1) = f(x_0) + f(x_1) + f(x_2) + f(x_3) = \cdots$