Per definition we have

\displaystyle \int_0^{B} f(x) \, dx = \lim_{n \to \infty} \left( \sum_{i=0}^{n-1} f(x_i) \Delta x \right).

We devide [0,B] in n partial intervals [x_i,x_{i+1}] of the same width \displaystyle \Delta x = \frac{B -0}n. Therefore x_i = i \cdot \dfrac{B}n.

Using this we determine the limit.

It follows:

\displaystyle \lim_{n \to \infty} \left( \sum_{i=0}^{n-1} f(x_i) \Delta x \right) = \lim_{n \to \infty} \left( \sum_{i=0}^{n-1} a \cdot i \cdot \frac{B}{n} \cdot \frac{B}{n} \right) = \lim_{n \to \infty} \left( \dfrac{a*B*B}{n^2} \sum_{i=0}^{n-1} i \right) .

Apply now that: \displaystyle \sum_{i=0}^{n-1} i = \frac{(n-1)n}{2}.

It yields:

\displaystyle \lim_{n \to \infty} \left( \frac{a*B*B}{{\color{blue}n^2}} \sum_{i=0}^{n-1} i \right) = \lim_{n \to \infty} \left( \frac{a*B*B}{{\color{blue}n^2}} \cdot \frac{(n-1)n}{2} \right) = \dfrac{a*B*B}{2} \cdot {\color{blue}\lim_{n \to \infty} \frac{(n-1)n}{n^2}}.

The limit is

\displaystyle {\color{blue}\lim_{n \to \infty} \frac{(n-1)n}{n^2}} = \lim_{n \to \infty} \frac{n^2-n}{n^2} = \lim_{n \to \infty} 1-\frac{1}{n} = \color{blue}1.

and eventually

\displaystyle \int_{0}^{B} a x \, dx = \dfrac{a*B*B}{2} \cdot {\color{blue}1} = fractionReduce(a*B*B,2).