find a limit equal to $int_{1}^{5} ln(2x - 1) dx.$

Explanation:

Step1: Recall the definition of definite - integral as a limit

The definite integral $\int_{a}^{b}f(x)dx=\lim_{n\rightarrow\infty}\sum_{i = 1}^{n}f(x_i)\Delta x$, where $\Delta x=\frac{b - a}{n}$ and $x_i=a + i\Delta x$. Here, $a = 1$, $b = 5$, and $f(x)=\ln(2x - 1)$. So, $\Delta x=\frac{5 - 1}{n}=\frac{4}{n}$ and $x_i=1+\frac{4i}{n}$.

Step2: Write the Riemann - sum

The Riemann - sum $\sum_{i = 1}^{n}f(x_i)\Delta x=\sum_{i = 1}^{n}\ln\left(2\left(1+\frac{4i}{n}\right)-1\right)\cdot\frac{4}{n}=\sum_{i = 1}^{n}\ln\left(1+\frac{8i}{n}\right)\cdot\frac{4}{n}$.

Step3: Write the limit

The definite integral $\int_{1}^{5}\ln(2x - 1)dx=\lim_{n\rightarrow\infty}\sum_{i = 1}^{n}\ln\left(1+\frac{8i}{n}\right)\cdot\frac{4}{n}$.

Answer:

$\lim_{n\rightarrow\infty}\sum_{i = 1}^{n}\ln\left(1+\frac{8i}{n}\right)\cdot\frac{4}{n}$