Question

evaluate the definite integral.

$$ int _ { - 2 } ^ { 0 } \frac { d x } { 6 x - 4 } $$

Explanation:

Step1: Use substitution

Let \(u = 6x-4\), then \(du=6dx\), \(dx=\frac{du}{6}\).
When \(x = - 2\), \(u=6\times(-2)-4=-16\). When \(x = 0\), \(u=6\times0 - 4=-4\).
The integral becomes \(\int_{-16}^{-4}\frac{1}{u}\cdot\frac{du}{6}=\frac{1}{6}\int_{-16}^{-4}\frac{du}{u}\).

Step2: Integrate \(\frac{1}{u}\)

We know that \(\int\frac{1}{u}du=\ln|u|+C\).
So \(\frac{1}{6}\int_{-16}^{-4}\frac{du}{u}=\frac{1}{6}[\ln|u|]_{-16}^{-4}\).

Step3: Evaluate the definite - integral

\(\frac{1}{6}(\ln|-4|-\ln|-16|)=\frac{1}{6}(\ln4-\ln16)\).
Using the property \(\ln a-\ln b=\ln\frac{a}{b}\), we have \(\frac{1}{6}\ln\frac{4}{16}=\frac{1}{6}\ln\frac{1}{4}\).
Since \(\ln\frac{1}{4}=\ln4^{-1}=-\ln4\), then \(\frac{1}{6}\ln\frac{1}{4}=-\frac{\ln4}{6}=-\frac{\ln2^{2}}{6}\).
Using the property \(\ln a^{b}=b\ln a\), we get \(-\frac{2\ln2}{6}=-\frac{\ln2}{3}\).

Answer:

\(-\frac{\ln2}{3}\)