Question

evaluate the definite integral.

$$\int_{0}^{2} \frac{d x}{3 x+3}$$

question help: video

Explanation:

Step1: Use substitution

Let \(u = 3x+3\), then \(du=3dx\), \(dx=\frac{1}{3}du\).
When \(x = 0\), \(u=3\times0 + 3=3\); when \(x = 2\), \(u=3\times2+3 = 9\).
The integral \(\int_{0}^{2}\frac{dx}{3x + 3}\) becomes \(\frac{1}{3}\int_{3}^{9}\frac{du}{u}\).

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

We know that \(\int\frac{du}{u}=\ln|u|+C\).
So \(\frac{1}{3}\int_{3}^{9}\frac{du}{u}=\frac{1}{3}[\ln u]_{3}^{9}\).

Step3: Evaluate the definite - integral

Using the fundamental theorem of calculus \(F(b)-F(a)\), where \(F(u)=\frac{1}{3}\ln u\), \(a = 3\), \(b = 9\).
\(\frac{1}{3}(\ln9-\ln3)=\frac{1}{3}\ln\frac{9}{3}\) (since \(\ln m-\ln n=\ln\frac{m}{n}\)).
\(\frac{1}{3}\ln3=\frac{\ln3}{3}\).

Answer:

\(\frac{\ln3}{3}\)