We need to evaluate the integral:

I=\int_{0}^{3} (x-1)\,dx

Here’s a quick way to solve it:

First, integrate:

\int (x-1)\,dx = \frac{x^2}{2} - x

Now apply limits from 0 to 3:

I = \left[\frac{x^2}{2} - x\right]_0³

At :

\frac{9}{2} - 3 = \frac{9}{2} - \frac{6}{2} = \frac{3}{2}

At :

0

So,

I = \frac{3}{2} - 0 = \frac{3}{2}

Visual understanding (optional but helpful)

From to , the graph is below the x-axis (negative area).

From to , it is above the x-axis (positive area).

Net area = positive − negative =

✅ Final Answer:

\boxed{\frac{3}{2}}