Transform the integral `int_(0)^(3) (x-2)^(2) dx ` by the substitution `(x - 2) = t`

Answer

To solve the integral \( \int_{0}^{3} (x-2)^{2} \, dx \) using the substitution \( (x - 2) = t \), we will follow these steps:

### Step 1: Substitute \( (x - 2) = t \)
Let \( t = x - 2 \). Then, we can express \( x \) in terms of \( t \):
\[ x = t + 2 \]

### Step 2: Find \( dx \) in terms of \( dt \)
Differentiating both sides with respect to \( t \):
\[ dx = dt \]

### Step 3: Change the limits of integration
We need to change the limits of integration according to our substitution:
- When \( x = 0 \):
\[ t = 0 - 2 = -2 \]
- When \( x = 3 \):
\[ t = 3 - 2 = 1 \]

So the new limits of integration are from \( -2 \) to \( 1 \).

### Step 4: Rewrite the integral
Now we can rewrite the integral in terms of \( t \):
\[ \int_{0}^{3} (x-2)^{2} \, dx = \int_{-2}^{1} t^{2} \, dt \]

### Step 5: Integrate
Now we will integrate \( t^{2} \):
\[ \int t^{2} \, dt = \frac{t^{3}}{3} + C \]
Thus, we evaluate:
\[ \int_{-2}^{1} t^{2} \, dt = \left[ \frac{t^{3}}{3} \right]_{-2}^{1} \]

### Step 6: Evaluate the definite integral
Now we substitute the limits:
\[ = \left( \frac{1^{3}}{3} \right) - \left( \frac{(-2)^{3}}{3} \right) \]
Calculating each term:
\[ = \frac{1}{3} - \frac{-8}{3} = \frac{1}{3} + \frac{8}{3} = \frac{9}{3} = 3 \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{3} (x-2)^{2} \, dx = 3 \]

---