Evaluate : `int_(0)^(3) f(x) dx`,
where `f(x)={(cos2x",",0 le x le (pi)/(2)),(3",",(pi)/(2) le x le 3):}`

Answer

To evaluate the integral \( \int_{0}^{3} f(x) \, dx \) where

\[ f(x) = \begin{cases} \cos(2x) & \text{for } 0 \leq x \leq \frac{\pi}{2} \\ 3 & \text{for } \frac{\pi}{2} < x \leq 3 \end{cases} \]

we can break the integral into two parts based on the definition of \( f(x) \):

1. From \( 0 \) to \( \frac{\pi}{2} \)
2. From \( \frac{\pi}{2} \) to \( 3 \)

Thus, we can write:

\[ \int_{0}^{3} f(x) \, dx = \int_{0}^{\frac{\pi}{2}} f(x) \, dx + \int_{\frac{\pi}{2}}^{3} f(x) \, dx \]

### Step 1: Evaluate the first integral

For the first integral, we have:

\[ \int_{0}^{\frac{\pi}{2}} f(x) \, dx = \int_{0}^{\frac{\pi}{2}} \cos(2x) \, dx \]

To solve this integral, we can use the substitution:

\[ \int \cos(2x) \, dx = \frac{1}{2} \sin(2x) + C \]

Now we evaluate the definite integral:

\[ \int_{0}^{\frac{\pi}{2}} \cos(2x) \, dx = \left[ \frac{1}{2} \sin(2x) \right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} \sin(\pi) - \frac{1}{2} \sin(0) = \frac{1}{2} \cdot 0 - \frac{1}{2} \cdot 0 = 0 \]

### Step 2: Evaluate the second integral

For the second integral, we have:

\[ \int_{\frac{\pi}{2}}^{3} f(x) \, dx = \int_{\frac{\pi}{2}}^{3} 3 \, dx \]

This integral can be computed as follows:

\[ \int_{\frac{\pi}{2}}^{3} 3 \, dx = 3 \left[ x \right]_{\frac{\pi}{2}}^{3} = 3 \left( 3 - \frac{\pi}{2} \right) = 9 - \frac{3\pi}{2} \]

### Step 3: Combine the results

Now we combine the results from both integrals:

\[ \int_{0}^{3} f(x) \, dx = 0 + \left( 9 - \frac{3\pi}{2} \right) = 9 - \frac{3\pi}{2} \]

Thus, the final answer is:

\[ \int_{0}^{3} f(x) \, dx = 9 - \frac{3\pi}{2} \]