if f(x) =`{[3x+4 , 0 le x le 2],[5x , 2 le x le 3]},`then evaluate `int_(0)^(3) f(x) dx.`

Answer

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

\[ f(x) = \begin{cases} 3x + 4 & \text{for } 0 \leq x \leq 2 \\ 5x & \text{for } 2 < x \leq 3 \end{cases} \]

we will split the integral into two parts based on the definition of \( f(x) \).

### Step 1: Split the Integral
We can split the integral from \( 0 \) to \( 3 \) into two parts:
\[ \int_{0}^{3} f(x) \, dx = \int_{0}^{2} f(x) \, dx + \int_{2}^{3} f(x) \, dx \]

### Step 2: Substitute the Function
Now we substitute the function \( f(x) \) into the respective intervals:
\[ \int_{0}^{2} (3x + 4) \, dx + \int_{2}^{3} (5x) \, dx \]

### Step 3: Evaluate the First Integral
Now we evaluate the first integral:
\[ \int_{0}^{2} (3x + 4) \, dx \]
We can break this down:
\[ = \int_{0}^{2} 3x \, dx + \int_{0}^{2} 4 \, dx \]

Calculating each part:
1. For \( \int_{0}^{2} 3x \, dx \):
\[ = 3 \left[ \frac{x^2}{2} \right]_{0}^{2} = 3 \left[ \frac{2^2}{2} - \frac{0^2}{2} \right] = 3 \left[ \frac{4}{2} \right] = 3 \times 2 = 6 \]

2. For \( \int_{0}^{2} 4 \, dx \):
\[ = 4 \left[ x \right]_{0}^{2} = 4 [2 - 0] = 8 \]

Combining these results:
\[ \int_{0}^{2} (3x + 4) \, dx = 6 + 8 = 14 \]

### Step 4: Evaluate the Second Integral
Now we evaluate the second integral:
\[ \int_{2}^{3} 5x \, dx \]
Calculating this:
\[ = 5 \left[ \frac{x^2}{2} \right]_{2}^{3} = 5 \left[ \frac{3^2}{2} - \frac{2^2}{2} \right] = 5 \left[ \frac{9}{2} - \frac{4}{2} \right] = 5 \left[ \frac{5}{2} \right] = \frac{25}{2} \]

### Step 5: Combine Both Integrals
Now we combine the results of both integrals:
\[ \int_{0}^{3} f(x) \, dx = 14 + \frac{25}{2} \]

To add these, convert \( 14 \) to a fraction:
\[ 14 = \frac{28}{2} \]
Thus:
\[ \int_{0}^{3} f(x) \, dx = \frac{28}{2} + \frac{25}{2} = \frac{53}{2} \]

### Final Answer
The value of the integral \( \int_{0}^{3} f(x) \, dx \) is:
\[ \frac{53}{2} \]