`"If "f(x) ={underset(x^(2)+1.2 le x le 3)(2x+1.1 le x le 2),` then evaluate `int_(1)^(3) f(x) dx.`

Answer

To evaluate the integral \( \int_{1}^{3} f(x) \, dx \) where the function \( f(x) \) is defined as:

\[ f(x) = \begin{cases} 2x + 1 & \text{for } 1 \leq x \leq 2 \\ x^2 + 1 & \text{for } 2 < x \leq 3 \end{cases} \]

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

### Step 1: Break the integral into two parts

We can split the integral from 1 to 3 into two intervals: from 1 to 2 and from 2 to 3.

\[ \int_{1}^{3} f(x) \, dx = \int_{1}^{2} f(x) \, dx + \int_{2}^{3} f(x) \, dx \]

### Step 2: Substitute the function for each interval

For the first interval \( [1, 2] \), \( f(x) = 2x + 1 \). For the second interval \( [2, 3] \), \( f(x) = x^2 + 1 \).

Thus, we have:

\[ \int_{1}^{3} f(x) \, dx = \int_{1}^{2} (2x + 1) \, dx + \int_{2}^{3} (x^2 + 1) \, dx \]

### Step 3: Evaluate the first integral

Now we evaluate \( \int_{1}^{2} (2x + 1) \, dx \):

\[ \int (2x + 1) \, dx = x^2 + x + C \]

Now we apply the limits from 1 to 2:

\[ \left[ x^2 + x \right]_{1}^{2} = (2^2 + 2) - (1^2 + 1) = (4 + 2) - (1 + 1) = 6 - 2 = 4 \]

### Step 4: Evaluate the second integral

Next, we evaluate \( \int_{2}^{3} (x^2 + 1) \, dx \):

\[ \int (x^2 + 1) \, dx = \frac{x^3}{3} + x + C \]

Now we apply the limits from 2 to 3:

\[ \left[ \frac{x^3}{3} + x \right]_{2}^{3} = \left( \frac{3^3}{3} + 3 \right) - \left( \frac{2^3}{3} + 2 \right) = \left( \frac{27}{3} + 3 \right) - \left( \frac{8}{3} + 2 \right) \]

Calculating this gives:

\[ = (9 + 3) - \left( \frac{8}{3} + 2 \right) = 12 - \left( \frac{8}{3} + \frac{6}{3} \right) = 12 - \frac{14}{3} \]

Finding a common denominator:

\[ = \frac{36}{3} - \frac{14}{3} = \frac{22}{3} \]

### Step 5: Combine the results

Now, we combine the results of both integrals:

\[ \int_{1}^{3} f(x) \, dx = 4 + \frac{22}{3} \]

To add these, convert 4 to a fraction:

\[ 4 = \frac{12}{3} \]

So,

\[ \int_{1}^{3} f(x) \, dx = \frac{12}{3} + \frac{22}{3} = \frac{34}{3} \]

### Final Answer

Thus, the value of the integral is:

\[ \int_{1}^{3} f(x) \, dx = \frac{34}{3} \]

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