`int_(1)^(3)f(x)dx`, equals where `f(x)={{:(2x+1","," when "1lexle2),(x^(2)+1","," when "2ltxle3):}`

Options

A. `(34)/(3)`
B. `(23)/(3)`
C. `(33)/(4)`
D. `(43)/(3)`

Answer

To solve the integral \( \int_{1}^{3} f(x) \, dx \), where the function \( f(x) \) is defined piecewise 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 can break the integral into two parts based on the definition of \( f(x) \).

### Step 1: Break the Integral
We can split the integral at the point where the definition of \( f(x) \) changes, which is at \( x = 2 \):

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

### Step 2: Substitute the Function
Now we substitute the appropriate expressions for \( f(x) \) in each interval:

1. For \( 1 \leq x \leq 2 \), \( f(x) = 2x + 1 \).
2. For \( 2 < x \leq 3 \), \( f(x) = x^2 + 1 \).

Thus, we have:

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

### Step 3: Calculate Each Integral
Now we calculate each integral separately.

**Integral from 1 to 2:**

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

Calculating this integral:

\[ = \left[ x^2 + x \right]_{1}^{2} \]

Calculating the limits:

\[ = \left[ (2^2 + 2) - (1^2 + 1) \right] = \left[ 4 + 2 - (1 + 1) \right] = 6 - 2 = 4 \]

**Integral from 2 to 3:**

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

Calculating this integral:

\[ = \left[ \frac{x^3}{3} + x \right]_{2}^{3} \]

Calculating the limits:

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

Simplifying:

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

Finding a common denominator:

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

### Step 4: Combine the Results
Now we combine the results of both integrals:

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

Converting 4 to a fraction:

\[ = \frac{12}{3} + \frac{22}{3} = \frac{34}{3} \]

### Final Answer
Thus, the value of the integral \( \int_{1}^{3} f(x) \, dx \) is:

\[ \boxed{\frac{34}{3}} \]