Evaluate: `int_1^3 (x^2+3x+e^x) dx`

Options

A. `(-e^2-1)`
B. `(e^2+1)`
C. `(e^2-1)`
D. None of these

Answer

To evaluate the integral \( \int_1^3 (x^2 + 3x + e^x) \, dx \), we will break it down into manageable parts.

### Step 1: Break down the integral
We can separate the integral into three parts:
\[ \int_1^3 (x^2 + 3x + e^x) \, dx = \int_1^3 x^2 \, dx + \int_1^3 3x \, dx + \int_1^3 e^x \, dx \]

### Step 2: Evaluate each integral separately

1. **Evaluate \( \int_1^3 x^2 \, dx \)**:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Now, we apply the limits from 1 to 3:
\[ \left[ \frac{x^3}{3} \right]_1^3 = \frac{3^3}{3} - \frac{1^3}{3} = \frac{27}{3} - \frac{1}{3} = \frac{26}{3} \]

2. **Evaluate \( \int_1^3 3x \, dx \)**:
\[ \int 3x \, dx = \frac{3x^2}{2} \]
Now, we apply the limits from 1 to 3:
\[ \left[ \frac{3x^2}{2} \right]_1^3 = \frac{3(3^2)}{2} - \frac{3(1^2)}{2} = \frac{3 \cdot 9}{2} - \frac{3 \cdot 1}{2} = \frac{27}{2} - \frac{3}{2} = \frac{24}{2} = 12 \]

3. **Evaluate \( \int_1^3 e^x \, dx \)**:
\[ \int e^x \, dx = e^x \]
Now, we apply the limits from 1 to 3:
\[ \left[ e^x \right]_1^3 = e^3 - e^1 = e^3 - e \]

### Step 3: Combine the results
Now we combine all the results:
\[ \int_1^3 (x^2 + 3x + e^x) \, dx = \frac{26}{3} + 12 + (e^3 - e) \]

To combine \( 12 \) with \( \frac{26}{3} \), we convert \( 12 \) into a fraction:
\[ 12 = \frac{36}{3} \]
Thus,
\[ \int_1^3 (x^2 + 3x + e^x) \, dx = \frac{26}{3} + \frac{36}{3} + (e^3 - e) = \frac{62}{3} + (e^3 - e) \]

### Final Answer
\[ \int_1^3 (x^2 + 3x + e^x) \, dx = \frac{62}{3} + e^3 - e \]