Evaluate `int_(1)^(4) (ax^(2)+bx+c)dx` .

Answer

To evaluate the integral \( \int_{1}^{4} (ax^2 + bx + c) \, dx \), we can break it down into simpler parts. Here’s how we can do it step by step:

### Step 1: Break down the integral
We can separate the integral into three parts:
\[ \int_{1}^{4} (ax^2 + bx + c) \, dx = \int_{1}^{4} ax^2 \, dx + \int_{1}^{4} bx \, dx + \int_{1}^{4} c \, dx \]

### Step 2: Factor out constants
Since \( a \), \( b \), and \( c \) are constants, we can factor them out of their respective integrals:
\[ = a \int_{1}^{4} x^2 \, dx + b \int_{1}^{4} x \, dx + c \int_{1}^{4} 1 \, dx \]

### Step 3: Evaluate each integral
Now, we will evaluate each integral separately.

1. **Integral of \( x^2 \)**:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Evaluating from 1 to 4:
\[ \left[ \frac{x^3}{3} \right]_{1}^{4} = \frac{4^3}{3} - \frac{1^3}{3} = \frac{64}{3} - \frac{1}{3} = \frac{63}{3} = 21 \]

2. **Integral of \( x \)**:
\[ \int x \, dx = \frac{x^2}{2} \]
Evaluating from 1 to 4:
\[ \left[ \frac{x^2}{2} \right]_{1}^{4} = \frac{4^2}{2} - \frac{1^2}{2} = \frac{16}{2} - \frac{1}{2} = 8 - \frac{1}{2} = \frac{15}{2} \]

3. **Integral of 1**:
\[ \int 1 \, dx = x \]
Evaluating from 1 to 4:
\[ \left[ x \right]_{1}^{4} = 4 - 1 = 3 \]

### Step 4: Combine the results
Now we can substitute back into our expression:
\[ = a \cdot 21 + b \cdot \frac{15}{2} + c \cdot 3 \]
This simplifies to:
\[ = 21a + \frac{15}{2}b + 3c \]

### Final Result
Thus, the evaluated integral is:
\[ \int_{1}^{4} (ax^2 + bx + c) \, dx = 21a + \frac{15}{2}b + 3c \]

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