`int_1^4 (x^2-x)dx`

Answer

To solve the integral \(\int_1^4 (x^2 - x) \, dx\), we will follow these steps:

### Step 1: Set up the integral
We need to evaluate the definite integral of the function \(x^2 - x\) from 1 to 4.

\[ \int_1^4 (x^2 - x) \, dx \]

### Step 2: Break down the integral
We can separate the integral into two parts:

\[ \int_1^4 x^2 \, dx - \int_1^4 x \, dx \]

### Step 3: Integrate each part
Now we will integrate each part separately.

1. **Integrate \(x^2\)**:
\[ \int x^2 \, dx = \frac{x^3}{3} \]

2. **Integrate \(x\)**:
\[ \int x \, dx = \frac{x^2}{2} \]

### Step 4: Evaluate the definite integrals
Now we will evaluate the definite integrals from 1 to 4.

1. **For \(\int_1^4 x^2 \, dx\)**:
\[ \left[ \frac{x^3}{3} \right]_1^4 = \frac{4^3}{3} - \frac{1^3}{3} = \frac{64}{3} - \frac{1}{3} = \frac{63}{3} = 21 \]

2. **For \(\int_1^4 x \, dx\)**:
\[ \left[ \frac{x^2}{2} \right]_1^4 = \frac{4^2}{2} - \frac{1^2}{2} = \frac{16}{2} - \frac{1}{2} = 8 - \frac{1}{2} = \frac{16}{2} - \frac{1}{2} = \frac{15}{2} \]

### Step 5: Combine the results
Now we will combine the results from both integrals:

\[ \int_1^4 (x^2 - x) \, dx = 21 - \frac{15}{2} \]

To subtract these, we can convert 21 into a fraction:

\[ 21 = \frac{42}{2} \]

Now we can perform the subtraction:

\[ \frac{42}{2} - \frac{15}{2} = \frac{42 - 15}{2} = \frac{27}{2} \]

### Final Answer
Thus, the value of the integral \(\int_1^4 (x^2 - x) \, dx\) is:

\[ \frac{27}{2} \]

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