`int_2^4 (x^2-1)dx`

Answer

To solve the integral \(\int_2^4 (x^2 - 1) \, dx\), we will follow these steps:

### Step 1: Break down the integral
We can split the integral into two separate integrals:
\[ \int_2^4 (x^2 - 1) \, dx = \int_2^4 x^2 \, dx - \int_2^4 1 \, dx \]

### Step 2: Calculate \(\int_2^4 x^2 \, dx\)
Using the power rule for integration, we have:
\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]
For \(n = 2\):
\[ \int x^2 \, dx = \frac{x^{2+1}}{2+1} = \frac{x^3}{3} \]
Now, we evaluate this from 2 to 4:
\[ \int_2^4 x^2 \, dx = \left[ \frac{x^3}{3} \right]_2^4 = \frac{4^3}{3} - \frac{2^3}{3} = \frac{64}{3} - \frac{8}{3} = \frac{64 - 8}{3} = \frac{56}{3} \]

### Step 3: Calculate \(\int_2^4 1 \, dx\)
The integral of 1 over an interval is simply the length of the interval:
\[ \int_2^4 1 \, dx = [x]_2^4 = 4 - 2 = 2 \]

### Step 4: Combine the results
Now, we can combine the results from Step 2 and Step 3:
\[ \int_2^4 (x^2 - 1) \, dx = \frac{56}{3} - 2 \]
To subtract 2, we convert it into a fraction:
\[ 2 = \frac{6}{3} \]
Thus,
\[ \int_2^4 (x^2 - 1) \, dx = \frac{56}{3} - \frac{6}{3} = \frac{56 - 6}{3} = \frac{50}{3} \]

### Final Answer
The value of the integral \(\int_2^4 (x^2 - 1) \, dx\) is:
\[ \frac{50}{3} \]

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