To solve the integral \( \int_1^3 (x^2 - 2x - 2) \, dx \), we will follow these steps:
### Step 1: Simplify the integrand
First, we rewrite the integrand \( x^2 - 2x - 2 \) in a more manageable form. We can complete the square for the quadratic expression.
\[
x^2 - 2x - 2 = (x^2 - 2x + 1) - 1 - 2 = (x - 1)^2 - 3
\]
### Step 2: Set up the integral
Now we can rewrite the integral using the completed square:
\[
\int_1^3 (x^2 - 2x - 2) \, dx = \int_1^3 ((x - 1)^2 - 3) \, dx
\]
### Step 3: Split the integral
We can split the integral into two parts:
\[
\int_1^3 ((x - 1)^2 - 3) \, dx = \int_1^3 (x - 1)^2 \, dx - \int_1^3 3 \, dx
\]
### Step 4: Calculate the first integral
Now we calculate \( \int_1^3 (x - 1)^2 \, dx \).
Let \( u = x - 1 \), then \( du = dx \). When \( x = 1 \), \( u = 0 \) and when \( x = 3 \), \( u = 2 \).
Thus,
\[
\int_1^3 (x - 1)^2 \, dx = \int_0^2 u^2 \, du
\]
Calculating this integral:
\[
\int u^2 \, du = \frac{u^3}{3} \Big|_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3}
\]
### Step 5: Calculate the second integral
Next, we calculate \( \int_1^3 3 \, dx \):
\[
\int_1^3 3 \, dx = 3 \cdot (3 - 1) = 3 \cdot 2 = 6
\]
### Step 6: Combine the results
Now we combine the results of both integrals:
\[
\int_1^3 (x^2 - 2x - 2) \, dx = \frac{8}{3} - 6
\]
To combine these, we convert \( 6 \) to a fraction with a denominator of \( 3 \):
\[
6 = \frac{18}{3}
\]
Thus,
\[
\int_1^3 (x^2 - 2x - 2) \, dx = \frac{8}{3} - \frac{18}{3} = \frac{8 - 18}{3} = \frac{-10}{3}
\]
### Final Answer
The value of the integral is:
\[
\int_1^3 (x^2 - 2x - 2) \, dx = -\frac{10}{3}
\]
