Evaluate the following integral given \( f(x)=\left\{\begin{array}{ll}x & \text { if } x<1 \\ \frac{1}{x} & \text { if } x \geq 1\end{array}\right. \) \( \int_{-3}^{3} f(x) d x=\square \)

To evaluate the integral \( \int_{-3}^{3} f(x) \, dx \) where \[ f(x) = \begin{cases} x & \text{if } x < 1 \\ \frac{1}{x} & \text{if } x \geq 1 \end{cases} \] we need to split the integral at the point where the definition of \( f(x) \) changes, which is at \( x = 1 \). Thus, we can write: \[ \int_{-3}^{3} f(x) \, dx = \int_{-3}^{1} f(x) \, dx + \int_{1}^{3} f(x) \, dx \] ### Step 1: Evaluate \( \int_{-3}^{1} f(x) \, dx \) For \( x < 1 \), we have \( f(x) = x \). Therefore, \[ \int_{-3}^{1} f(x) \, dx = \int_{-3}^{1} x \, dx \] Calculating this integral: \[ \int x \, dx = \frac{x^2}{2} + C \] Now we evaluate from \(-3\) to \(1\): \[ \left[ \frac{x^2}{2} \right]_{-3}^{1} = \frac{1^2}{2} - \frac{(-3)^2}{2} = \frac{1}{2} - \frac{9}{2} = \frac{1 - 9}{2} = \frac{-8}{2} = -4 \] ### Step 2: Evaluate \( \int_{1}^{3} f(x) \, dx \) For \( x \geq 1 \), we have \( f(x) = \frac{1}{x} \). Therefore, \[ \int_{1}^{3} f(x) \, dx = \int_{1}^{3} \frac{1}{x} \, dx \] Calculating this integral: \[ \int \frac{1}{x} \, dx = \ln |x| + C \] Now we evaluate from \(1\) to \(3\): \[ \left[ \ln |x| \right]_{1}^{3} = \ln(3) - \ln(1) = \ln(3) - 0 = \ln(3) \] ### Step 3: Combine the results Now we combine the results of the two integrals: \[ \int_{-3}^{3} f(x) \, dx = \int_{-3}^{1} f(x) \, dx + \int_{1}^{3} f(x) \, dx = -4 + \ln(3) \] Thus, the final result is: \[ \int_{-3}^{3} f(x) \, dx = -4 + \ln(3) \] Therefore, the answer is: \[ \boxed{-4 + \ln(3)} \]