Find \( \int_{0}^{6} f(x) d x \) if \( f(x)=\left\{\begin{array}{ll}4 & \text { if } x<4 \\ x & \text { if } x \geq 4\end{array}\right. \)

To evaluate the integral \( \int_{0}^{6} f(x) \, dx \), we need to break it down based on the piecewise definition of the function \( f(x) \). The function \( f(x) \) is defined as follows: - \( f(x) = 4 \) for \( x < 4 \) - \( f(x) = x \) for \( x \geq 4 \) We will split the integral at the point where the definition of \( f(x) \) changes, which is at \( x = 4 \): \[ \int_{0}^{6} f(x) \, dx = \int_{0}^{4} f(x) \, dx + \int_{4}^{6} f(x) \, dx \] 1. For \( \int_{0}^{4} f(x) \, dx \): Since \( f(x) = 4 \) for \( x < 4 \): \[ \int_{0}^{4} f(x) \, dx = \int_{0}^{4} 4 \, dx = 4x \bigg|_{0}^{4} = 4(4) - 4(0) = 16 \] 2. For \( \int_{4}^{6} f(x) \, dx \): Since \( f(x) = x \) for \( x \geq 4 \): \[ \int_{4}^{6} f(x) \, dx = \int_{4}^{6} x \, dx = \frac{x^2}{2} \bigg|_{4}^{6} = \frac{6^2}{2} - \frac{4^2}{2} = \frac{36}{2} - \frac{16}{2} = 18 - 8 = 10 \] Now combine the results: \[ \int_{0}^{6} f(x) \, dx = 16 + 10 = 26 \] Thus, the value of the integral is \[ \int_{0}^{6} f(x) \, dx = 26. \]