We are given the integral: \[ \int_{-1}^1 |x - 3| \, dx \] Since \( |x - 3| \) represents the absolute value, we split the integral at the point where \( x = 3 \). In this case, the function \( |x - 3| \) will behave differently depending on whether \( x<3 \) or \( x \geq 3 \). However, since we are integrating from \( -1 \) to \( 1 \), the expression \( |x - 3| \) is always positive, and the integral can be simplified: \[ \int_{-1}^1 (x - 3) \, dx \] Evaluating the integral: \[ = \left[\frac{x^2}{2} - 3x\right]_{-1}^1 = \left(\frac{1^2}{2} - 3(1)\right) - \left(\frac{(-1)^2}{2} - 3(-1)\right) \] \[ = \left(\frac{1}{2} - 3\right) - \left(\frac{1}{2} + 3\right) \] \[ = \left(\frac{-5}{2}\right) - \left(\frac{7}{2}\right) = -6 \]
Thus, the final answer is \( -6 \).
If, \( I_n = \int_{-\pi}^{\pi} \frac{\cos(nx)(1+2^x)}{dx} \), where \( n = 0, 1, 2, \dots \), then which of the following are correct?
A. \( I_n = I_{n+2} \), for all \( n = 0, 1, 2, \dots \)
B. \( I_n = 0 \), for all \( n = 0, 1, 2, \dots \)
C. \( \sum_{n=1}^{10} I_n = 2^{10} \)
D. \( \sum_{n=1}^{10} I_n = 0 \)