\( =\begin{array}{l}\text { Use symmetry to evaluate the following imtegral, } \\ \int_{-B}^{B}(8-|x|) d x \\ 0\end{array}(8) \)
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To solve the integral \( \int_{-B}^{B}(8-|x|) \, dx \) using symmetry, we can break it down. The function \( 8 - |x| \) is even because \( |x| \) is an even function, meaning that the integral from \(-B\) to \(B\) can be simplified to two times the integral from \(0\) to \(B\): \[ \int_{-B}^{B}(8-|x|) \, dx = 2 \int_{0}^{B}(8 - x) \, dx. \] Now, we can calculate this integral from 0 to \(B\): \[ \int_{0}^{B}(8 - x) \, dx = \left[ 8x - \frac{x^2}{2} \right]_{0}^{B} = \left( 8B - \frac{B^2}{2} \right) - 0 = 8B - \frac{B^2}{2}. \] Multiplying by 2 gives us: \[ \int_{-B}^{B}(8 - |x|) \, dx = 2 \left( 8B - \frac{B^2}{2} \right) = 16B - B^2. \] Thus, the evaluated integral is: \[ 16B - B^2. \]
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