\( \begin{array}{c}\text { Obtenga las } \\ \text { siguientes } \\ \text { integrales definidas }\end{array} \) \( \int_{-3}^{3}|1-x| d x \)
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Para resolver la integral \( \int_{-3}^{3}|1-x| d x \), primero identificamos dónde la función cambia de forma. La función \( |1-x| \) se iguala a cero en \( x = 1 \). Esto divide la integral en dos partes: 1. De \( -3 \) a \( 1 \), donde \( |1-x| = 1-x \). 2. De \( 1 \) a \( 3 \), donde \( |1-x| = x-1 \). Así que la integral se convierte en: \[ \int_{-3}^{1} (1-x) dx + \int_{1}^{3} (x-1) dx \] Calculando cada parte: 1. \( \int_{-3}^{1} (1-x) dx = [x - \frac{x^2}{2}]_{-3}^{1} = (1 - \frac{1}{2}) - (-3 + \frac{9}{2}) = \frac{1}{2} + \frac{3}{2} = 2 \). 2. \( \int_{1}^{3} (x-1) dx = [\frac{x^2}{2} - x]_{1}^{3} = (4.5 - 3) - (0.5 - 1) = 1.5 + 0.5 = 2 \). Finalmente, sumamos las dos partes: \[ 2 + 2 = 4 \] Por lo tanto, \( \int_{-3}^{3}|1-x| d x = 4 \).