int_{-1}^{1} |1 - x| ,dx = | vedclass

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(C) We need to evaluate the definite integral $I = \int_{-1}^{1} |1 - x| \,dx$. Since the interval of integration is $[-1, 1]$,we observe that for all

Vedclass · Accepted Answer

$2$

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(C) We need to evaluate the definite integral $I = \int_{-1}^{1} |1 - x| \,dx$. Since the interval of integration is $[-1, 1]$,we observe that for all $x \in [-1, 1]$,$1 - x \ge 0$. Specifically,when $x = 1$,$1 - x = 0$,and for $x  0$. Therefore,$|1 - x| = 1 - x$ for all $x \in [-1, 1]$. Thus,the integral becomes: $I = \int_{-1}^{1} (1 - x) \,dx$ $I = \left[ x - \frac{x^2}{2} \right]_{-1}^{1}$ $I = \left( 1 - \frac{1^2}{2} \right) - \left( -1 - \frac{(-1)^2}{2} \right)$ $I = \left( 1 - \frac{1}{2} \right) - \left( -1 - \frac{1}{2} \right)$ $I = \frac{1}{2} - \left( -\frac{3}{2} \right)$ $I = \frac{1}{2} + \frac{3}{2} = \frac{4}{2} = 2$. Hence,the correct option is $C$.

$\int_{-1}^{1} |1 - x| \,dx = $

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