If f(x) = |x - 1|,then int_0^2 {f(x)dx} is | vedclass

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(A) Given $f(x) = |x - 1|$.We know that $|x - 1| = (1 - x)$ for $x < 1$ and $|x - 1| = (x - 1)$ for $x \ge 1$.Therefore,$\int_0^2 |x - 1| dx = \int_0^

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(A) Given $f(x) = |x - 1|$.We know that $|x - 1| = (1 - x)$ for $x Therefore,$\int_0^2 |x - 1| dx = \int_0^1 (1 - x) dx + \int_1^2 (x - 1) dx$.Evaluating the first integral: $\int_0^1 (1 - x) dx = [x - \frac{x^2}{2}]_0^1 = (1 - \frac{1}{2}) - (0 - 0) = \frac{1}{2}$.Evaluating the second integral: $\int_1^2 (x - 1) dx = [\frac{x^2}{2} - x]_1^2 = (\frac{4}{2} - 2) - (\frac{1}{2} - 1) = (2 - 2) - (-\frac{1}{2}) = \frac{1}{2}$.Adding both parts: $\frac{1}{2} + \frac{1}{2} = 1$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

If $f(x) = |x - 1|$,then $\int_0^2 {f(x)dx}$ is

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