The value of int_1^5 (|x - 3| + |1 - x|) , dx | vedclass

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(D) Let $I = \int_1^5 (|x - 3| + |1 - x|) \, dx$. Since $x \in [1, 5]$,$|1 - x| = x - 1$. So,$I = \int_1^5 |x - 3| \, dx + \int_1^5 (x - 1) \, dx$. Fo

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) Let $I = \int_1^5 (|x - 3| + |1 - x|) \, dx$. Since $x \in [1, 5]$,$|1 - x| = x - 1$. So,$I = \int_1^5 |x - 3| \, dx + \int_1^5 (x - 1) \, dx$. For the first part,$\int_1^5 |x - 3| \, dx = \int_1^3 -(x - 3) \, dx + \int_3^5 (x - 3) \, dx$. $= [-\frac{(x - 3)^2}{2}]_1^3 + [\frac{(x - 3)^2}{2}]_3^5 = (0 - (-2)) + (2 - 0) = 2 + 2 = 4$. For the second part,$\int_1^5 (x - 1) \, dx = [\frac{x^2}{2} - x]_1^5 = (\frac{25}{2} - 5) - (\frac{1}{2} - 1) = 7.5 - (-0.5) = 8$. Therefore,$I = 4 + 8 = 12$.

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

The value of $\int_1^5 (|x - 3| + |1 - x|) \, dx$ is

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