$\int_{0}^{4} (| x - l | + 3 | x - 2 | + 4 | x - 3 |) d x =$

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Solution

The correct option is A 37
Note: $| x - a | = {\begin{matrix} - (x - a); & f o r & x < a x - a; & f o r & x \geq a \end{matrix}}$ .

$\int_{0}^{4} (| x - 1 | + 3 | x - 2 | + 4 | x - 3 |) d x$

$= \int_{0}^{1} - (x - 1 + 3 x - 6 + 4 x - 12) d x$
+

$\int_{1}^{2} (x - 1 - 3 x + 6 - 4 x + 12) d x$

+
$\int_{2}^{3} (x - 1 + 3 x - 6 - 4 x + 12) d x$

+
$\int_{3}^{4} (x - 1 + 3 x - 6 + 4 x - 12) d x$

$= \int_{1}^{0} (19 - 8 x) d x + \int_{1}^{2} (17 - 6 x) d x + \int_{2}^{3} 5 d x + \int_{3}^{4} (8 x - 19) d x$

$= [19 x - 4 x^{2}]_{0}^{1} + [17 x - 3 x^{2}]_{1}^{2} + [5 x]_{2}^{3} + [4 x^{2} - 19 x]_{3}^{4}$

$= 15 + (22 - 14) + (15 - 10) + (- 12 - (- 21))$

$= 37$

$∴ \int_{0}^{4} (| x - 1 | + 3 | x - 2 | + 4 | x - 3 |) d x = 37$

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