int_{1}^{5} (|x-3| + |1-x|) dx = | vedclass

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(C) We need to evaluate the integral $I = \int_{1}^{5} (|x-3| + |1-x|) dx$. Since $x$ ranges from $1$ to $5$,we have $|1-x| = x-1$ because $x \ge 1$.

Vedclass · Accepted Answer

$12$

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(C) We need to evaluate the integral $I = \int_{1}^{5} (|x-3| + |1-x|) dx$. Since $x$ ranges from $1$ to $5$,we have $|1-x| = x-1$ because $x \ge 1$. Now,we split the integral at $x=3$ due to the modulus $|x-3|$: For $1 \le x For $3 \le x \le 5$,$|x-3| = x-3$. Thus,$I = \int_{1}^{3} (3-x + x-1) dx + \int_{3}^{5} (x-3 + x-1) dx$. $I = \int_{1}^{3} 2 dx + \int_{3}^{5} (2x-4) dx$. $I = [2x]_{1}^{3} + [x^2-4x]_{3}^{5}$. $I = (6-2) + ((25-20) - (9-12))$. $I = 4 + (5 - (-3)) = 4 + 8 = 12$.

$\int_{1}^{5} (|x-3| + |1-x|) dx =$

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