int_1^5 (|x-3| + |1-x|) , dx = | vedclass

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(A) We need to evaluate the integral $I = \int_1^5 (|x-3| + |1-x|) \, dx$. Since $x \in [1, 5]$,we have $|1-x| = x-1$. Thus,$I = \int_1^5 |x-3| \, dx

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) We need to evaluate the integral $I = \int_1^5 (|x-3| + |1-x|) \, dx$. Since $x \in [1, 5]$,we have $|1-x| = x-1$. Thus,$I = \int_1^5 |x-3| \, dx + \int_1^5 (x-1) \, dx$. For the first part,$|x-3| = 3-x$ for $x \in [1, 3]$ and $|x-3| = x-3$ for $x \in [3, 5]$. $\int_1^3 (3-x) \, dx = [3x - \frac{x^2}{2}]_1^3 = (9 - 4.5) - (3 - 0.5) = 4.5 - 2.5 = 2$. $\int_3^5 (x-3) \, dx = [\frac{x^2}{2} - 3x]_3^5 = (12.5 - 15) - (4.5 - 9) = -2.5 - (-4.5) = 2$. For the second part,$\int_1^5 (x-1) \, dx = [\frac{x^2}{2} - x]_1^5 = (12.5 - 5) - (0.5 - 1) = 7.5 - (-0.5) = 8$. Adding these results: $I = 2 + 2 + 8 = 12$.

$\int_1^5 (|x-3| + |1-x|) \, dx =$

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