The value of int_{0}^{4}|x-1| dx is | vedclass

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(B) To evaluate $\int_{0}^{4}|x-1| dx$,we first identify the point where the expression inside the modulus changes sign,which is $x = 1$.We split the

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$5$

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(B) To evaluate $\int_{0}^{4}|x-1| dx$,we first identify the point where the expression inside the modulus changes sign,which is $x = 1$.We split the integral at $x = 1$:$\int_{0}^{4}|x-1| dx = \int_{0}^{1}-(x-1) dx + \int_{1}^{4}(x-1) dx$Evaluating the first part:$\int_{0}^{1}(-x+1) dx = [-\frac{x^2}{2} + x]_{0}^{1} = (-\frac{1}{2} + 1) - (0) = \frac{1}{2}$Evaluating the second part:$\int_{1}^{4}(x-1) dx = [\frac{x^2}{2} - x]_{1}^{4} = (\frac{16}{2} - 4) - (\frac{1}{2} - 1) = (8 - 4) - (-\frac{1}{2}) = 4 + \frac{1}{2} = \frac{9}{2}$Adding both parts:$\frac{1}{2} + \frac{9}{2} = \frac{10}{2} = 5$

The value of $\int_{0}^{4}|x-1| dx$ is

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