By using the properties of definite integrals | vedclass

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(C) Let $I = \int_{0}^{4}|x-1| d x$.We know that the absolute value function $|x-1|$ changes its sign at $x = 1$.Specifically,$(x-1) \leq 0$ for $0 \l

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$5$

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(C) Let $I = \int_{0}^{4}|x-1| d x$.We know that the absolute value function $|x-1|$ changes its sign at $x = 1$.Specifically,$(x-1) \leq 0$ for $0 \leq x \leq 1$ and $(x-1) \geq 0$ for $1 \leq x \leq 4$.Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{c} f(x) d x + \int_{c}^{b} f(x) d x$,we can split the integral as:$I = \int_{0}^{1} -(x-1) d x + \int_{1}^{4} (x-1) d x$.Now,evaluate the integrals:$I = \int_{0}^{1} (1-x) d x + \int_{1}^{4} (x-1) d x$.$I = \left[ x - \frac{x^2}{2} \right]_{0}^{1} + \left[ \frac{x^2}{2} - x \right]_{1}^{4}$.For the first part: $\left( 1 - \frac{1}{2} \right) - (0 - 0) = \frac{1}{2}$.For the second part: $\left( \frac{16}{2} - 4 \right) - \left( \frac{1}{2} - 1 \right) = (8 - 4) - \left( -\frac{1}{2} \right) = 4 + \frac{1}{2} = \frac{9}{2}$.Therefore,$I = \frac{1}{2} + \frac{9}{2} = \frac{10}{2} = 5$.

By using the properties of definite integrals,evaluate the integral $\int_{0}^{4}|x-1| d x$.

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