By using the properties of definite integrals | vedclass

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(A) Let $I = \int_{2}^{8} |x-5| dx$.Since the integrand $|x-5|$ changes its sign at $x=5$,we split the integral at $x=5$ using the property $\int_{a}^

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$9$

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(A) Let $I = \int_{2}^{8} |x-5| dx$.Since the integrand $|x-5|$ changes its sign at $x=5$,we split the integral at $x=5$ using the property $\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx$.$I = \int_{2}^{5} -(x-5) dx + \int_{5}^{8} (x-5) dx$.Evaluating the first integral:$\int_{2}^{5} (-x+5) dx = [-\frac{x^2}{2} + 5x]_{2}^{5} = (-\frac{25}{2} + 25) - (-\frac{4}{2} + 10) = \frac{25}{2} - 8 = \frac{9}{2}$.Evaluating the second integral:$\int_{5}^{8} (x-5) dx = [\frac{x^2}{2} - 5x]_{5}^{8} = (\frac{64}{2} - 40) - (\frac{25}{2} - 25) = (32 - 40) - (-12.5) = -8 + 12.5 = \frac{9}{2}$.Adding both parts:$I = \frac{9}{2} + \frac{9}{2} = 9$.

By using the properties of definite integrals,evaluate the integral $\int_{2}^{8}|x-5| d x$.

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