int_0^4 |2x - 5| , dx = | vedclass

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(D) To evaluate the integral $\int_0^4 |2x - 5| \, dx$,we first find the point where the expression inside the absolute value changes sign.Set $2x - 5

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$\frac{17}{2}$

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(D) To evaluate the integral $\int_0^4 |2x - 5| \, dx$,we first find the point where the expression inside the absolute value changes sign.Set $2x - 5 = 0$,which gives $x = \frac{5}{2}$.Since $\frac{5}{2}$ lies between $0$ and $4$,we split the integral at $x = \frac{5}{2}$:$\int_0^4 |2x - 5| \, dx = \int_0^{\frac{5}{2}} -(2x - 5) \, dx + \int_{\frac{5}{2}}^4 (2x - 5) \, dx$$= \int_0^{\frac{5}{2}} (5 - 2x) \, dx + \int_{\frac{5}{2}}^4 (2x - 5) \, dx$$= [5x - x^2]_0^{\frac{5}{2}} + [x^2 - 5x]_{\frac{5}{2}}^4$$= (5(\frac{5}{2}) - (\frac{5}{2})^2) - (0) + ((4)^2 - 5(4)) - ((\frac{5}{2})^2 - 5(\frac{5}{2}))$$= (\frac{25}{2} - \frac{25}{4}) + ((16 - 20) - (\frac{25}{4} - \frac{25}{2}))$$= \frac{25}{4} + (-4 - (-\frac{25}{4}))$$= \frac{25}{4} + (-4 + \frac{25}{4}) = \frac{25}{4} + \frac{9}{4} = \frac{34}{4} = \frac{17}{2}$.

$\int_0^4 |2x - 5| \, dx = $

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