The greatest value of the function F(x) = int | vedclass

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(C) The function is defined as $F(x) = \int_1^x {|t| \, dt}$.By the Fundamental Theorem of Calculus,$F'(x) = |x|$.Since $|x| \ge 0$ for all $x \in \le

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$-\frac{3}{8}$

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(C) The function is defined as $F(x) = \int_1^x {|t| \, dt}$.By the Fundamental Theorem of Calculus,$F'(x) = |x|$.Since $|x| \ge 0$ for all $x \in \left[ -\frac{1}{2}, \frac{1}{2} \right]$,the function $F(x)$ is non-decreasing on the given interval.Therefore,the maximum value occurs at the right endpoint $x = \frac{1}{2}$.$F\left( \frac{1}{2} \right) = \int_1^{1/2} {|t| \, dt} = -\int_{1/2}^1 {|t| \, dt}$.Since $t > 0$ in the interval $\left[ \frac{1}{2}, 1 \right]$,we have $|t| = t$.$F\left( \frac{1}{2} \right) = -\int_{1/2}^1 {t \, dt} = -\left[ \frac{t^2}{2} \right]_{1/2}^1 = -\left( \frac{1}{2} - \frac{1}{8} \right) = -\left( \frac{4-1}{8} \right) = -\frac{3}{8}$.

The greatest value of the function $F(x) = \int_1^x {|t| \, dt}$ on the interval $\left[ -\frac{1}{2}, \frac{1}{2} \right]$ is:

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