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Question: The greatest value of the function \(F(x) = \int_{1}^{x}{|t|dt}\) on the interval \(\left\lbrack - \...

The greatest value of the function F(x)=∫1x∣t∣dtF(x) = \int_{1}^{x}{|t|dt} on the interval [−12,12]\left\lbrack - \frac{1}{2},\frac{1}{2} \right\rbrack is given by

A

38\frac{3}{8}

B

−12- \frac{1}{2}

C

−38- \frac{3}{8}

D

25\frac{2}{5}

Answer

−38- \frac{3}{8}

Explanation

Solution

F′(x)=∣x∣>0∀x∈[−12,12]F'(x) = |x| > 0\forall x \in \left\lbrack - \frac{1}{2},\frac{1}{2} \right\rbrack

Hence the function is increasing on [−12,12]\left\lbrack - \frac{1}{2},\frac{1}{2} \right\rbrack and therefore F(x)F(x) has maxima at the right end point of [−12,12]\left\lbrack - \frac{1}{2},\frac{1}{2} \right\rbrack

⇒ MaxF(x)=F(12)=∫11/2∣t∣dt=−38\text{Max}F(x) = F\left( \frac{1}{2} \right) = \int_{1}^{1/2}{|t|}dt = - \frac{3}{8}.