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The greatest value of the function $$f(x) =\int_1^x | t |dt$$ on the interval$[{-1\over2},{1\over2}]$is given by:

(1)$\frac 38$

(2)${-1\over2}$

(3)${-3\over 8}$

(4)${2\over 5}$

I have quite a problem dealing with this question.Should I take$f'(x)>0$?if yes then how should I proceed thereafter?

Yash Lekhwani
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    something seems missing from question. – S L May 11 '14 at 08:19
  • The question, as stated now, would imply that the greatest value for $f$ is $+\infty$, the integrand is positive, so the area under the curve is increasing and as $x$ goes to $\infty$ it goes to $\infty$ as well. – math.n00b May 11 '14 at 08:21
  • Yes thanks..@SantoshLinkha the interval was missing. – Yash Lekhwani May 11 '14 at 08:28

2 Answers2

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You can also solve this without any calculations:

Note that by the given interval the upper integration limit $x$ is always smaller than the lower limit $1$. For this reason, since the integrand is non-negative, the value of the integral is always negative. You can make it least negative (i.e. maximize it) by choosing the upper limit $x$ which differs least from the lower limit $1$. That's why you need to choose $x=1/2$ which gives $f(1/2)=-3/8$ as the answer.

Matt L.
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After noticing that $\displaystyle\int |x|dx = \frac{1}{2}x|x| + C$, rewrite $f(x)$ as: $$f(x) = \int_1^x |t|dt = \left [ \frac{1}{2}t \cdot |t|\right ]_1^x = \frac{1}{2}x|x| - \frac{1}{2}$$

Now: $$f'(x) = \frac{1}{2} \left (|x| + x \frac{|x|}{x} \right ) = |x|$$ Since $f'(x) \ge 0~~\forall x \in \mathbb{R}$, $f(x)$ is increasing. We therefore conclude that in the interval $\left [ -\frac{1}{2}, \frac{1}{2}\right ]$ it assumes the maximum value at its extremum. Since $f(1/2) = -\frac{3}{8}$ the answer is $(3)$.

rubik
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