Definite Integration Question 247
Question: The greatest value of the function $ F(x)=\int_1^{x}{|t|dt} $ on the interval $ [ -\frac{1}{2},\frac{1}{2} ] $ is given by
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Options:
A) $ \frac{3}{8} $
B) $ -\frac{1}{2} $
C) $ -\frac{3}{8} $
D) $ \frac{2}{5} $
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Answer:
Correct Answer: C
Solution:
$ F’(x)=|x|>0\forall x\in [ -\frac{1}{2},\frac{1}{2} ] \setminus {0} $
Hence the function is increasing on $ [ -\frac{1}{2},\frac{1}{2} ] $ and therefore $ F(x) $ has maxima at the right end point of $ [ -\frac{1}{2},\frac{1}{2} ] $ .
therefore $ MaxF(x)=F( \frac{1}{2} )=\int_{1/2}^{1}{|t|}dt=\frac{3}{8} $ .
BETA
