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Values of $stretchy integral subscript negative 1 divided by 2 end subscript superscript plus 1 divided by 2 end superscript c o s invisible function application x l o g invisible function application fraction numerator 1 plus x over denominator 1 minus x end fraction$ is :

$\frac{1}{2}$
– 1/2
0
none of these

hint Hint:

$f open parentheses x close parentheses space i s space a n space o d d space f u n c t i o n.$

The correct answer is: 0

$f open parentheses x close parentheses equals integral subscript negative 1 divided by 2 end subscript superscript plus 1 divided by 2 end superscript c o s x l o g open parentheses fraction numerator 1 plus x over denominator 1 minus x end fraction close parentheses d x f open parentheses negative x close parentheses equals integral subscript negative 1 divided by 2 end subscript superscript plus 1 divided by 2 end superscript c o s open parentheses negative x close parentheses l o g open parentheses fraction numerator 1 minus x over denominator 1 plus x end fraction close parentheses d x f open parentheses negative x close parentheses equals integral subscript negative 1 divided by 2 end subscript superscript plus 1 divided by 2 end superscript c o s open parentheses x close parentheses l o g open parentheses fraction numerator 1 plus x over denominator 1 minus x end fraction close parentheses d x f open parentheses negative x close parentheses equals negative 1 integral subscript negative 1 divided by 2 end subscript superscript plus 1 divided by 2 end superscript c o s open parentheses x close parentheses l o g open parentheses fraction numerator 1 minus x over denominator 1 plus x end fraction close parentheses d x equals negative f open parentheses x close parentheses therefore f open parentheses negative x close parentheses plus f open parentheses x close parentheses equals 0 I equals integral subscript negative 1 divided by 2 end subscript superscript plus 1 divided by 2 end superscript f open parentheses negative x close parentheses plus f open parentheses x close parentheses d x space equals space 0$

$I f space f space i s space a n space o d d space c o n t i n u o u s space f u n c t i o n space o n space open square brackets negative a comma a close square brackets comma space integral subscript negative a end subscript superscript a f open parentheses x close parentheses d x equals 0.$

A rectangular tank of depth 8 meter is full of water ( $mu equals 4 divided by 3$ ), the bottom is seen at the depth

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Which is correct relationship:

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Values of is :

– 1/2 0 none of these

The correct answer is: 0

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Values of $stretchy integral subscript negative 1 divided by 2 end subscript superscript plus 1 divided by 2 end superscript c o s invisible function application x l o g invisible function application fraction numerator 1 plus x over denominator 1 minus x end fraction$ is :

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none of these

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