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Easy

Question

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 :

  1. 1 half    
  2. – 1/2    
  3. 0    
  4. none of these    

hintHint:

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.

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