Question

Evaluate $integral subscript negative 1 end subscript superscript 1 space log space open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses dx.$

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Solution

integral subscript negative straight a end subscript superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space minus integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space space plus integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx

rightwards double arrow space space space integral subscript negative straight a end subscript superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space 0

Let

straight I space equals space integral subscript negative 1 end subscript superscript 1 space log open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses dx

Let

straight f left parenthesis straight x right parenthesis space equals space log open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses

therefore space space space straight f left parenthesis negative straight x right parenthesis space equals space log open parentheses fraction numerator 2 minus straight x over denominator 2 plus straight x end fraction close parentheses space equals space log open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses to the power of negative 1 end exponent space equals space minus log open parentheses fraction numerator 2 plus straight x over denominator 2 minus straight x end fraction close parentheses space equals space minus straight f left parenthesis straight x right parenthesis therefore space space space space space space space straight f left parenthesis straight x right parenthesis space is space an space odd space function space of space straight x

therefore space space space straight I space equals space 0 space space space space space space space space space space space space open square brackets because space space integral subscript negative straight a end subscript superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space 0 space space if space straight f left parenthesis straight x right parenthesis space is space an space odd space function space of space straight x close square brackets

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