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Integrals

Question
CBSEENMA12032586

Evaluate integral subscript 2 superscript 3 fraction numerator square root of straight x over denominator square root of 5 minus straight x end root plus square root of straight x end fraction dx

Solution

Let I = integral subscript 2 superscript 3 fraction numerator square root of straight x over denominator square root of 5 minus straight x end root plus square root of straight x end fraction dx                              ...(1)
therefore space space space space space straight I space equals space integral subscript 2 superscript 3 fraction numerator square root of 2 plus 3 minus straight x end root over denominator square root of 5 minus left parenthesis 2 plus 3 minus straight x right parenthesis plus square root of 2 plus 3 minus straight x end root end root end fraction dx
                                                      open square brackets because space integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript straight a superscript straight b straight f left parenthesis straight a plus straight b minus straight x right parenthesis space dx close square brackets
therefore space space space space space straight I space equals space integral subscript 2 superscript 3 fraction numerator square root of 5 minus straight x end root over denominator square root of straight x plus square root of 5 minus straight x end root end fraction dx                           ...(2)
Adding (1) and (2), we get,
         2 space straight I space equals space integral subscript 2 superscript 3 fraction numerator square root of straight x plus square root of 5 minus straight x end root over denominator square root of straight x plus square root of 5 minus straight x end root end fraction dx space equals space integral subscript 2 superscript 3 1 dx space equals space open square brackets straight x close square brackets subscript 2 superscript 3 space equals space 3 space minus space 2 space space equals space 1
therefore space space space space straight I space equals space 1 half

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