Evaluate $integral subscript negative 1 end subscript superscript 1 5 straight x to the power of 4 square root of straight x to the power of 5 plus 1 end root dx.$

Solution

Let I = $integral subscript negative 1 end subscript superscript 1 left parenthesis straight x to the power of 5 plus 1 right parenthesis to the power of 1 half end exponent left parenthesis 5 straight x to the power of 4 right parenthesis space dx$
Put x⁵ + 1 = y, ∴ 5 x⁴ dx = dy When x = –1 , y = – 1 + 1 = 0 When x = 1, y = 1 + 1 = 2
$therefore$ $straight I space equals space integral subscript 0 superscript 2 straight y to the power of 1 half end exponent dy space equals space open square brackets fraction numerator straight y to the power of begin display style 3 over 2 end style end exponent over denominator begin display style 3 over 2 end style end fraction close square brackets subscript 0 superscript 2 space equals space 2 over 3 open square brackets straight y to the power of 3 over 2 end exponent close square brackets subscript 0 superscript 2 space equals space 2 over 3 open parentheses 2 to the power of 3 over 2 end exponent minus 0 close parentheses space equals space 2 over 3 open parentheses 2 square root of 2 close parentheses space equals space fraction numerator 4 square root of 2 over denominator 3 end fraction$

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Integrals

Evaluate $integral subscript negative 1 end subscript superscript 1 5 straight x to the power of 4 square root of straight x to the power of 5 plus 1 end root dx.$

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