Evaluate the following integrals
$integral subscript negative 1 end subscript superscript 1 space straight x cubed space left parenthesis straight x to the power of 4 plus 1 right parenthesis cubed space dx$

Solution

Let I = $integral subscript negative 1 end subscript superscript 1 straight x cubed left parenthesis straight x to the power of 4 plus 1 right parenthesis cubed dx space equals space integral subscript negative 1 end subscript superscript 1 left parenthesis straight x to the power of 4 plus 1 right parenthesis cubed space straight x cubed dx$
Put $straight x to the power of 4 plus 1 space equals space straight y comma space space space space space therefore space space space space 4 space straight x cubed space dx space space equals dy space space space space space space rightwards double arrow space space space space straight x cubed dx space equals space 1 fourth dy$
When x = – 1, y = 1 + 1 = 2 When x = 1, y = 1 + 1 = 2
$therefore space space space space space straight I space equals space 1 fourth integral subscript 2 superscript 2 straight y cubed dy space equals 1 fourth open square brackets straight y to the power of 4 over 4 close square brackets subscript 2 superscript 2 space equals space 1 over 16 open square brackets straight y to the power of 4 close square brackets subscript 2 superscript 2 space equals space 1 over 16 left square bracket 16 minus 16 right square bracket space equals space 1 over 16 cross times 0 space equals space 0$

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Integrals

Evaluate the following integrals
$integral subscript negative 1 end subscript superscript 1 space straight x cubed space left parenthesis straight x to the power of 4 plus 1 right parenthesis cubed space dx$

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