Evaluate the following integrals
$integral subscript 0 superscript 1 straight x square root of 1 minus straight x squared end root dx$

Solution

Let I = $integral subscript 0 superscript 1 straight x square root of 1 minus straight x squared end root space dx space equals space integral subscript 0 superscript 1 left parenthesis 1 minus straight x squared right parenthesis to the power of 1 half end exponent. space straight x space dx$
Put $1 minus straight x squared space equals space straight y comma space space space space space therefore space space space minus 2 space straight x space dx space equals space dy space space space space rightwards double arrow space space space straight x space dx space equals space minus 1 half dy$
When x = 0, y = 1 – 0 = 1 When x = 1, y = 1 – 1 = 0
$therefore$ I = $negative 1 half integral subscript 1 superscript 0 straight y to the power of 1 half end exponent dy space equals space minus 1 half open square brackets fraction numerator straight y to the power of begin display style 3 over 2 end style end exponent over denominator begin display style 3 over 2 end style end fraction close square brackets subscript 1 superscript 0 space equals space minus 1 third open square brackets straight y to the power of 3 over 2 end exponent close square brackets subscript 1 superscript 0 space equals space minus 1 third left square bracket 0 minus 1 right square bracket space equals space 1 third$

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Integrals

Evaluate the following integrals
$integral subscript 0 superscript 1 straight x square root of 1 minus straight x squared end root dx$

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