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Integrals

Short Answer Type

221.

Show that $integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space straight g left parenthesis straight x right parenthesis space dx space equals space 2 space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx comma space space if space straight f$ and g are defined as f(x) = f(a - x) and g(x) + g(a-x) = 4

146 Views

Long Answer Type

222.

Evaluate:
$integral subscript 0 superscript 1 cot to the power of negative 1 end exponent left parenthesis 1 minus straight x plus straight x squared right parenthesis dx.$

142 Views

Multiple Choice Questions

223.

The value of $integral subscript negative straight pi over 2 end subscript superscript straight pi over 2 end superscript left parenthesis straight x cubed plus straight x space cosx space plus space tan to the power of 5 straight x plus 1 right parenthesis space dx space is$

0
2
$straight pi$
$straight pi$

125 Views

224.

The value of $integral subscript 0 superscript straight pi over 2 end superscript space log space space open parentheses fraction numerator 4 plus 3 space sinx over denominator 4 plus 3 space cosx end fraction close parentheses dx$ is

2
$3 over 4$
0
0

139 Views

225.

If $straight f left parenthesis straight a plus straight b minus straight x right parenthesis space equals space straight f left parenthesis straight x right parenthesis comma space space then space integral subscript straight a superscript straight b straight x space straight f left parenthesis straight x right parenthesis space dx$ is equal to

$fraction numerator straight a plus straight b over denominator 2 end fraction integral subscript straight a superscript straight b straight f left parenthesis straight b minus straight x right parenthesis space dx$
$fraction numerator straight a plus straight b over denominator 2 end fraction integral subscript straight a superscript straight b straight f left parenthesis straight b plus straight x right parenthesis space dx$
$fraction numerator straight b minus straight a over denominator 2 end fraction integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx$
$fraction numerator straight b minus straight a over denominator 2 end fraction integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx$

108 Views

226.
The value of $integral subscript 0 superscript 1 space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x minus 1 over denominator 1 plus straight x minus straight x squared end fraction close parentheses dx$ is

1

0

-1

-1

Let I = $integral subscript 0 superscript 1 tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x minus 1 over denominator 1 plus straight x minus straight x squared end fraction close parentheses dx$

$equals space integral subscript 0 superscript 1 tan to the power of negative 1 end exponent open curly brackets fraction numerator straight x plus left parenthesis straight x minus 1 right parenthesis over denominator 1 minus straight x left parenthesis straight x minus 1 right parenthesis end fraction close curly brackets space dx space equals space integral subscript 0 superscript 1 left square bracket tan to the power of negative 1 end exponent straight x space plus space tan to the power of negative 1 end exponent left parenthesis straight x minus 1 right parenthesis right square bracket space dx$

$therefore space space space space space straight I space equals integral subscript 0 superscript 1 space open square brackets tan to the power of negative 1 end exponent minus space tan to the power of negative 1 end exponent left parenthesis 1 minus straight x right parenthesis close square brackets dx$ ...(1)

Again, $straight I space equals space integral subscript 0 superscript 1 open square brackets tan to the power of negative 1 end exponent left parenthesis 1 minus straight x right parenthesis space minus space tan to the power of negative 1 end exponent open curly brackets 1 minus left parenthesis 1 minus straight x close curly brackets close square brackets space dx$
$open square brackets because space space integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript straight a superscript straight b straight f left parenthesis straight a plus straight b minus straight x right parenthesis space dx close square brackets$
$equals space integral subscript 0 superscript 1 left square bracket tan to the power of negative 1 end exponent left parenthesis 1 minus straight x right parenthesis space minus space tan to the power of negative 1 end exponent straight x right square bracket dx space equals space integral subscript 0 superscript 1 left square bracket tan to the power of negative 1 end exponent straight x space minus space tan to the power of negative 1 end exponent left parenthesis 1 minus straight x right parenthesis right square bracket space dx$

$therefore space space space space space space straight I space equals space minus straight I space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 1 right parenthesis close square brackets$
$therefore space space space space space space 2 space space straight I space equals space 0 space space space space space space space space space space space rightwards double arrow space space space space space straight I space equals space 0 therefore space space space integral subscript 0 superscript 1 tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x minus 1 over denominator 1 plus straight x minus straight x squared end fraction close parentheses dx space equals space 0$
$therefore space space space left parenthesis straight B right parenthesis space is space correct space answer.$

137 Views

Long Answer Type

227.

Evaluate $integral subscript 0 superscript straight pi over 2 end superscript log space sinx space dx.$

111 Views

Short Answer Type

228.

Evaluate the definite integral:
$integral subscript 2 superscript 3 fraction numerator 1 over denominator straight x squared minus 1 end fraction dx$

124 Views

229.

Evaluate the definite integral:
$integral subscript 0 superscript 1 divided by square root of 2 end superscript fraction numerator sin to the power of negative 1 end exponent straight x over denominator left parenthesis 1 minus straight x squared right parenthesis to the power of begin display style 3 over 2 end style end exponent end fraction dx$

181 Views

Long Answer Type

230.

Evaluate:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator dx over denominator straight a space cosx plus straight b space sinx end fraction. space straight a comma space straight b space greater than space 0$