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Integrals

Long Answer Type

211.

Evaluate:
$integral subscript 0 superscript straight pi fraction numerator straight x over denominator 1 plus sin squared straight x end fraction dx.$

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Short Answer Type

212. By using the properties of definite integrals, evaluate the following:

integral subscript 0 superscript straight pi log space left parenthesis 1 plus cosx right parenthesis space dx

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213. By using the properties of definite integrals, evaluate the following:

integral subscript 0 superscript straight pi over 2 end superscript fraction numerator straight x space sinx space cosx over denominator sin to the power of 4 straight x plus cos to the power of 4 straight x end fraction dx

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Long Answer Type

214. By using the properties of definite integrals, evaluate the following:

integral subscript 0 superscript straight pi fraction numerator straight x space dx over denominator straight a squared space cos squared straight x plus straight b squared space sin squared straight x end fraction

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Short Answer Type

215.

Show that:
$integral subscript 0 superscript straight a fraction numerator square root of straight x over denominator square root of straight x plus square root of straight a minus straight x end root end fraction space equals space straight a over 2$

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216.

Show that:
$integral subscript 0 superscript straight a fraction numerator 1 over denominator straight x plus square root of straight a squared minus straight x squared end root end fraction dx space equals space straight pi over 4$

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217.

Evaluate $integral subscript 1 superscript 4 straight f left parenthesis straight x right parenthesis space dx space space where space straight f left parenthesis straight x right parenthesis space equals space open vertical bar straight x minus 1 close vertical bar space plus space open vertical bar straight x minus 2 close vertical bar space plus space open vertical bar straight x minus 3 close vertical bar$

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218.
Evaluate: $integral subscript negative 5 end subscript superscript 0 space straight f left parenthesis straight x right parenthesis space dx comma space space where space straight f left parenthesis straight x right parenthesis space equals space open vertical bar straight x close vertical bar plus open vertical bar straight x plus 2 close vertical bar plus open vertical bar straight x plus 5 close vertical bar.$

Here $straight f left parenthesis straight x right parenthesis space equals space open vertical bar straight x close vertical bar plus open vertical bar straight x plus 2 close vertical bar plus open vertical bar straight x plus 5 close vertical bar$
Let I = $integral subscript negative 5 end subscript superscript 0 space straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript negative 5 end subscript superscript negative 2 end superscript straight f left parenthesis straight x right parenthesis space dx space plus space integral subscript negative 2 end subscript superscript 0 space straight f left parenthesis straight x right parenthesis space dx$
$equals space integral subscript negative 5 end subscript superscript negative 2 end superscript open square brackets open vertical bar straight x close vertical bar plus open vertical bar straight x plus 2 close vertical bar plus open vertical bar straight x plus 5 close vertical bar close square brackets dx plus integral subscript negative 2 end subscript superscript 0 open square brackets open vertical bar straight x close vertical bar plus open vertical bar straight x plus 2 close vertical bar space plus open vertical bar straight x plus 5 close vertical bar close square brackets dx$
$equals space integral subscript negative 5 end subscript superscript negative 2 end superscript left square bracket negative straight x minus left parenthesis straight x plus 2 right parenthesis plus left parenthesis straight x plus 5 right parenthesis right square bracket space dx space plus space integral subscript negative 2 end subscript superscript 0 open square brackets negative straight x plus left parenthesis straight x plus 2 right parenthesis plus left parenthesis straight x plus 5 right parenthesis close square brackets space dx equals space integral subscript negative 5 end subscript superscript negative 2 end superscript left parenthesis negative straight x minus straight x minus 2 plus straight x plus 5 right parenthesis space dx plus integral subscript negative 2 end subscript superscript 0 left parenthesis negative straight x plus straight x plus 2 plus straight x plus 5 right parenthesis space dx equals space integral subscript negative 5 end subscript superscript negative 2 end superscript left parenthesis negative straight x plus 3 right parenthesis space dx plus integral subscript negative 2 end subscript superscript 0 left parenthesis straight x plus 7 right parenthesis space dx space equals space open square brackets negative straight x squared over 2 plus 3 straight x close square brackets subscript negative 5 end subscript superscript 2 plus open square brackets straight x squared over 2 plus 7 straight x close square brackets subscript negative 2 end subscript superscript 0 equals left parenthesis negative 2 minus 6 right parenthesis minus open parentheses negative 25 over 2 minus 15 close parentheses plus left parenthesis 0 plus 0 right parenthesis minus left parenthesis 2 minus 14 right parenthesis space equals space minus 8 plus 55 over 2 plus 12 space equals 63 over 2.$

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219.

Evaluate: $integral subscript negative 5 end subscript superscript 0 straight f left parenthesis straight x right parenthesis space dx comma space space space where space straight f left parenthesis straight x right parenthesis space equals space open vertical bar straight x close vertical bar plus open vertical bar straight x plus 3 close vertical bar plus open vertical bar straight x plus 6 close vertical bar.$

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220.

If [ x ] stands for integral part of x, then show that $integral subscript 0 superscript 1 left square bracket 5 space straight x right square bracket space dx space equals space 2.$

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Previous Year Papers

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Long Answer Type

Short Answer Type

Long Answer Type

Short Answer Type

218. Evaluate: