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Integrals

Short Answer Type

191.

Show that:
$integral subscript straight pi over 4 end subscript superscript straight pi over 4 end superscript open vertical bar sin space straight x close vertical bar dx space equals space 2 minus square root of 2$

120 Views

Long Answer Type

192.
Show that:
$integral subscript negative 1 end subscript superscript 3 over 2 end superscript open vertical bar straight x space sinπ space straight x close vertical bar space dx space equals space 3 over straight pi plus 1 over straight pi squared$

We have
   $negative 1 less or equal than straight x less or equal than 0 space space space rightwards double arrow space space straight x space less or equal than space 0 comma space space sinπx less or equal than 0 space space rightwards double arrow space space straight x space sinπx space greater or equal than 0 0 space less or equal than space straight x space less or equal than space 1 space space rightwards double arrow straight x greater or equal than 0 comma space space sin space πx space greater or equal than 0 space space rightwards double arrow space straight x space sinπx greater or equal than 0 1 less or equal than straight x less or equal than 3 over 2 space rightwards double arrow space straight x greater or equal than 0 comma space space sin space straight pi space straight x space less or equal than 0 space space rightwards double arrow space straight x space sinπx space less or equal than 0$

$therefore space space space open vertical bar straight x space sin space straight pi space straight x close vertical bar space equals space open curly brackets table attributes columnalign left end attributes row cell straight x space sinπx comma space space space space space minus 1 less or equal than straight x less or equal than 1 end cell row cell negative straight x space sinπx comma space space space space space 1 less or equal than straight x less or equal than 3 over 2 space space space space end cell end table close$
$therefore space space space integral subscript 1 superscript 3 over 2 end superscript open vertical bar straight x space sinπ space straight x close vertical bar dx space equals space integral subscript negative 1 end subscript superscript 1 open vertical bar straight x space sinπ space straight x close vertical bar dx plus integral subscript 1 superscript 3 over 2 end superscript open vertical bar straight x space sinπx close vertical bar dx space space space space space space space space space space space equals space integral subscript negative 1 end subscript superscript 1 space straight x space sinπ space straight x space dx space space minus integral subscript 1 superscript 3 over 2 end superscript straight x space sinπx space dx$
   $equals space open square brackets straight x. space fraction numerator negative cos space πx over denominator straight pi end fraction close square brackets subscript negative 1 end subscript superscript 1 space minus space integral subscript negative 1 end subscript superscript 1 1. space fraction numerator negative cosπx over denominator straight pi end fraction dx minus open square brackets straight x fraction numerator negative cosπ space straight x over denominator straight pi end fraction close square brackets subscript 1 superscript 3 over 2 end superscript plus integral subscript 1 superscript 3 over 2 end superscript 1. fraction numerator negative cos space πx over denominator straight pi end fraction dx$
$equals space 1 over straight pi left square bracket straight x space cos space straight pi space straight x right square bracket subscript negative 1 end subscript superscript 1 space plus space 1 over straight pi squared left square bracket sin space straight pi space straight x right square bracket subscript negative 1 end subscript superscript 1 plus 1 over straight pi left square bracket straight x space cosπ space straight x right square bracket subscript 1 superscript 3 over 2 end superscript minus 1 over straight pi squared left square bracket sin space straight pi space straight x right square bracket subscript 1 superscript 3 over 2 end superscript equals space minus 1 over straight pi left square bracket cosπ space plus space cosπ right square bracket space plus space 1 over straight pi squared left square bracket sin space straight pi space plus space sin space straight pi right square bracket$
   $plus 1 over straight pi open square brackets 3 over 2 cos fraction numerator 3 straight pi over denominator 2 end fraction minus cosπ close square brackets minus 1 over straight pi squared open square brackets sin space fraction numerator 3 straight pi over denominator 2 end fraction minus sin space straight pi close square brackets$
$equals negative 1 over straight pi left square bracket negative 1 minus 1 right square bracket space plus space 1 over straight pi squared left square bracket 0 plus 0 right square bracket space plus 1 over straight pi left square bracket negative 1 minus 0 right square bracket space equals space 2 over straight pi plus 0 plus 1 over straight pi plus 1 over straight pi squared$
   $equals space 3 over straight pi plus 1 over straight pi squared$

152 Views

Short Answer Type

193.

Show that:
$integral subscript 0 superscript 2 straight x square root of 2 minus straight x end root space equals fraction numerator 16 square root of 2 over denominator 15 end fraction$

147 Views

194.

By using the properties of definite integrals, evaluate the following integral:
$integral subscript 0 superscript 1 space straight x space left parenthesis 1 minus straight x right parenthesis to the power of straight n space dx$

108 Views

195.

By using the properties of definite integrals, evaluate the following integral:
$integral subscript 0 superscript straight pi over 2 end superscript fraction numerator sinx minus cosx over denominator 1 plus sinxcosx end fraction dx$

134 Views

196.

Show that:
$integral subscript 0 superscript straight pi over 2 end superscript 2 straight x space log space tanx space dx space equals space 0$

110 Views

197.

By using the properties of definite integrals, evaluate the following integral:
$integral subscript 0 superscript straight pi over 2 end superscript left parenthesis 2 space log space sinx space minus space log space sin space 2 straight x right parenthesis space dx$

410 Views

198.

Show that:
$integral subscript 0 superscript straight pi fraction numerator straight x space tanx over denominator secx space plus space cosx end fraction dx space equals space straight pi squared over 4$

100 Views

199.

Show that:
$integral subscript 0 superscript infinity log space open parentheses straight x plus 1 over straight x close parentheses. space fraction numerator dx over denominator 1 plus straight x squared end fraction space equals space straight pi space log space 2$

132 Views

200.

Show that:
$integral subscript 0 superscript 1 fraction numerator log space left parenthesis 1 plus straight x right parenthesis over denominator 1 plus straight x squared end fraction space equals space straight pi over 8 space log space 2$