Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Integrals

Long Answer Type

211.

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

123 Views

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

115 Views

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

153 Views

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

170 Views

Short Answer Type

215.
Show that:
$integral subscript 0 superscript straight a fraction numerator square root of straight x over denominator square root of straight x plus square root of straight a minus straight x end root end fraction space equals space straight a over 2$

Let I = $integral subscript 0 superscript straight a fraction numerator square root of straight x over denominator square root of straight x plus square root of straight a minus straight x end root end fraction dx$ ...(1)
$therefore space space space straight I space equals space integral subscript 0 superscript straight a fraction numerator square root of straight a minus straight x end root over denominator square root of straight a minus straight x end root plus square root of straight a minus left parenthesis straight a minus straight x right parenthesis end root end fraction dx$ $open square brackets because space space integral subscript 0 superscript straight a straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript 0 superscript straight a straight f left parenthesis straight a minus straight x right parenthesis space dx close square brackets$
$rightwards double arrow$ $straight I space equals space integral subscript 0 superscript straight a fraction numerator square root of straight a minus straight x end root over denominator square root of straight a minus straight x end root plus square root of straight x end fraction dx$ ...(2)
Adding (1) and (2), we get,
$2 space straight I space equals space integral subscript 0 superscript straight a open parentheses fraction numerator square root of straight x over denominator square root of straight x plus square root of straight a minus straight x end root end fraction plus fraction numerator square root of straight a minus straight x end root over denominator square root of straight a minus straight x end root plus square root of straight x end fraction close parentheses dx space equals space integral subscript 0 superscript straight a fraction numerator square root of straight x plus square root of straight a minus straight x end root over denominator square root of straight x plus square root of straight a minus straight x end root end fraction dx space equals integral subscript 1 superscript straight a 1. space dx$
$equals space open square brackets straight x close square brackets subscript 0 superscript straight a space equals space straight a space minus space 0$
$rightwards double arrow space space space 2 space straight I space equals space straight a space space space space space space rightwards double arrow space space space space straight I space equals space straight a over 2$

104 Views

216.

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

131 Views

217.

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

231 Views

218.

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

149 Views

219.

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

171 Views

220.

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