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Integrals

Short Answer Type

181.

By using the properties of definite integrals, evaluate the following integral:
$integral subscript 0 superscript 4 open vertical bar straight x minus 1 close vertical bar space dx$

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182.
By using the properties of definite integrals, evaluate the following integral:
$integral subscript 2 superscript 5 open vertical bar straight x minus 1 close vertical bar dx$

Let I = $integral subscript 2 superscript 5 open vertical bar straight x minus 1 close vertical bar space dx$
For $2 less or equal than straight x less or equal than 5 comma space space straight x minus 1 greater than 0 space space space space space space rightwards double arrow space space space open vertical bar straight x minus 1 close vertical bar space equals space straight x minus 1$
$therefore space space space space space straight I space equals space integral subscript 2 superscript 5 left parenthesis straight x minus 1 right parenthesis space dx space equals space open square brackets straight x squared over 2 minus straight x close square brackets subscript 2 superscript 5 space equals space open parentheses 25 over 2 minus 5 close parentheses minus left parenthesis 0 minus 0 right parenthesis space equals space 15 over 2$

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183.

Show that:
$integral subscript 1 superscript 5 open vertical bar straight x minus 3 close vertical bar space dx space equals space 4$

90 Views

184.

Show that:
$integral subscript 0 superscript 1 open vertical bar 3 straight x minus 1 close vertical bar space dx space equals space 5 over 6$

99 Views

185.

By using the properties of definite integrals, evaluate the following integral:

$integral subscript negative 5 end subscript superscript 5 open vertical bar straight x plus 2 close vertical bar dx$

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186.

By using the properties of definite integrals, evaluate the following integral:
$integral subscript 2 superscript 8 open vertical bar straight x minus 5 close vertical bar space dx$

119 Views

Long Answer Type

187.

Show that:
$integral subscript negative 1 end subscript superscript 2 open vertical bar straight x cubed minus straight x close vertical bar space dx.$

171 Views

Short Answer Type

188.

Evaluate: $integral subscript 0 superscript 1 open vertical bar 2 straight x minus 1 close vertical bar space dx.$

151 Views

189.

Show that:
$integral subscript 0 superscript straight pi open vertical bar cos space straight x close vertical bar space dx space equals 2$

103 Views

190.

Show that:
$integral subscript 0 superscript straight pi over 2 end superscript space open vertical bar sin space straight x space cosx close vertical bar space dx space equals space 1 half$