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Mathematics : 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:
$integral subscript 2 superscript 5 open vertical bar straight x minus 1 close vertical bar dx$

93 Views

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:
$integral subscript 2 superscript 8 open vertical bar straight x minus 5 close vertical bar space dx$

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

114 Views