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

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183.
Show that:
$integral subscript 1 superscript 5 open vertical bar straight x minus 3 close vertical bar space dx space equals space 4$

Let I = $integral subscript 1 superscript 5 open vertical bar straight x minus 3 close vertical bar space dx$
For $1 less or equal than straight x less or equal than 3 comma space space straight x minus 3 space less or equal than space 0 space space space space space space space space rightwards double arrow space space space space open vertical bar straight x minus 3 close vertical bar space equals space minus left parenthesis straight x minus 3 right parenthesis$
and for $3 less or equal than straight x less or equal than 5 comma space space space straight x minus 3 greater or equal than 0 space space space space space rightwards double arrow space space space open vertical bar straight x minus 3 close vertical bar space equals space straight x minus 3$
$therefore space space space space straight I space equals space integral subscript 1 superscript 3 open vertical bar straight x minus 3 close vertical bar dx space space plus integral subscript 3 superscript 5 open vertical bar straight x minus 3 close vertical bar space dx space equals space minus integral subscript 1 superscript 3 left parenthesis straight x minus 3 right parenthesis dx space plus integral subscript 3 superscript 5 left parenthesis straight x minus 3 right parenthesis space dx$
$equals negative open square brackets straight x squared over 2 minus 3 straight x close square brackets subscript 1 superscript 3 space plus space open square brackets straight x squared over 2 minus 3 straight x close square brackets subscript 3 superscript 5 equals space minus open square brackets open parentheses 9 over 2 minus 9 close parentheses space minus open parentheses 1 half minus 3 close parentheses close square brackets plus open square brackets open parentheses 25 over 2 minus 15 close parentheses minus open parentheses 9 over 2 minus 9 close parentheses close square brackets space equals space 9 over 2 minus 5 over 2 minus 5 over 2 plus 9 over 2 equals 4.$

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

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Short Answer Type

188.

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

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189.

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

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190.

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