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Integrals

Short Answer Type

181.

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

86 Views

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

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

We have
$1 less or equal than straight x less or equal than 0 space space rightwards double arrow space straight x cubed minus straight x space equals space straight x left parenthesis straight x squared minus 1 right parenthesis space greater or equal than 0 space space rightwards double arrow space space open vertical bar straight x cubed minus straight x close vertical bar space equals space straight x cubed minus straight x 0 less or equal than straight x less or equal than 1 space space space rightwards double arrow space space straight x cubed minus straight x space equals space straight x left parenthesis straight x squared minus 1 right parenthesis less or equal than 0 space space rightwards double arrow space space open vertical bar straight x cubed minus straight x close vertical bar space equals space minus left parenthesis straight x cubed minus straight x right parenthesis 1 less or equal than straight x less or equal than 2 space space space rightwards double arrow space straight x cubed minus straight x space equals space straight x left parenthesis straight x squared minus 1 right parenthesis space greater or equal than space 0 space rightwards double arrow space space open vertical bar straight x cubed minus straight x close vertical bar space equals space straight x cubed minus straight x$
Let $straight I space equals space integral subscript negative 1 end subscript superscript 2 open vertical bar straight x cubed minus straight x close vertical bar space dx space equals space integral subscript negative 1 end subscript superscript 0 open vertical bar straight x cubed minus straight x close vertical bar space dx space space plus integral subscript 0 superscript 1 open vertical bar straight x cubed minus straight x close vertical bar space dx plus integral subscript 1 superscript 2 open vertical bar straight x cubed minus straight x close vertical bar space dx$
$equals space integral subscript negative 1 end subscript superscript 0 left parenthesis straight x cubed minus straight x right parenthesis dx space minus space integral subscript 0 superscript 1 left parenthesis straight x cubed minus straight x right parenthesis space dx space plus space integral subscript 1 superscript 2 left parenthesis straight x cubed minus straight x right parenthesis space dx$
$equals space open square brackets straight x to the power of 4 over 4 minus straight x squared over 2 close square brackets subscript negative 1 end subscript superscript 0 space minus open square brackets straight x to the power of 4 over 4 minus straight x squared over 2 close square brackets subscript 0 superscript 1 plus open square brackets straight x to the power of 4 over 4 minus straight x squared over 2 close square brackets subscript 1 superscript 2 equals space open square brackets left parenthesis 0 minus 0 right parenthesis minus open parentheses 1 fourth minus 1 half close parentheses close square brackets space minus open square brackets open parentheses 1 fourth minus 1 half close parentheses minus left parenthesis 0 minus 0 right parenthesis close square brackets space plus space open square brackets left parenthesis 4 minus 2 right parenthesis minus open parentheses 1 fourth minus 1 half close parentheses close square brackets equals space minus open parentheses 1 fourth minus 1 half close parentheses minus open parentheses 1 fourth minus 1 half close parentheses plus 2 minus open parentheses 1 fourth minus 1 half close parentheses space equals space minus 3 open parentheses 1 fourth minus 1 half close parentheses plus 2 equals negative 3 open parentheses fraction numerator 1 minus 2 over denominator 4 end fraction close parentheses plus 2 space equals space minus 3 open parentheses negative 1 fourth close parentheses space plus 2 space equals space 3 over 4 plus 2 space equals space fraction numerator 3 plus 8 over denominator 4 end fraction space equals space 11 over 4$

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

103 Views

190.

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