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

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184.

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

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

Let I = $integral subscript 2 superscript 8 open vertical bar straight x minus 5 close vertical bar space dx$
For 2 ≤ r ≤ 5, x – 5 ≤ 0 ⇒ | x – 5 | = – (x – 5) and for 5 ≤ x ≤ 8, x – 5 ≥ 0 ⇒ | x – 5 | = x – 5
$therefore space space space space straight I space equals space integral subscript 2 superscript 5 open vertical bar straight x minus 5 close vertical bar dx space plus space integral subscript 5 superscript 8 open vertical bar straight x minus 5 close vertical bar space dx$
$equals negative integral subscript 2 superscript 5 left parenthesis straight x minus 5 right parenthesis dx plus integral subscript 5 superscript 8 left parenthesis straight x minus 5 right parenthesis dx space equals space minus open square brackets straight x squared over 2 minus 5 straight x close square brackets subscript 2 superscript 5 plus open square brackets straight x squared over 2 minus 5 straight x close square brackets subscript 5 superscript 8$
$equals negative open square brackets open parentheses 25 over 2 minus 25 close parentheses space minus space open parentheses 4 over 2 minus 10 close parentheses close square brackets space plus space open square brackets open parentheses 64 over 2 minus 40 close parentheses minus open parentheses 25 over 2 minus 25 close parentheses close square brackets equals negative open square brackets negative 25 over 2 plus 8 close square brackets space plus open square brackets negative 8 plus 25 over 2 close square brackets space equals space 9 over 2 plus 9 over 2 space equals 9$

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Long Answer Type

187.

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

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