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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:

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

LetI = $integral subscript negative 5 end subscript superscript 5 open vertical bar straight x plus 2 close vertical bar dx$
For - 5 ≤ x ≤ – 2, x + 2 ≤ 0 ⇒ | x + 2 | = – (x + 2) and for – 2 ≤ x ≤ 5, x + 2 ≥ 0 ⇒ | x + 2 | = x + 2
$therefore space space space space space space straight I space equals space integral subscript 2 superscript negative 2 end superscript open vertical bar straight x plus 2 close vertical bar space dx space plus space integral subscript negative 2 end subscript superscript 5 open vertical bar straight x plus 2 close vertical bar space dx space space equals integral subscript negative 5 end subscript superscript negative 2 end superscript minus left parenthesis straight x plus 2 right parenthesis space dx plus integral subscript negative 2 end subscript superscript 5 left parenthesis straight x plus 2 right parenthesis space dx$
$equals negative integral subscript 5 superscript negative 2 end superscript left parenthesis straight x plus 2 right parenthesis space dx plus integral subscript negative 2 end subscript superscript 5 left parenthesis straight x plus 2 right parenthesis space dx space equals space minus open square brackets straight x squared over 2 plus 2 straight x close square brackets subscript negative 5 end subscript superscript negative 2 end superscript plus open square brackets straight x squared over 2 plus 2 straight x close square brackets subscript negative 2 end subscript superscript 5 equals negative open square brackets open parentheses 4 over 2 minus 4 close parentheses minus open parentheses 25 over 2 minus 10 close parentheses close square brackets space plus space open square brackets open parentheses 25 over 2 plus 10 close parentheses minus open parentheses 4 over 2 minus 4 close parentheses close square brackets equals negative open square brackets negative 2 minus 5 over 2 close square brackets plus open square brackets 45 over 2 plus 2 close square brackets space equals space 9 over 2 plus 49 over 2 equals 58 over 2 equals space space 29$

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

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