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

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

Let I = $integral subscript 0 superscript straight pi over 2 end superscript open vertical bar sinx space cosx close vertical bar dx space equals space 1 half integral subscript 0 superscript straight pi over 2 end superscript open vertical bar 2 space sinx space cosx close vertical bar dx$
$equals space 1 half integral subscript 0 superscript straight pi over 2 end superscript open vertical bar sin space 2 straight x close vertical bar dx space equals space 1 half integral subscript 0 superscript straight pi over 2 end superscript sin space 2 straight x space dx space equals space 1 half open square brackets fraction numerator negative cos space 2 straight x over denominator 2 end fraction close square brackets subscript 0 superscript straight pi over 2 end superscript space equals space minus 1 fourth open square brackets cos space 2 straight x close square brackets subscript 0 superscript straight pi over 2 end superscript$
$equals negative 1 fourth left square bracket cos space straight pi space minus space cos space 0 right square bracket space equals space minus 1 fourth left square bracket negative 1 minus 1 right square bracket space space equals space 1 half$