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Mathematics 16 Online
OpenStudy (anonymous):

definite integrals? integral of (x^2-4x+4)^1/2 dx?

jimthompson5910 (jim_thompson5910):

luckily x^2 - 4x + 4 = (x-2)^2

OpenStudy (anonymous):

\[\int\limits_{0}^{2}\sqrt{x^2-4x+4}dx\]

jimthompson5910 (jim_thompson5910):

so what you effectively have is sqrt( (x-2)^2 )

OpenStudy (anonymous):

i tried it that way, but i got it wrong

jimthompson5910 (jim_thompson5910):

sqrt( (x-2)^2 ) = |x-2|

OpenStudy (anonymous):

oh absolute value

jimthompson5910 (jim_thompson5910):

so you want the area under the curve from x = 0 to x = 2

jimthompson5910 (jim_thompson5910):

the graph looks like this |dw:1365376757955:dw|

jimthompson5910 (jim_thompson5910):

and you want this area |dw:1365376776887:dw|

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