anonymous
  • anonymous
How do I solve the definite integral of xabs(49-x^2) from x=-14 to 1?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Luigi0210
  • Luigi0210
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Luigi0210
  • Luigi0210
Is this your question? \[\int\limits_{-14}^{1} x|49-x^2| dx\]
phi
  • phi
you have to break up the integral into regions where you know sign of the expression inside the | | operator. in this case, 49-x^2 = (7-x)(7+x) over what interval is this expression negative ? once we find that interval we can replace | (7-x)(7+x) | with - (7-x)(7+x) (because - a negative quantity will be + ) also, we find the interval over which (7-x)(7+x) is positive... in that interval we can replace |(7-x)(7+x)| with (7-x)(7+x)

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phi
  • phi
next, find the interval over which (7-x)(7+x) > 0 this occurs if both expressions are Case 1: both negative 7-x < 0 and 7+x < 0 --> 7< x and x < -7....cannot occur Case 2: both positive 7-x > 0 and 7+x> 0 ---> 7> x and x > -7, i.e. -7
phi
  • phi
when is (7-x)(7+x) negative? i.e. (7-x)(7+x) < 0 it will be negative if Case 1: (7-x) is neg and 7+x is + 7-x < 0 and 7+x>0 ---> 7 < x and x > -7, i.e x > 7 over this interval replace |49-x^2| with x^2 -49 or Case 2: 7-x is positive and 7+x is negative 7-x> 0 and 7+x < 0 --> 7> x and x < -7, i.e. x < -7 over this interval replace |49-x^2| with x^2 -49
phi
  • phi
putting it all together we have \[ \int\limits_{-14}^{1} x|49-x^2| dx \\ \int\limits_{-14}^{-7} x(x^2-49) \ dx + \int\limits_{-7}^{1} x(49-x^2)\ dx\]
phi
  • phi
I think I got that straight... but it is easy to flip a sign when doing this...
anonymous
  • anonymous
Thank you phi :) that helps

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