Question

How do I solve the definite integral of xabs(49-x^2) from x=-14 to 1?

Answer 1

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Answer 2

Is this your question?
\[\int\limits_{-14}^{1} x|49-x^2| dx\]

Answer 3

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)

Answer 4

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<x < 7

we find that when -7<x < 7, we replace |49-x^2| with 49-x^2

Answer 5

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

Answer 6

putting it all together we have
\[ \int\limits_{-14}^{1} x|49-x^2| dx \\ \int\limits_{-14}^{-7} x(x^2-49) \ dx + \int\limits_{-7}^{1} x(49-x^2)\ dx\]

Answer 7

I think I got that straight... but it is easy to flip a sign when doing this...

Answer 8

Thank you phi :) that helps