I suddenly ran into a problem I've never seen before...a variation I suppose? I have no idea how to work this one out, but some other guys say they found a solution, but right now, I'm thinking no solution...here's the question:
BTW: ~ is the integration sign (sorry I don't know how to do it on comp) so 1~2 would mean integrating from 1 to 2.
Let:
1~2 f(x) dx = -3,
1~5 f(x) dx = 5, and
1~5 g(x) dx = 9.
Find: 1~2 (3 f(x) +1) dx.
I've done tons of just integration with these kinds of problems, but I've never seen one where there is a constant inside...how can you solve this? Others say the answer is -8.
2007-01-04
14:16:02
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4 answers
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asked by
Moosehead
2
in
Mathematics