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Integration question AREA C4

I was wondering if the way I found Area 3 was correct. A3 is the area under the graph under the x axis .60782356_447470759359428_656704687151513600_n[1].jpg
60492140_432773810833964_8243542228287356928_n[1].jpg
Since it's under the x axis I thought it'd be -ve but I got a +ve area and the overall answer is correct to so I'm slightly confused.
(edited 4 years ago)
Reply 1
Avatar for mqb2766
mqb2766
21
Answer looks Ok to me.
The only minor bit, which you seem to comment on is the lower curve. I think you have the limits 0,1? That would be for the upper curve. What would be the (similar) values for the lower one?
Reply 2
Avatar for JacobBob
JacobBob
OP
10
Original post by mqb2766
Answer looks Ok to me.
The only minor bit, which you seem to comment on is the lower curve. I think you have the limits 0,1? That would be for the upper curve. What would be the (similar) values for the lower one?


oh 0 and -1 ?
But would it be from 0 to -1 or -1 to 0
If it was in terms of x the limits would be from 0 to 3 but translating that directly to t would make it from 0 to -1 which is from a bigger to a smaller .. is that okay ?
Reply 3
Avatar for mqb2766
mqb2766
21
Original post by JacobBob
oh 0 and -1 ?
But would it be from 0 to -1 or -1 to 0
If it was in terms of x the limits would be from 0 to 3 but translating that directly to t would make it from 0 to -1 which is from a bigger to a smaller .. is that okay ?


0 to -1 would be the most sensible and that would give you the negative area (under the x-axis).
Flipping the order of the limits flips the sign of the integral. You could define a new variable u=-t and work it through if you really wanted.

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