General Question on area

  • Thread starter Miike012
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In the book it says that if f is say positive in [a,b) and negative from (b,c] where f(x) is continuous on [a,c]. Letting the area of [a,b) = A1 and (b,c] = A2 then the net area is
A1 - A2.

Question: This does not make sense to me... how can area be negative? For instance what if f < 0 in its entire domain [a,b] then its entire area is negative?
 

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Dick
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If f<0 and you integrate it over [a,b] you will get the negative of the area between the curve and the x-axis. That's what they mean. That the integral will be negative. Not the area.
 
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cepheid
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In the book it says that if f is say positive in [a,b) and negative from (b,c] where f(x) is continuous on [a,c]. Letting the area of [a,b) = A1 and (b,c] = A2 then the net area is
A1 - A2.

Question: This does not make sense to me... how can area be negative?
It's just a matter of definition. The definition of the "area under the curve" over an inteval is in terms of the integral of the function over that interval, which can come out negative or positive.

For instance what if f < 0 in its entire domain [a,b] then its entire area is negative?
Yes, the entire area would be negative.

Granted, this is sometimes a very useful definition. For instance what if the electric current at a point in space as a function of time is given by i(t) = I0sin(ωt)? What would be the net amount of charge passing across that point in space over one period of the sine function? Ans: This would be given by q(t) = ∫i(t)dt = 0. Just by glancing at the function over an interval equal to one period, you can see that the current spends just as much time in the positive direction as in the negative direction. The area under the first half is positive, and the area under the second half is negative and equal to the first area in magnitude, meaning that just as much charge flowed past the point from left to right in the first half as flowed from right to left in the second half, and the net flow was 0.
 

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