There;s a property that says that if f(x) is greater than or equal to 0, on an interval [a,b] then the definite integral of f(x) from a to b, is also greater than or equal to zero. Is the opposite of this property true? The opposite being:"If f(x) is less than or equal o zero then the definite integral is also less than or equal to zero."

Question

anonymous · Answer

Yes, here's an example: f(x)=-x^2 from a=1 to b=2

integrate: -x^3/3 from 1 to 2 =-7/3

The reason is that you are calculating area below the curve.

anonymous · Answer

Where did you get the 3? So the integral is negative, but the area is positive?

anonymous · Answer

Yes, that is because you read it from left to right. It may be easier to explain the process by going backwards: when you integrate, you anti-differentiate. So take the derivative:
Drop the ^3 down and subtract 1 from the exponent. Then the 3/3 cancels to 1 and we get -x^2. If the integral were from 2 to 1 (rather than 1 to 2) it would be positive.

anonymous · Answer

Oh I see, got it. Thanks!

anonymous · Answer

Np :)