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."

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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."

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

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Oh I see, got it. Thanks!
Np :)

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