integral

[hallow oxide]

how is the integral function (blue) positive in the negative side if the function (green) is always negative ?

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[sonic plank]

If a function is always negative, its antiderivative will always be decreasing.

[hallow oxide]

but doesnt the graph imply that the area under the curve in the negative side is positive?

[sonic plank]

No.
Recall that the area is found using the fundamental theorem of calculus - as the difference of values of the antiderivative.

And since the antiderivative is decreasing, then for x2 > x1 we have F(x2) - F(x1) < 0.

[hallow oxide]

i understand that the rules end up with this , but the way from where i can't fully understand is using the definition of integrals , if they are the area under a curve then why going from right to left here change the sign of the area while the main function is a constent

[sonic plank]

The antiderivative by itself isn't an area of anything.

The area is found as the difference of values of an antiderivative.

By itself, the antiderivative F(x) of a function f(x) is just such a function that F'(x) = f(x).

[hallow oxide]

thanks

+close

[livid wing]

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[sonic plank]

You're welcome!

[livid wing]

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