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# Negative definite integrals

We learned that definite integrals give us the area under the curve and above the x-axis. But what if the curve itself is below the x-axis? In this case, the definite integral is still related to area, but it's negative. See how this works and get some intuition for why this is so.

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• what happen when a is greater than b?
• The function will become negative because you are going in the reverse order.
• How the unit of the area is the only meter?
shouldn't it be Meter squared? at 3.40
(1 vote)
• We're multiplying velocity, measured in meters per second (m/s), and time, measured in seconds (s), so the resulting unit will be
(m/s) ∙ s = m ∙ s/s = m ∙ 1 = m (meters).
• This question has always bugged me: which is greater: -8m/s or 4m/s? Or does it not make sense to ask this question at all?
(1 vote)
• I think the difference between speed and velocity are key here.

Speed is pure numbers. There can be no negative speed. if a car is moving to the right at 4 m/s and another os moving left at 8 m/s, the left one may be moving in the negative direction, but speed doesn't account for that.

Velocity DOES take into account direction. That being said, it can have parts that are negative and parts that are positive. for instance if you had a space ship moving tot he right and down, right is the positive x directiona dn down is the negative y direction So it is very hard to compare things like that. It all depends on the question you are asking
• is it possible to know if the integral is below the curve if im not given a graph?
(1 vote)
• If your integral comes out to be negative, it would imply that the area is under the x-axis, not under the curve.
• Can we just simply say that if the quantity is vectorial then under the x-axis there must be a minus ?? For example in this video velocity is a vectorial quantity.
(1 vote)
• The quantity needn't be a vector. For example, you can have a force-displacement curve which extends below the x-axis, making the area under the curve negative. The area under the curve here is work, a scalar quantity.

In general, area under the x axis is treated as negative area, irrespective of what the quantity is.
• what if a=-6; b=-2; integral(f(x)=6), what would be the area?+24 or -24?
(1 vote)
• If you're integrating from -6 to -2, you're taking the positive area because -6 is less than -2. f(x) = 6 is always above the x-axis, so this means that your area will be positive, as you're taking the integral in the normal direction of a function that has a positive area.
• at , if a is greater than b, what's the difference and why?
(1 vote)
• If a becomes greater than b, two possibilities come into play:

1. If the function is strictly below the x axis, the area will be negative. But, as your bounds are going from a higher number to lower number, on reversing them, a negative sign appears which negates the sign of the area, hence, giving a positive answer.

2. If the function is above the x axis, the area is positive. But, as we're going from a higher bound to a lower one, on reversing it, a negative sign comes and makes the area negative.
• Here is my understanding, could you please let me know if this sounds about correct
1. g(x) is the curve of the area under the curve and x-axis
2. g'(x) is the curve of f(x) which is the curve of the slope
3. f(x) is the curve of the actual function
(1 vote)