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Switching bounds of definite integral

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.A (LO)
,
FUN‑6.A.1 (EK)
,
FUN‑6.A.2 (EK)
What happens when you swap the bounds on an integral?

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• The area under the curve stays the same but if we switch the bounds we label it a negative area. What is a negative area?
Thanks
(55 votes)
• When we are doing Euclidean geometry, we always think of heights and widths and sides as having non-negative values. In fact, the "triangle inequality", which asserts that any side of a triangle is less than the sum of the other two sides, would break down if we allowed a side to have a negative value. But when we do calculus, we want those Reimann rectangles to be able to have negative width, if our delta x goes from a greater value to a lesser value, or negative height, if the value of the function is less than 0. To help see why, consider the definite integral of a non-negative function f between a and b, with a less than b. The function -f is non-positive between a and b. If we didn't allow those Reimann rectangles to have negative area, then the area of -f between a and b would be the same as the area of f between a and b. So the area between -f and f would simply be twice the area under f. But we want the area of the sum of functions to be the sum of their areas. The sum of f and -f is everywhere 0, so we want the area of their sum between a and b to be 0. For that to work, we need the concept of a negative area. It's a bit confusing, but not all that different from recognizing positive and negative numbers. (What, after all, is a negative dollar?)
(53 votes)
• Is it possible to have a Riemann Sum with sequentially increasing widths?
(10 votes)
• Yes, but your area approximation is going to be more and more inaccurate as it goes on.
(11 votes)
• i dont understand one thing. Sal so far said that integrals means the anti-derivative BUT now he says its the area under a curve. totally confused
(4 votes)
• To add to what Yamanqui stated,

An antiderivative can be interpreted as just one type of integral (it is also called an indefinite integral or a primitive integral). There are other types of integrals such definite integrals.
(11 votes)
• I was curious about what happened to the effect of the new delta x=((b-a)/n) that appeared in the x sub i formula then I realized that the x sub i formula would change too, in this case, from (x sub i = (a+((b-a)i/n))) to (x sub i = (b+((a-b)i/n))). After searching, this led me to wonder why the x sub i formula is rarely, if ever, expanded in terms of a, b, i, and n or a, i, and delta x when the definite integral formula is presented...
(6 votes)
• Shouldn´t the notation of the Sum be i=a instead of i=1 ?
(3 votes)
• No, the 'i' in this case refers only to the rectangle number. So when i = 1 we have f ( x sub 1 ) which is f of the first rectangle, not f ( 1 ).

The only thing you could represent in terms of a is the f of something. We could say that the statement shown is equivalent to: The limit of ( The sum of f ( a+i*Δ x ) * Δ x , from i = 1 to n ) as n approaches infinity.
(6 votes)
• How (integral(a to b)f(x)*dx)=negative((integral(b to a)f(x)*dx))?Both of them represents
same area.For example area under the function y=3 calculated in the direction from x=0 and x=3 is same as area under y=3 calculated in the direction from x=3 to x=0 which is equal to 9 units.I shall be grateful if my doubt is clarified!
(3 votes)
• It's a consequence of the way we use the Fundamental Theorem of Calculus to evaluate definite integrals. In general, take `int(a=>b) [ f(x) dx ]`. If the function `f(x)` has an antiderivative `F(x)`, then the integral is equal to `F(b) - F(a) + C`. Now take the reverse: `int(b=>a) [ f(x) dx ] = F(a) - F(b) + C = - ( F(b) - F(a) ) + C`.

Effectively, this just means we have to consider direction when we evaluate integrals in addition to considering whether the area is above or below the axis. It's another way of adding a sign to the area, just like stating that area under the x-axis counts as negative (even though, really, it's still the same area). Integrals evaluated from left to right are "normal" (that is, area above the axis is positive, and area below is negative), and integrals evaluated from right to left are the opposite of "normal".
(3 votes)
• Does the negative area mean only an opposite direction of a vector? As multiplying it by the scalar `-1`, or something more and different?
(3 votes)
• It does only mean that the area is in an opposite direction. Area is always positive, when calculating an area using integrals the absolute value would be used to produce a final answer.
(3 votes)
• It's true that if b>a, then (a-b)/n will give a negative value. But since ∆x is a length, how can it be negative?
(3 votes)
• ∆x is not actually a distance, it's a displacement (the vector equivalent). Since we are moving in the negative x direction in that case, ∆x is negative.
(3 votes)
• We know that definite integration is nothing but calculating the area of a curve.
So, then why is integral of a function from a to b equal to the negative of the integral of the function from b to a. Well, does this mean the area is negative when u swap the lower and upper bounds? I'm quite confused here.
(3 votes)
• Go back and watch the previous videos.

What you taking when you integrate is the area of an infinite number of rectangles to approximate the area. When f(x) < 0 then area will be negative as f(x)*dx <0 assuming dx>0.

Switch bound rule can be proved with some theorem, which was mention in one of the previous videos.
(1 vote)
• i understand how he got to the final answer, but shouldn't the area under the curve be the same, no matter which side you start summing the rectangles from? Please correct where i went wrong. Thanks
(2 votes)
• The whole idea of lower and upper bounds in Integration is that the lower bound represents the smallest value from which we start summing areas(smallest value of the interval) and upper bound is the value to which we sum to(maximum value of the interval). For example: you're asked to calculate the area under the curve between x values ranging from [1,5](from x=1 to x=5). In this case, lower bound would be 1 and upper-bound would be 5.
So, what we're doing is,summing up the area from the lowest to the highest 'x' value. But, when you swap the bounds, we're no more summing from lowest to highest, instead, from highest to lowest. In other words, In when you're integrating, you're following the positive x-direction, but, when you swap bounds, you're calculating the area in the negative x-direction and thus, negative sign! Hope this helps!(I know I'm 3 years late! I'm just starting now!)
(3 votes)

Video transcript

- [Voiceover] We've already seen one definition of the definite integral, and many of them are closely related to this definition that we've already seen is the definite integral from a to b of f of x d of x is this area shaded in blue, and we can approximate it by splitting it into n rectangles. So let's say that's the first rectangle, one. That's the second rectangle, two, and you're going to go all the way to the nth rectangle, so this would be the n minus oneth rectangle. For the sake of this argument I'm going to make in this video, we're going to assume that they're all the same width. So this is the nth rectangle. They all have the same width, and we see they're definitions of integration where you don't have to have the same width here, but let's say that each of those widths are delta x, and the way that we calculate delta x is we take b minus a and we divide it by n, which is common sense, or this is what you learned in division. We're just taking this length and dividing it by n to get n equals spacings of delta x. So if you do this, you'll say, okay, well we see this multiple times, you can approximate it. You can approximate this area using these rectangles as the sum from i equals one to n. So you're summing n of these rectangle's areas where the height of each of these rectangles are going to be f of x sub i, where x sub i is the point at which you're taking the function value to find out its height. So that could be x of one, x of two, x of three, so on and so forth, and you're multiplying that times your delta x. So you take x sub two, f of x sub two is that height right there. You multiply it times delta x. You get the area. We saw that when we looked at Riemann sums and using that to approximate. We said, hey the one definition of the definite integral is that since this is the area this is going to be the limit as n approaches infinity of this where delta x is defined as that. So let me just copy and paste that. So that's one way to think about it. Now, given this definition, what do you think this, or maybe another way to think about it, how do you think this expression that I'm writing right over here based on this definition should relate to this expression? So notice, all I've done is I've segued from a to b. I'm now going from b to a. How do you think these two things should relate? I encourage you to look at all of this to come to that conclusion, and pause the video to do so. Well let's just think about what's going to happen. This is going to be, if I were to literally just take this and copy and paste it, which is exactly what I'm going to do, if I just took this, by definition, since I swapped these two bounds, I'm going to want to swap these two. Instead of b minus a it's going to be a minus b now. It's going to be a minus b. This value right over here. Let me make these color-coded maybe. So this orange delta x is going to be the negative of this green delta x. This is the negative of that right over there, and everything else is the same. So what am I going to end up doing? Well I'm essentially going to end up having the negative value of this. So this is going to be equal to the negative of the integral from a to b of f of x dx. So this is the result we get, which is another really important integration property, that if you swap the bounds of integration, and it really just comes from this idea, instead of delta x being b minus a, if you swap the bounds of integration, it's going to be a minus b. We're going to get the negative delta x, or the negative of your original delta x, which is going to give you the negative of this original value right over here. Once again, this is a really, really useful integration property where you're trying to make sense of some integrals and even sometimes solve some of them.