If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Switching bounds of definite integral

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.A (LO)
,
FUN‑6.A.1 (EK)
,
FUN‑6.A.2 (EK)

## Video transcript

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 like 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 into n rectangles so let's say that's the first rectangle 1 that's the second rectangle 2 and you're going to go all the way to the enth rectangle so this would be the N minus 1 threat tangle and for the sake of this argument I'm going to make in this video we're going to sue them that they're all the same width so this is the nth rectangle and they all have the same width and we see their 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 we take B minus a and we divide it by n 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 equal spacings which is or n equal spacings of Delta X and so if you do this you say okay we we see this multiple times you can approximate it you can approximate this area using this rec these rectangles as the sum from I equals 1 to N so you're summing your each of the you're summing n of these rectangles areas where the height of the the air 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 sub 1 X sub 2 X sub 3 so on and so forth and you're multiplying that times your Delta X times your Delta X so you take X sub 2 f of X sub 2 is that height right there f of X sub 2 is that height right there you multiply it times Delta X you get the area and we saw that when we looked at Riemann sums and using that to approximation we said hey the definite the one definite definition of the definite integral is that since this is the area this is going to be the limit and approaches infinity of this where where delta x is defined as that so let me just copy and paste that so copy and paste where that so that's one way to think about it now given this definition what do you think what do you think this or maybe another way to think about 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 into going away from a and a B I'm now going from B to a I'm now going from B to a how do you think these two things should relate and 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 this is going to be if I were to literally just take this 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 am 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 it's going to be a minus B so each of these are going this value right over here let me make these color-coded maybe so this orange Delta X this orange Delta X is going to be the negative of this green of this green Delta X this is the negative of that right over there so 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 and so this is the result we get which is another really important integration property that if you swap if you swap the bounds of integration 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 you'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 a year and 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