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.

Main content
Current time:0:00Total duration:5:53
AP.CALC:
FUN‑6 (EU)
,
FUN‑6.D (LO)
,
FUN‑6.D.1 (EK)
,
FUN‑6.D.2 (EK)

Video transcript

what we're going to do in this video is get some practice applying u-substitution to definite integrals so let's say we have the integral so we're going to go from x equals 1 to x equals 2 and the integral is 2x times x squared plus 1 to the 3rd power DX so I already told you that we're going to apply use substitution but it's interesting to be able to recognize when to use it and the key giveaway here is well I have this x squared plus 1 business of 3rd power but then I also have the derivative of x squared plus 1 which is 2x right over here so I could do the substitution I could say U is equal to x squared plus 1 in which case the derivative of U with respect to X is just 2x plus 0 or just 2x I could write that in differential form a mathematical hand wavy way of thinking about it is multiplying both sides by DX and so you get D U is equal to 2 X DX and so at least this part of the integral I can rewrite so let me at least write so this is going to be I'll write the integral we're going to think about the bounds in a second so we have U to the third power you do the same orange color U to the third power that's this stuff right over here and then 2x times DX remember you could just view this as 2x times x squared plus 1 to the third power times DX so 2x times DX well 2 x times DX that is d u so that and that together that is d u now an interesting question because this isn't an indefinite integral we're not just trying to find the antiderivative this is a definite integral so what happens to our bounds of integration well there's two ways that you can approach this you can change your bounds of integration because this one is x equals 1/2 x equals 2 but now we're integrating with respect to U so one way if you want to keep your bounds of integration or if you want to keep this a definite integral I guess you could say you would change your bounds from U is equal to something to U is equal to something else and so your bounds of integration let's see when X is equal to 1 what is U well when X is equal to 1 you have 1 squared plus 1 so you have 2 u is equal to 2 in that situation when X is equal to 2 what is U well you have 2 squared which is 4 plus 1 which is 5 so U is equal to 5 and you won't typically see someone writing u equal to or u equals 5 it's often just from 2 to 5 because we're integrating with respect to u you assume it's u equals to 2 u equals 5 and so we could just rewrite this as being equal to the integral from 2 to 5 of U to the third d u but it's really important to realize why we changed our bounds we are now integrating with respect to U and the way we did it is we used our substitution right over here what x equals 1 u is equal to 2 when X is equal to 2 2 squared plus 1 u is equal to 5 and then we can just evaluate this right over here let's see this is going to be equal to the antiderivative of U to the third power is U to the 4th over 4 and we're going to evaluate that at 5 & 2 and so this is going to be 5 to the 4th over 4 minus 2 to the 4th over 4 and then we could simplify this if we like but we've just evaluated this definite integral now another way to do it is to think about the is to try to solve the indefinite integral in terms of X and use u substitution as an intermediate so one way to think about this is to say well let's just try to evaluate what the indefinite integral of 2x times x squared plus 1 to the third power DX is and then whatever this expression ends up being algebraically I'm going to evaluate it at x equals two and at x equals one and so then you use the u substitution right over here and you would get this would simplify using the same substitution as the integral of you you to the third power do you use the third power D you and once again you're going to evaluate this whole thing from x equals two and then subtract from that it evaluated at x equals one and so this is going to be equal to well let's see this is U to the fourth power over 4 and once again evaluating it at x equals two and then subtracting from that at x equals one and then now we can just back substitute we could say hey look you is equal to x squared plus one so this is the same thing as x squared plus one to the fourth power over 4 and we're now going to evaluate that at x equals two and it x equals one and you will notice that you'll get the exact same thing when you put x equals two here you get two squared plus 1 which is five to the fourth power over 4 that right over there and then minus one squared plus one is two to the fourth power over 4 that right over there so either way you'll get the same result you can either keep it a definite integral and then change your bounds of integration and express them in terms of U that's one way to do it the other way is to try to evaluate the indefinite integral use u substitution as an intermediary step then back substitute back and then evaluate at your bounds
AP® is a registered trademark of the College Board, which has not reviewed this resource.