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Current time:0:00Total duration:5:11

AP.CALC:

FUN‑6 (EU)

, FUN‑6.D (LO)

, FUN‑6.D.1 (EK)

let's say that we have the indefinite integral and the function is 3x squared plus 2x times e to the X to the third plus x squared DX so how would we go about solving this so first when you look at it seems like a really complicated integral we have this polynomial right over here being multiplied by this exponential expression and over here in the exponent we essentially have another polynomial it seems kind of crazy and the key intuition here the key insight is that you might want to use a technique here called u substitution substitution and I'll tell you in a second how I would recognize that we have to use u substitution and then over time you might even be able to do this type of thing in your head u substitution is essentially unwinding the chain rule in the chain well I'll do I'll go in more depth in another video where I really talk about that intuition but the way I would think about it is well I have this crazy exponent right over here I have the X to the third plus x squared and this thing right over here happens to be the derivative of X to the third plus x squared the derivative of X to the third is 3x squared derivative of x squared is 2x which is a huge clue to me that I could use u-substitution so what I do here is this thing where this little this little expression here where I also see its derivative being multiplied I can set that equal to U so I can say U is equal to X to the third plus x squared now what is going to be the derivative of U with respect to X D u DX well we've done this multiple times it's going to be 3x squared plus 2x and now we can write this in differential form and D U DX this isn't really a fraction of a differential of D U divided by a differential of DX it really is a form of notation but it is often useful to kind of pretend that is a fraction and you could kind of view this if you wanted to just get a DUI if you just wanted to get a differential form over here how much does you change for given change in X you could multiply you could multiply both sides times a DX so both sides times a DX and so if we were to pretend that they were fractions and it will give you the correct differential form you're going to be left with you're going to be left with D U is equal to is equal to 3x squared 3x squared plus 2x DX now why is this over here why did I go through the trouble of doing that well we see we have a 3x squared plus 2x and then it's being multiplied by a DX right over here I could rewrite this original integral I could rewrite this as the integral of let me do in that color of 3x squared plus 2 x times DX x times e let me do that in that other color times e to the X to the third plus x squared now what's interesting about this well the stuff that I have in magenta here is exactly equal to D u this is exactly equal to D U and then this stuff I have up here X to the third plus x squared that is what I set u equal to that is going to be equal to u so I can rewrite my entire integral and now you might recognize why this is going to simplify things a good bit it's going to be equal to and what I'm going to do is I'm going to change the order I'm going to put the D you this entire D you I'm going to stick it on the other side here so it looks like more of the standard form that we're used to seeing our indefinite integrals in so it's going to be we're going to have our D U our D u @ times e times e to the U times e to the U and so what would the antiderivative of this be in terms of U well the derivative of e to the U is e to the U the antiderivative of either is e to the U so it's going to be equal to e to the U e to the u now there's a possibility that there was some type of a constant factor here so let me write that so plus C and now to get it in terms of X we just have to unsubstituted oh what u is equal to so we could say that this is going to be equal to e instead of writing u we could say u is x to the third plus x squared U is X to the third plus x squared X to the third plus x squared and then we have our plus C and we are done we have found the antiderivative and I encourage you to take the derivative of this and I think you will find yourself using the chain rule and getting right back to what we had over here

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