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𝘶-substitution: defining 𝘶

FUN‑6 (EU)
FUN‑6.D (LO)
FUN‑6.D.1 (EK)
A common challenge when performing 𝘶-substitution is to realize which part should be our 𝘶.

Video transcript

- [Tutor] What we're going to do in this video is give ourselves some practice in the first step of u substitution, which is often the most difficult for those who are first learning it and that's recognizing when u substitution is appropriate and then defining an appropriate u. So let's just start with an example here, so let's say we wanna take the indefinite integral of two x plus one times the square root of x squared plus x, dx, does u substitution apply here and if it does, how would you define the u? Pause the video and try to think about that. Well, we just have to remind ourselves, that u substitution is really trying to undo the chain rule, if we remind ourselves what the chain rule tells us, it says look, if we have a composite function, let's say f of g of x, f of g of x and we take the derivative of that with respect to x, that that is going to be equal to the derivative of the outside function with respect to the inside function, so f prime of g of x times the derivative of the inside function, times the derivative of the inside function and so u substitution is all about well, do we see a pattern like that inside the integral? Do we see a potential inside function, a g of x, where I see its derivative being multiplied? Well, we see that over here, if I look at x squared plus x, if I make that the u, what's the derivative of that? Well, the derivative of x squared plus x is two x plus one, so we should make that substitution, if we say u is equal to x squared plus x, then we could say du dx, the derivative of u with respect to x is equal to two x plus one, if we treat our differentials like variables or numbers, we can multiply both sides by dx, which is a little bit of hand wavy mathematics, but it's appropriate here, so we could say two x plus one times dx and now what's really interesting here is we have our u right over there and notice we have our two x plus one times dx, in fact it's not conventional to see an integral rewritten the way I'm about to write it, but I will, I could rewrite this integral, you should really view this is the product of three things, oftentimes people will just view the dx as somehow part of the integral operator, but you could rearrange it, this would actually be legitimate, you could say the integral of the square root of x squared plus x times two x plus one dx and if you wanted to be really clear, you could even put all of those things in parentheses or something like that and so here, this is our u and this right over here is our du and so we could rewrite this as being equal to the integral of the square root of u, 'cause x squared plus x is u, times du, which is much easier to evaluate. If you are still confused there, you might recognize it if I rewrite this as u to the one half power, because now we could just use the reverse power rule to evaluate this and then we would have to undo the substitution, once we figure out what this anti derivative is, we then would then reverse substitute the x expression back in for the u.