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Studying for a test? Prepare with these 9 lessons on Antiderivatives and the fundamental theorem of calculus.
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Video transcript
Let's say that we have the indefinite integral, and the function is 3x squared plus 2x times e to x to the third plus x squared dx. So how would we go about solving this? So first when you look at it, 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. 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. And the chain rule-- 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, or 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? du 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 du dx, this isn't really a fraction of the differential of du divided by differential of dx. It really is a form of notation, but it is often useful to kind of pretend that it is a fraction, and you could kind of view this if you wanted to just get a du, if you just wanted to get a differential form over here, how much does u change for a given change in x? 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 du is equal to 3x squared plus 2x dx. Now why is this over here? Why did I go to 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-- and let me do it in that color-- of 3x squared plus 2x times dx 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 du. This is exactly equal to du. 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 du, this entire du, I'm gonna 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 we're going to have our du 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 e to the u is e to the u. So it's going to be equal to e to the u. Now, there is 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 unsubstitute the u. We know 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. 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.