 We've seen already several examples of computing antiderivatives aka indefinite integrals using this technique of u-substitution What I want to do now is change directions a little bit and look at Definite integrals and see how u-substitution can help us in that regard So if we want to integrate from 0 to 5 x times the square root of x or 25 minus x squared dx How could one go about doing this well for for any problem with? Involving indefinite integral or sorry involving definite integrals you always start off by just approaching as a indefinite problem, right? So we want to integrate x times 20 the square root of 25 minus x squared dx And this is basically because of the fundamental theorem calculus We can compute the area of the curve using antiderivatives And so when you look at this And to go right here, uh, we are inclined to try to use some type of u-substitution We take u to be 25 minus x squared in which case then du Equals negative 2x dx. Uh, we have the x right here. We have the dx right here We need a negative 2 so make sure you divide by negative 2 right there And so then our integral would look something like negative one half The integral of u to the one half power because the square root of 25 minus x squared would give us The the u to the one half power and then the negative 2x dx becomes a du So it becomes something like that there And then continuing on By the power rule for antiderivatives The antiderivative u to the one half becomes u to the three halves. You raise the power by one divide everything by three halves plus a constant Then we see that since you're dividing by two thirds You're going to get the negative one half, but then you're going to if you divide by sorry Do you divide by three halves? You should times by two thirds like that the one the two should cancel u to the three halves plus a constant And so then in the end you get this antiderivative of negative one half I guess actually negative one sixth Uh that nope the two's cancelled out negative one third is what we get there And then u was 25 minus x squared Three halves plus a constant. So this gives us our antiderivative Sorry that gives us our antiderivative there uh, and so Then to evaluate the indefinite integral, right? We're going from zero to five x times the square root of 25 minus x squared dx Well, what we need to do is find an antiderivative which what we have one now We're going to get negative one third 25 minus x squared Three halves We evaluate it from zero to five. You don't have to worry about the constant with you do these definite integrals here and so when we plug in The five and the zero we end up with a negative one third We get 25 minus well five squared is 25 uh raised to the three halves and then we subtract from that 25 minus zero square which is zero three halves there And so notice 25 take away 25 is zero So you get zero to the three halves, of course, it's just zero and then you're going to get 25 to the three halves When dealing with these rational exponents, I usually like to take the radicals first the one half power so the one half power of So when you look at this the one half power of 25 will give you a five so you get five cubed, uh, which is going to be 125 And so we end up with negative one third times negative 125 Which would then simplify to become 125 over three Which is the area under this curve right here And so this is a perfectly acceptable way of computing These antiderivatives that is calculating these indefinite integrals. Sorry definite integrals. That's what we're working on right now What I want to do is actually approach this problem from a second perspective. So if we were to do the remake Of this problem right here, we're going from zero to five x times the square root of 25 minus x squared dx like so What if we don't bother with the indefinite integral first could we sort of start with the With the definite integral because after all we still get this u substitution We had before we could do u equals 25 minus x squared and then du Would equal negative 2x dx And so we're still in a position where we can have a negative one half in front and a negative two right there Uh, but the issue that I want to address is what can you do with these right here? Because after all when you look at the limits here the bounds of the integral These are x coordinates when x equals zero and x equals four these could potentially be adapted as well and By adapting them can we turn can we change the bounds to be Into the variable x instead because if I can you sort of use the following analogy as we're looking for areas under the curve Right, we've been mostly focused with this approach where we're trying to find the area into the curve using rectangles We use rectangles to find the area of the curve and by taking you know Sufficiently small rectangles. We can do a pretty good job with that. But why do we have to use rectangles? Why couldn't we not use something that maybe fits the curve a little bit better? And we've seen this example where using trapezoids Can actually be a much better approach To finding the area under the curve That is one of there's a slant to it Uh, that is if we switch up the shape we don't have to use rectangles at time And this u substitution Technique in some regards the change of the variable is like changing the shape that the original problem started off with Rectangles, but why not switch to trapezoids or why not switch to circles or parabolas things that might fit better under the curve? and so Because of this relationship u equals 25 minus x squared we could think of this as a function relationship What happens when x equals five? What happens when x equals zero well when x equals five and we plug this into the function u would equal 25 minus Five squared Which if we simplify that we end up with 25 minus 25 which equals zero And then likewise what is u when x is zero so u minus u equals 25 minus zero squared here This would be 25 minus zero which is equal to 25 And so if we approach the integral with this perspective the original integral Negative one half the integral from zero to five Of negative two x times the square root of 25 x squared dx We could switch the limits of the integral when we switch the variable So the negative two x dx that just becomes a du and the square root of 25 minus x squared That becomes u to the one half power But then if we change the limits when u equals Sorry when x equals zero u will equal 25 and when x equals five u will equal Zero and so we can change the balance and so now we have a simpler A simpler definite integral here Because after all if we're looking for the integral the definite integral This is the area into the curve the area of the curve is not a function It is a number we're trying to find a number And the fundamental theorem of calculus says we can find an antiderivative to help us find that number But our goal is not to find the definite integral the goal is to find this number Now here if you don't like that the 25 is on the bottom you can switch the order This is something we do all the time But if you switch the order of the limits you have to negate it And that's actually great because we have this negative sign that's already in front So it's like oh, that's a win-win situation for me So you're going to get one half the integral from zero to 25 Of u to the one half du And then the antiderivative like we kind of similar to what we saw before the antiderivative You'll take one to the power so that gives you three halves. You'll times it by two-thirds There was already a two right there. There was a one-half in front. Sorry, so When you take one half times two-thirds you end up with one-third And then you evaluate this from zero to 25 Plug it in zero will give you zero plug in in 25 will give you 125 so you have one-third 25 To the two-thirds. Sorry three halves Minus is zero to the three halves and you end up and again with the exact same number 125 over three So the technique doesn't change. I should say the answer doesn't change based upon the technique But we have these two different approaches We can find the antiderivative in terms of x and then plug those in in terms of x or We can turn away from x face towards this new variable u and then solve the problem with you Either approach is quite successful And in fact, I'm a big fan of this second approach as as a person who somehow missed this the first time I saw it probably I probably just didn't understand what my professor was talking about the first time I saw this type of problem in cacus and so these numbers sort of mysteriously changed from zero to five to 25 and zero and so for A good long time I would do the first approach and then like in cacus three I finally figured out what my professor was doing Honestly, I should was the goober who should just ask the question But I was too embarrassed to do so don't don't be like me right you should if you have a question You should ask ask right Either if you're in a class ask your question If you're watching this video post the comment ask your question So you don't get trapped in some misinformation for a long time But once I finally figured out how this this change of limits worked out. I love this new one over here. It's my favorite It's so nice to switch these variables and the more and more we do with these things The more and more you're gonna see that switching these variables is a good friend of ours Let's like a look at another one of these let's Integrate from one to e the natural log of x over x dx and you might wonder why would e be my upper bound But for this function that's actually quite generous Now in this situation, we're looking for a function whose derivative is present in the integrand And we might be interested in selecting the bottom But if you just set u equal to x that really doesn't do you a lick of good here Instead, let's actually choose u to be the natural log of x If we take u to be the natural log of x then its derivative is equal to 1 over x dx And you'll see that's exactly what we have in this integral We can take the integral here the natural log becomes a u and then the dx over x becomes a du So we're almost there except there are these bounds here. How did the limits change? Well, if x equals e or x equals 1 Well when x equals e you're going to get u equals the natural log of e a k 1 And if x equals 1 then u will equal the natural log of 1 a k a 0 And so for our function here our lower bound will be u equals 0 and the upper bound will be u equals 1 So in the original coordinate system you had 1 and e e was sort of obnoxious But when you switch from the x coordinates to the natural log of x coordinates a k a u coordinates This actually works out really nicely Anti-derivative of u becomes u squared over 2 Plug in 0 and 1 you end up with 1 squared over 2 minus the 0 squared over 2 And you end up with the final answer of 1 half, which is pretty slick changing the limits It's something you want to do As you're working with Anti-derivatives that might involve a u substitution If you're doing a definite integral and you have to use u substitution to find the anti-derivative I say just switch to u coordinates and never ever turn back To x don't be like lots wife and look back and turn into salt. You don't want to do that Instead look forward switch to your new variables and just go from there Bye everyone. We'll talk some more about this next time. See you