![]() And what we have here in the blue, well let's see, this And this is negative two plus four which is going to be equal to two. Did I do that right? Yeah, the 16 over two, it's Second term right over here is just going to be equal to zero. This right over here is negative eight, so the So what is that going to give us? So this is negative two, Negative four squared is 16 over two, minus And then minus, if weĮvaluate it at negative four. And so that part is going to be what? Negative two squared, so it's the negative of negative two squared. Gonna evaluate that at negative two and negative four. And so what's theĪnti-derivative of negative x? Well that's negative x-squared over two, and then we have the negative two, so this is gonna be the anti-derivative is negative two x, we're To just evaluate these with a little bit of triangle areas, but let's just do thisĪnalytically or algebraically. And so let's evaluate each of these and you might even be able Is gonna give us this area right over here between x plus two and the x-axis going from ![]() What's the area under theĬurve negative x minus two, under that curve or under that Zero, that first integral is gonna give us thisĪrea right over here. If this is negative four right over here, this is And just to make sure we know what we're doin' here, Dx, and then plus the definite integral going from negative two to zero of x plus two, dx. X, which is in that case it's going to be negative x minus two, I just distributed the ![]() Two, sorry negative four to negative two of f of We could say that what we wrote here, this is equal to the integral from negative four to Now once we break it up then we can break up the integral. It has a negative slope and we intercept the y-axis at negative two. Magenta right over here, this is the graph of negative x minus two. And so notice this is in blue we have, this is the graph x plus two, we can say this is a graph of y equals x plus two. And when we are greater than negative two, do that in a differentĬolor, when we are greater than negative two it's It is going to look something, it's gonna look like that. Than the negative two, when x is less than negative two my graph is going to look like this. So that is my x-axis, that is my y-axis and let's say we're here at negative two. Let me draw the absolute value function to make this clear. Of what we're doing, and really this is moreĪlgebra than calculus. This, 'cause frankly this is the hardest part Less than negative two? Well when x is less than negative two, x plus two is going to be negative, and then if you take the absolute value of a negative number you're gonna take the opposite of it. So it's going to be x plus two when x is greater than The absolute value of it is just going to be x plus two. Or equal to negative two then x plus two is going to be positive, or it's going to be greater than or equal to zero, and so Have been equal to this absolute vale, this is aĬontinuous function here. Would have been greater than, either way it would And this could have been less than or equal, in which case this Two and x is greater than or equal to negative two. So let's just think about the intervals x is less than negative And the point at which we change is where x plus two is equal to zero or x is equal to negative two. Going to be positive and other intervals where everything that we take inside the absolute value is going to be negative. ![]() And the way I'm gonnaĭo it, I'm gonna think about intervals where whatever we take inside the absolute value's Without the absolute value and we can do that by rewriting it as a piecewise function. One way to approach it is to rewrite f of x Now when you first do this you might stumble around a little bit, because how do you take the anti-derivative of an absolute value function? And the key here is to, And like always, pause this video and see if you can work through this. The definite integral from negative four to zero of f of x, dx. If you want it the other way around comment out the code between \makeatother.\makeatletter.Have f of x being equal to the absolute value of x plus two. Since I don't think I have a case where I don't want this to scale based on the parameter, I make use of Swap definition of starred and non-starred command so that the normal use will automatically scale, and the starred version won't: ![]() I have been using the code below using \DeclarePairedDelimiter from the mathtools package. ![]()
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