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Welcome back.

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I'm now going to use definite integrals to figure out the

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areas under a bunch of curves and, if we have time, maybe

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even between some curves.

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So let me right down the fundamental theorem

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of calculus.

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I know I covered it really fast in the last presentation.

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Just to make sure you understand this formula.

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The last couple presentations were really to give you an

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intuition for this exact formula.

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Let's say that big f prime of x is equal to f of x, right?

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That's also like saying that the -- that's equivalent to

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saying that f of x -- big f of x -- is equal to the

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antiderivative of f of x, right?

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Well, let's just that it's one of the possible antiderivatives

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of f of x, right?

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Because there's always a constant term here and you're

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not sure whether it is.

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And this is why people tend to use this standard because we

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know that f of x is the derivative of big f prime of x.

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Big f of x is just one of the antiderivatives of f of x.

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So this is a little bit not true, but I think

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you get the idea.

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But the fundamental theorem of calculus tells us if this top

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line is true, then the definite integral from a to b of f of x

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d x is equal to its antiderivative evaluated at b

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minus its antiderivative evaluated at a.

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And I know I said here that big f isn't the only

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antiderivative, right?

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Because you could at any constant to this and that would

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also be the antiderivative.

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But when you subtract here, the constants will cancel out.

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So it really doesn't matter which of the

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constants you pick.

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The constant actually doesn't matter.

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So that's why I actually said the antiderivative.

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But let's apply this.

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You might be confused right now.

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So let me draw a graph.

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There you go.

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Look how straight that is.

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Draw the x-axis.

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Not perfect but it'll do.

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Let's say that my f of x is equal to x squared plus 1.

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So f of x looks like this.

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This is 1.

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So it'll start at 1.

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It'll just be a parabola.

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Let me see how good I can draw this.

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I've done worse.

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OK.

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So that's f of x.

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It's a parabola. y-intercept at 1.

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And let's say I want to figure out the area under the curve --

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between the curve and, really, the x-axis.

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Let's say I want to figure out the area between the curve and

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the x-axis from x equals negative 1 to, I don't

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know, x equals 3.

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So this is the area I want to figure out.

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I'm going to shade it in.

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So this is the area.

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All of this stuff.

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I want to figure out this area.

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And you could imagine, before you knew calculus, figuring out

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an area of something with a curve -- it's kind

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of top boundary.

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It would have been very difficult.

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But we will now use the fundamental theorem of calculus

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and hopefully you have an intuition of why this works and

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how the integral is really just a sum of a bunch of little,

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little, small squares with infinitely small bases.

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But if you watched the last videos, hopefully that

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hit the point home.

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But now we'll just mechanically compute because, actually,

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understanding it is a bit harder than just doing it.

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But let's just mechanically compute it.

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So we are essentially just going to figure out the

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integral from minus 1 to 3 of f of x, which is x

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squared plus 1 d x.

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What's the antiderivative of x squared plus 1?

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This just equals the antiderivative.

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So it's just x to the third -- we could say 1/3 x to the third

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or x to the third over 3 -- plus x, right?

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The derivative of x is 1.

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And then we don't have to worry about plus c because we're

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going to subtract out the c's.

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You'll see.

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I think you'll get the point.

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It doesn't matter.

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You could pick an arbitrary c right here and it'll

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just cancel out.

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And we're going to evaluate that at 3 and negative 1 and

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we're going to subtract out big f of negative

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1 from big f of 3.

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This is just the notation they use.

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You figure out the antiderivative and

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you say where you're going to evaluate it.

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And then this is equal to -- So if I evaluate 3.

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3 to the third power is what?

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That's 27.

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27 divided by 3 is 9.

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And then 9 plus 3 is 12.

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Right?

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This is just big f of 3, right?

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Because I figured out the end -- This is big f of x.

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You can kind of view this as big f of x.

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But not to be confused with small, cursive f of x.

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This is big f of x.

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So this big f of 3.

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And then, from that we'll subtract big f

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of negative 1, right?

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Minus big f of negative 1.

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And if we put minus 1 here.

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Let's see, minus 1 to the third power is minus 1.

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So it's minus 1/3 and then plus minus 1, right?

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So minus 1/3 plus minus 1.

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I think that equals minus 4/3, correct?

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I think so.

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Maybe I'm making a mistake with negative signs.

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Minus 1/3 minus 4/3.

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And I'm going to subtract that, right?

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So if I'm subtracting minus 4/3, it's the same thing

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as adding minus 4/3.

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And then we have our answer.

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Actually, it's 12 and 4/3 -- whatever -- units.

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Squared units.

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12 and 4/3 squared units.

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We could write this as a mixed number as well.

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Let's do another one.

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I'll do a slight variation.

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OK.

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Let me draw again.

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Some coordinates.

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I don't know if I'm going to have time to do it in

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this video but I'll try.

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I always try.

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Let's say I have f of x is equal to the square root of x.

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So it looks something like this.

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That's actually a pretty nice looking, kind of sideways

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parabola, I think.

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This is f of x.

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And let's say I have another function.

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g of x which equals x squared.

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So g of x is actually going to look something like this.

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Whoops!

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I was doing well and then something happened.

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And, of course, it'll continue on this side as well.

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Because it is defined for negative numbers.

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But anyway, my question to you, or my question to myself,

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really, is what is the area between the curves

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where they intersect?

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What is this?

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What is this area?

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Well, the first thing you have to figure out is just what

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are the boundary points?

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What is this point?

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And what is this point?

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Well this point, I think, is pretty clear.

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It's 0, 0, right?

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They both intersect at 0, 0.

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And even this point, you could probably do it from intuition.

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But if you don't, I guess, want to do it through intuition, you

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could just set these 2 equations equal to

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each other, right?

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You could say x squared is equal to the square

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root of x, right?

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And then you could do a bunch of things.

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You could square both sides or -- well, actually, this is the

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same thing as doing it by intuition.

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But I think it's pretty obvious that the only places where x

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squared is equal to the square root of x are the points x

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equals 0, which you already know, and x equals 1.

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So this is the point 1, 1.

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Which is true for both of them.

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And this is more algebra, so I won't go into that

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in too much detail.

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I'm kind of running out of time.

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So we want to figure out the area between these 2 curves.

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So what we can do is -- maybe you want to pause it and think

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about it yourself -- we can figure out the area

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under the grey curve.

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We could figure out this area.

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So we want to figure out -- this is a boundary, right?

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Between 0 and 1.

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We could figure out this area and then we could figure out

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the entire area under the green curve separately and then we

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could subtract the difference.

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Which is exactly how we're going to do it in the next

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video because I have run out of time.

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