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Welcome back.

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Where I had just left off we were trying to figure out this

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area between these two curves, and we figured out that it's

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really the area between the curves between the point 0,

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x equals 0 and x equals 1.

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And I was proposing of a way to do it.

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Let's figure out the entire area under the square root of

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x from 0 to 1, and we can subtract from that, this

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purple area, which is the area under x squared.

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So just based on the last example we did, we could just

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write the indefinite integral, and I'm not going to rewrite

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the fundamental theorem from calculus, because I think

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you know that by now.

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Let me do it in a loud color.

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Magenta.

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So I want to know the larger area, right.

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The area under just the square root of x.

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That's Kind of like the combined area.

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Well that's just from 0 to 1, the integral of square root of

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x, because square root of x is the green function, dx.

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And I want to subtract from that the area from 0 to 1

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what's under x squared.

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x squared dx.

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And I just want to make a point.

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This could have just been rewritten this way.

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You could just rewrite a new function, which is the

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difference of these two functions, and it would

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have been equivalent.

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You could have said this.

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This isn't kind of a step of the problem, but you could

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have done it this way.

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In fact, some people start this way.

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See those are the same thing as the integral from 0 to 1

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of square root of x minus x squared dx.

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So you could do this two separate problems, two separate

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indefinite integrals, or you could do it as one

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indefinite integral.

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Actually that might be even simpler because when you

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evaluate it from 1 to 0 it simplify things a little bit.

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So let's stick with the second one.

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So first of all we just have to figure out what the

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antiderivative of this inner expression is.

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So you haven't seen square root of x yet.

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Do you think you know how to do it?

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Well, I think you do.

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Let's say that equals square root of x is just x to

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the 1/2 power, right?

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So we just use the same antiderivative rules

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we've always used.

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We raise it one more power so it becomes x to the 3/2.

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Right, it was 1/2, we added 1 to it.

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And then we divide by this new exponent.

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So dividing by a fraction is like multiplying

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by its reciprocal.

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So it's 2/3 x to the 3/2 and then minus-- I think the second

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term is pretty easy for you-- minus 1/3 x to the third.

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That's the antiderivative of minus x squared, minus

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1/3 x to the third.

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And we're going to have to evaluate this thing at 0 and 1

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and subtract the difference.

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Subtract this expression evaluated at 0 from this

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expression evaluated at 1.

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I think I'm running out of space.

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What happens when x equals 1?

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1 to the 3/2 is 1.

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1 to the third is 1.

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So it's 2/3 minus 1/3.

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Well that's easy.

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It's 1/3.

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I just put 1 in for x.

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And then when x is equal to 0 what is this expression equal?

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Well that's easy too.

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That's 0.

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So there you go.

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1/3 minus 0 or 1/3.

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That's kind of neat.

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You know I find is this of exciting because if just my

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intuition I was like, oh I have these two curves, and I mean

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they do intersect at the nice integer number, but you know

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what it's probably going to be some really messy number

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of what the areas between these two curves, right.

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Who knows, maybe it'll involve some you know-- a circle

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involves pi, which is this really messy number, so maybe

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all curves have these kind of messy areas.

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But this one it's just one of those neat things about math.

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The area between the square root of x and x squared is 1/3,

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which is a pretty clean number.

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Actually let me do one more problem since I have time.

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It's a bit of a trick problem.

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I mean, you might actually find this easy, but let's figure out

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the area between f of x is equal to-- I don't

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know, x to the fifth.

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I'm going to do something simple.

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Let me draw it actually.

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OK, I'm going to draw the x-axis.

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x to the fifth is going to go up super fast,

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something like that.

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It's going to go up real fast.

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Let's say I wanted to figure out the areas-- this side's

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going to go real fast too-- between that, and instead of

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figuring out the area between x-axis and that, f of x, I want

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to figure out the area between f of x and-- Instead

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of figuring out this bottom area, right?

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Like the normal problems we've done, we would figure out this

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type of area, you know, between two points.

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Let's say I want to figure out the area inside of the curve

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where the height here is 32.

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So I want to figure out this area inside the curve.

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How do we do that?

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Well one way we could do it is just like we did the last one,

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we can figure out some function that's essentially a line, a

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horizontal line that goes straight across here.

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And we'll essentially just be figuring out the area

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between the two functions.

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So what's a function that's a line that just goes

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straight at y equals 32?

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I think I just gave you the answer.

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Exactly.

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Let me stick with that greenish color.

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So we could say g of x is equal to 32.

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It's just a constant function.

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It just goes straight across.

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And then we need to figure out what the area is between the

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two, so we need to figure out what are these two points.

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So when does x to the fifth equal 32?

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I mean you could solve it algebraically, you know,

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you could say x to the fifth equals 32.

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x is equal to 2, and actually, you know what?

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I made a mistake.

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Let's say that this is not equal to x to the fifth.

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Let's say that f of x is equal to x to the absolute

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value of x to the fifth.

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Because the mistake, obviously x to the fifth is not a

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parabola looking thing. x to the fifth goes

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negative like this.

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But I have committed so much to this cup shape that I'll make

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it the absolute value of x to the fifth.

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So if I say the absolute value of x to the fifth is equal to

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32, I think you see where I realized my mistake.

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But if I say the absolute value of x to the fifth is 32,

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there's two places where that's true.

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It's x is equal to plus or minus 2.

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These are the two points.

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I should have done something with an even exponent so I

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could have had this cup shape, but anyway the absolute

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value solved my problem.

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So what is this area?

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We know it's between negative 2 and 2, so we just set up

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the indefinite integral.

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It's the indefinite integral for minus 2 to 2 of the top

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function, the top boundary, 32, minus the bottom boundary.

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Well, this will be a little bit tricky, but minus the absolute

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value of x to the fifth dx.

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And actually instead of doing this, I think you could see

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that there's symmetry here, so we could just figure out

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this area and multiply by 2.

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This problem's a little hairy, just because I had a bad

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choice of initial function.

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Not exactly what I wanted, but we'll work on forward.

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So instead of doing that, let's do the integral from 0 to 2 of

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32 minus x to the fifth dx.

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And then multiply that by 2.

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So what is that?

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That's 32x minus x to the sixth over 6.

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And we're going to evaluate it from 2 and 0 at 64 minus 64/6.

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2 to the sixth is 64, and then 32 times 0 is 0 and the

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next is 6, that's 0, so the answer is 64 minus 64/6.

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I'm about to run out of time.

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That's just a fraction problem there.

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Oh, and that's half of it, right?

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So we want to multiply that by 2.

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So if we multiply that by 2, we get 128 minus 128/6.

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I haven't figured out what it is.

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Well I guess we could figure it out.

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It's 128 times 1 minus 1/6 or 128 times 5/6.

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And I don't know what that is.

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I can multiply if I wanted to, but I have 10 seconds left

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so I'll leave you there.

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Hope I didn't confuse you.

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See you soon.

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