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Welcome back.

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I'm just continuing on with hopefully giving you, one, how

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to actually solve indefinite integrals and also giving you a

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sense of why you solve it the way you do.

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And I think that's often missing in some textbooks.

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But anyway, let's say that this is the distance and let me give

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you a formula, actually, for the distance, just for fun.

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Oh, my phone is ringing.

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Let me lower the volume, because you're more important.

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So, let's say that the distance, s -- this time I'll

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write it as a function -- let's say the distance is -- I said

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it started at five, so let's say it's 2t -- let's say this

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is actually a cubic function.

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You're not only accelerating, your rate of acceleration

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is increasing.

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I think, actually, the rate of acceleration, if I'm not

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mistaken, is actually called jerk, but I might have

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to Wikipedia that.

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Let's say its 2t to the third plus 5.

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And let's say I wanted to know how far I travel between t

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equals 2 seconds and t equals 5 seconds.

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Right?

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I'm not looking for the total distance I've traveled.

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I just want to know how far do I travel between time equals 2

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seconds and time equal 5 seconds.

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Right?

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So this might be 2 and this is 5.

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So an easy way to do that is I could just evaluate this

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function at t equals 5 -- let me use a different color.

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I think it's getting messy -- I could just evaluate this

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function at t equals 5.

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If t equals 5, 5 to the third power is 125, 250, it's 255,

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so the object has gone 255 feet at t equals 5, right?

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And then, at time equals 2, the object has gone how far?

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2 to the third is 8.

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16.

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It's gone 21 feet.

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Right?

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To figure out how far I travelled between time equals 2

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and time equals 5, I just say s of 5 minus s of 2, right?

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How far did I go after 5 seconds minus how far I

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already was after 2 seconds.

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And this is just 255 minus 21 and that's, what, 234.

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234 feet is how far I travelled between 2

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seconds and 5 seconds.

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Interesting.

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And I think you're starting to get a little intuition about

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why we evaluated that previous indefinite integral in the

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previous video the way we did.

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So let's actually draw the derivative of this function.

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So what's the derivative?

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So let me just call that v of t, I guess.

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v of t is just the derivative, right, because it's the rate

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of change of distance with respect to time.

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3 times 2 is 6t squared and the constant disappears, right?

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So it's just 6t squared.

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And that makes sense, right?

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Because your velocity doesn't care about where you

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started off from, right?

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You're going to be the same velocity if you started

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from 10 feet or if you started from 2 feet.

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Your velocity doesn't really matter about where your

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starting position is.

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So let's graph this.

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See?

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You're actually learning a little physics while

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you're learning calculus.

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Actually, I think it's silly that they're taught as

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two separate classes.

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I think physics and calculus should just be one

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fun 2-hour class.

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But I'll talk about that at another time.

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So, going back to this.

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Let me graph that.

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6t squared.

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Well, that's just going to look like a parabola.

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Right?

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It's going to look something like this.

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This is t.

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This is the velocity.

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And now, if we just had this velocity graph, if we didn't

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know all of this over here and I asked you the

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same question, though.

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I said, how far does this thing travel between 2

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seconds and 5 seconds?

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Right?

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Well, I could do it the way that we learned in the previous

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video where I draw a bunch of small rectangles, each of a

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really small width, and I multiply it times its

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instantaneous velocity at that exact moment, right?

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And then I sum up all of those rectangles -- look how pretty

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that is -- I sum up all of the rectangles.

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And I'll get a pretty good approximation for how far

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I've travelled between 2 and 5 seconds.

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Because remember, the area of each of these rectangles

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represents how far I traveled in that little

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amount of time, dt.

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Because time times a constant velocity is equal to distance.

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But as you can see this also tells me the area between

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t equals 2 and t equals 5.

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So, not only did I figure out the distance between how far I

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traveled from 2 seconds to 5 seconds, I also figured out the

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area under this curve from 2 seconds to 5 seconds.

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So, interestingly enough, if I just changed this from a to b,

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and, in general, if you want to figure out the area under a

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curve from a to b, it's just the indefinite integral from a

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to b -- actually, from b to a.

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The b should be the larger one.

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b to a.

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I guess a to b, depending on how you say it.

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Let me write that in a different color because I

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think I'm making it messier.

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From a to b of this velocity function.

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So, in this case, 6t squared d t, right?

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If these weren't 2 and 5, if this was just a and b.

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And the way you evaluate this is you figure out the

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antiderivative of this inside function, and then you evaluate

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the antiderivative at b, and then from that, you

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subtract it out at a.

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So in this case, the antiderivative of this is 2t to

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the third and we evaluated at b, and we evaluated at a.

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Actually, let me stick to the old numbers.

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We evaluated it at 5 and you evaluated it at 2.

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So if you evaluated it at 5, that's 255.

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If you evaluate it at 2, that's 21.

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So you're doing the exact same thing we did here when we

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actually had this graph.

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So I did all of this, not to confuse you further, but really

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just to give you an intuition of why one, why the

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antiderivative is the area under the curve, and then two,

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why-- let's say that a, b-- and then why we evaluate

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it this way.

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You might see this in your books.

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This is just saying, if I want to figure out the area under a

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curve from a to b of f of x, that we figure out

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the antiderivative.

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This capital F is just the antiderivative.

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We just figure out the antiderivative and we evaluated

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at b and we evaluated at a, and then we subtract

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the difference.

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And that's what we did here, right?

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This is what we did here intuitively when we

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worked with distance.

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The derivative and the antiderivative don't only apply

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to distance and velocity.

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But I did this to give you an intuition of why this works

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and why the antiderivative is the area under a curve.

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So let me clear this up and just rewrite that last thing

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I wrote, but maybe a little bit cleaner.

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OK.

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So let's say that F of x with a big, fat capital F is equal to

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-- actually, let me do it a better way -- let me say that

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the derivative of big fat F of x is equal to f of x.

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Right?

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I think, actually, this is the fundamental theorem of

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calculus, but I don't want to throw out things without

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knowing for sure.

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I have to go make sure.

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See, I haven't done math in a long time.

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I'm giving you all this based on intuition, not necessarily

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what I'm reading.

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So the derivative of big F is small f, and all we're saying

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is that if we take the integral of small f of x from a to b,

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dx, that this is big F, it's antiderivative, at b minus

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the antiderivative at a.

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In the next presentation, I'll use this.

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This is actually pretty easy to use once you know how

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to use antiderivatives.

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And we did these three videos really just to give you -- or

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actually, is this the third or the second -- just to give you

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an intuition of why this is, because I think that's really

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important if you're ever going to really use calculus in your

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life or write a computer program or whatever.

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And in the next couple videos I'll actually apply this to a

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bunch of problems and you'll hopefully see that it's a

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pretty straightforward thing to actually compute.

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I'll see you in the next presentation.

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