Definite Integrals (area under a curve) (part III) More on why the antiderivative and the area under a curve are essentially the same thing.
Definite Integrals (area under a curve) (part III)
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- Welcome back.
- I'm just continuing on with hopefully giving you, one, how
- to actually solve indefinite integrals and also giving you a
- sense of why you solve it the way you do.
- And I think that's often missing in some textbooks.
- But anyway, let's say that this is the distance and let me give
- you a formula, actually, for the distance, just for fun.
- Oh, my phone is ringing.
- Let me lower the volume, because you're more important.
- So, let's say that the distance, s -- this time I'll
- write it as a function -- let's say the distance is -- I said
- it started at five, so let's say it's 2t -- let's say this
- is actually a cubic function.
- You're not only accelerating, your rate of acceleration
- is increasing.
- I think, actually, the rate of acceleration, if I'm not
- mistaken, is actually called jerk, but I might have
- to Wikipedia that.
- Let's say its 2t to the third plus 5.
- And let's say I wanted to know how far I travel between t
- equals 2 seconds and t equals 5 seconds.
- I'm not looking for the total distance I've traveled.
- I just want to know how far do I travel between time equals 2
- seconds and time equal 5 seconds.
- So this might be 2 and this is 5.
- So an easy way to do that is I could just evaluate this
- function at t equals 5 -- let me use a different color.
- I think it's getting messy -- I could just evaluate this
- function at t equals 5.
- If t equals 5, 5 to the third power is 125, 250, it's 255,
- so the object has gone 255 feet at t equals 5, right?
- And then, at time equals 2, the object has gone how far?
- 2 to the third is 8.
- It's gone 21 feet.
- To figure out how far I travelled between time equals 2
- and time equals 5, I just say s of 5 minus s of 2, right?
- How far did I go after 5 seconds minus how far I
- already was after 2 seconds.
- And this is just 255 minus 21 and that's, what, 234.
- 234 feet is how far I travelled between 2
- seconds and 5 seconds.
- And I think you're starting to get a little intuition about
- why we evaluated that previous indefinite integral in the
- previous video the way we did.
- So let's actually draw the derivative of this function.
- So what's the derivative?
- So let me just call that v of t, I guess.
- v of t is just the derivative, right, because it's the rate
- of change of distance with respect to time.
- 3 times 2 is 6t squared and the constant disappears, right?
- So it's just 6t squared.
- And that makes sense, right?
- Because your velocity doesn't care about where you
- started off from, right?
- You're going to be the same velocity if you started
- from 10 feet or if you started from 2 feet.
- Your velocity doesn't really matter about where your
- starting position is.
- So let's graph this.
- You're actually learning a little physics while
- you're learning calculus.
- Actually, I think it's silly that they're taught as
- two separate classes.
- I think physics and calculus should just be one
- fun 2-hour class.
- But I'll talk about that at another time.
- So, going back to this.
- Let me graph that.
- 6t squared.
- Well, that's just going to look like a parabola.
- It's going to look something like this.
- This is t.
- This is the velocity.
- And now, if we just had this velocity graph, if we didn't
- know all of this over here and I asked you the
- same question, though.
- I said, how far does this thing travel between 2
- seconds and 5 seconds?
- Well, I could do it the way that we learned in the previous
- video where I draw a bunch of small rectangles, each of a
- really small width, and I multiply it times its
- instantaneous velocity at that exact moment, right?
- And then I sum up all of those rectangles -- look how pretty
- that is -- I sum up all of the rectangles.
- And I'll get a pretty good approximation for how far
- I've travelled between 2 and 5 seconds.
- Because remember, the area of each of these rectangles
- represents how far I traveled in that little
- amount of time, dt.
- Because time times a constant velocity is equal to distance.
- But as you can see this also tells me the area between
- t equals 2 and t equals 5.
- So, not only did I figure out the distance between how far I
- traveled from 2 seconds to 5 seconds, I also figured out the
- area under this curve from 2 seconds to 5 seconds.
- So, interestingly enough, if I just changed this from a to b,
- and, in general, if you want to figure out the area under a
- curve from a to b, it's just the indefinite integral from a
- to b -- actually, from b to a.
- The b should be the larger one.
- b to a.
- I guess a to b, depending on how you say it.
- Let me write that in a different color because I
- think I'm making it messier.
- From a to b of this velocity function.
- So, in this case, 6t squared d t, right?
- If these weren't 2 and 5, if this was just a and b.
- And the way you evaluate this is you figure out the
- antiderivative of this inside function, and then you evaluate
- the antiderivative at b, and then from that, you
- subtract it out at a.
- So in this case, the antiderivative of this is 2t to
- the third and we evaluated at b, and we evaluated at a.
- Actually, let me stick to the old numbers.
- We evaluated it at 5 and you evaluated it at 2.
- So if you evaluated it at 5, that's 255.
- If you evaluate it at 2, that's 21.
- So you're doing the exact same thing we did here when we
- actually had this graph.
- So I did all of this, not to confuse you further, but really
- just to give you an intuition of why one, why the
- antiderivative is the area under the curve, and then two,
- why-- let's say that a, b-- and then why we evaluate
- it this way.
- You might see this in your books.
- This is just saying, if I want to figure out the area under a
- curve from a to b of f of x, that we figure out
- the antiderivative.
- This capital F is just the antiderivative.
- We just figure out the antiderivative and we evaluated
- at b and we evaluated at a, and then we subtract
- the difference.
- And that's what we did here, right?
- This is what we did here intuitively when we
- worked with distance.
- The derivative and the antiderivative don't only apply
- to distance and velocity.
- But I did this to give you an intuition of why this works
- and why the antiderivative is the area under a curve.
- So let me clear this up and just rewrite that last thing
- I wrote, but maybe a little bit cleaner.
- So let's say that F of x with a big, fat capital F is equal to
- -- actually, let me do it a better way -- let me say that
- the derivative of big fat F of x is equal to f of x.
- I think, actually, this is the fundamental theorem of
- calculus, but I don't want to throw out things without
- knowing for sure.
- I have to go make sure.
- See, I haven't done math in a long time.
- I'm giving you all this based on intuition, not necessarily
- what I'm reading.
- So the derivative of big F is small f, and all we're saying
- is that if we take the integral of small f of x from a to b,
- dx, that this is big F, it's antiderivative, at b minus
- the antiderivative at a.
- In the next presentation, I'll use this.
- This is actually pretty easy to use once you know how
- to use antiderivatives.
- And we did these three videos really just to give you -- or
- actually, is this the third or the second -- just to give you
- an intuition of why this is, because I think that's really
- important if you're ever going to really use calculus in your
- life or write a computer program or whatever.
- And in the next couple videos I'll actually apply this to a
- bunch of problems and you'll hopefully see that it's a
- pretty straightforward thing to actually compute.
- I'll see you in the next presentation.
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