Name: 積分。Crash Course Physics #3 (Integrals: Crash Course Physics #3)
Uploaded: 2021-01-14T08:01:28.000Z
Duration: 10 min 9 s
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程度副詞

If you watched our last episode --
and really, if you haven't, you should.

...you now know all about derivatives, and how to use them, to describe the way an equation is changing.

情態副詞

Which means that now we can talk about the
other main part of calculus -- basically,

the inverse of derivatives, called integrals.

Integrals are useful because they also tell
you a lot about an equation: if you plotted

an equation on a graph, the integral is equal
to the area between the curve and the horizontal axis.

Finding an integral is a little less straightforward
than finding a derivative, but, as with derivatives,

there are shortcuts we can use to make things
a little easier.

We'll even be able to use integrals to talk
about how things move -- specifically, the

equation we've been calling the displacement
curve, and why it looks the way it does.

Say you want to know how high your bedroom
window is above the ground outside below.

But you don't have anything to measure it with --

just a ball, the stopwatch app on your phone, and your impressive knowledge of physics.

The force of gravity is what makes the ball fall,

so you know that its acceleration is small g
9.81 ms^2, downward.

But you're trying to find its change in
position -- how FAR it falls.

We've spent a lot of time talking about
the connection between the qualities of motion:

But so far, we've been describing that connection in a particular direction: velocity is the derivative of position

-- that is, a measure of its change --
and acceleration is the derivative of velocity.

To figure out how far the ball falls, you
need to use the reverse of that connection.

Expressed mathematically, that means that
velocity is the integral of acceleration,

and position is the integral of velocity.

In other words: if you draw these equation
on a graph, velocity is equal to the area

under the acceleration curve, and position
is equal to the area under the velocity curve.

There are simple ways to find the area of
pretty much any shape, as long as that shape

is made of nothing but straight lines and
corners.

And when you think about it, a curve is just
a set of infinitely tiny lines.

So the area under a curve can be divided into
a set of infinitely tiny rectangles.

Integrals tell you what happens when you divide
the area under a curve into those infinitely

tiny rectangles, take the area of each of
them, and add up those areas.

Well, you start by using the fact that integrals
are basically the OPPOSITE of derivatives.

If you know that your velocity is equal to twice time, for example, then you know that's the derivative of the position.

So, to find the equation for your position,
you just have to look for an equation whose DERIVATIVE is 2t … like x = t^2, for example.

It's kind of a roundabout way of doing things, compared to the neat equation we were able to use to find derivatives.

But there is no tidy equation that we can
use to calculate any integral we want.

But! As with derivatives, there ARE shortcuts
for finding certain, useful ones.

For instance, you can take the power rule
that we used for derivatives, and run it backward.

Basically: you add one to the exponent, then
divide the variable by that number.

So the integral of 2t -- which is written like this -- becomes t^2.

In the same way, the integral of 42t^5 is 7t^6.

You can take the trigonometric derivatives
that we talked about, and do those backward, too.

The integral of cos(x) is sin(x), and
so on. And the integral of e^x is just e^x.

But there's one complication that we haven't talked about yet -- maybe you've already spotted it: constants.

A constant is basically just a number. It can LITERALLY be a number -- like 2, or a half, or negative 4.

Or it can be a placeholder for a number, like the small g, we've been using to represent the acceleration caused
by gravity.

And constants pose a problem when it comes to integrals because: the derivative of a constant is just 0.

A derivative is a rate of change, after all,
so a constant, which by definition DOESN'T

change, will always have a derivative of zero.

Which means that lots of different equations - an infinite number, in fact - can all have the same derivative.

Like, the derivative of t^2 is 2t. But you
can add ANY number -- or a letter representing

a number -- to it, and the derivative will
STILL be 2t. So the derivative of t^2 + 1

is also 2t. And the same is true for t^2 - 7.

Which means: If you're looking for the INTERGRAL
of an equation like x = 2t, you have INFINITE

CHOICES, all of which are equally correct.

t^2 would work, but so would t^2 + 1 … or
t^2 - 7 … or t^2 + 0.256.

In these cases, we might know what the SHAPE
of the integral should look like on a graph

-- like, whether it's a straight line, or how it curves -- but we don't know where to put it along the vertical axis.

So we need to know what the constant is, in
order to know where to start drawing its integral.

Whatever the constant is equal to, that's
where the curve will intersect with the vertical axis.

So t^2 would intersect at 0, but t^2 - 7
would intersect at -7. You get the idea.

Mathematicians had to figure out how to get
around the problem of having infinite integrals

to choose from, so they came up with a way to represent ALL of them: just add a C at the end of the integral.

That C stands for all of the constants that
we know we COULD put there.

So if we say that the integral of 2t is t^2+ C,

then we're including t^2 + 1 and t^2
- 7 and all those other infinite options --

But sometimes you don't need the C at all,
because you CAN figure out where your integral

is supposed to be on the y axis. Like if you
have what's known as the initial value.

In the case of a position graph, for instance,
the initial value would be where you started out,

so you'd know to draw the rest of
the graph's shape from there.

If you started at the 2 meter mark, say, and
moved one meter every second, you'd put

the graph here. But if you started at the
4 meter mark, you'd shift it up a little.

Basically, it gives you the point where your integral intersects the vertical axis -- which is the value of C.

Let's try it, and at the same time, we might
as well figure out the height of your bedroom window.

You're standing in your room, holding a
tennis ball out the window with your arm resting on the sill.

Now you drop the ball and start your
stopwatch app at the same time.

Turns out that the ball takes 1.7
seconds to hit the ground.

Like we said earlier, we know the ball's acceleration -- 9.81 ms^2 -- and we know the time involved.

Somehow, we have to get from
there to the equation for the ball's position.

So, first, let's find its velocity -- the
middle step -- by taking the integral of its acceleration.

Take a look at this graph of the ball's
acceleration over time.

It's just a flat line, which means that it should be pretty easy to find the area between it and the horizontal axis.

It's a rectangle! And the area of a rectangle
is just its base times its height.

In this case, the base is t, the amount of
time the ball took to fall.

So, the area between the acceleration graph and the horizontal axis is just (a) times (t).

And the integral is (a) times (t), plus that constant
we add -- C.

For now, we need the C, because we know the
general shape of the velocity graph:

It's a diagonal line slanted in such a way that every second, it rises by an amount equal to the acceleration.

But still we don't know where to PUT that
line on the vertical axis. Not yet, anyway.

Now, we could have figured out the integral
of acceleration just as easily by using the power rule:

The acceleration, a, is a constant, but we
could also say that it's (a) x (t^0) --

because anything raised to the power 0 is just 1.

So, according to the power rule, the integral
of acceleration -- which is the velocity -- would

be equal to the acceleration, multiplied by
time, plus C.