Area between curves with multiple boundariesYouTube↗  IA↗(long shot)
Education  264.8K views 389 likes 17 dislikes   28 Jan 2013
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Thai Tee Ponsukcharoen
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What I want to do in this video is find the area of this region that I'm shading in yellow.
V tomto videu chci zjistit plochu této oblasti, kterou stínuji žlutě.
And what might seem challenging is that throughout this region, I have the same lower function.
Výzva je v tom, že oblast je ohraničena funkcí pod osou x.
Or I guess the lower boundary is y is equal to x squared over 4 minus 1.
Funkce níže je y se rovná (x na druhou lomeno 4) minus 1.
But I have a different upper boundary.
Ale mám jinou horní hranici.
And the way that we can tackle this is by dividing this area into two sections,
Způsob, jak se s tím vypořádáme, je, že tuto plochu rozdělíme na dvě sekce
or dividing this region into two regions, the region on the left
neboli rozdělíme tuto oblast na dvě oblasti,
and the region on the right, where
čili na oblast vlevo a oblast vpravo,
for this first region, which I'll do--
kde pro první oblast, kterou...
I'll color even more in yellow-- for this first region,
...vybarvím ji žlutě... první oblast celého intervalu x.
over that entire interval in x. And it looks like x is going between 0 and 1.
A vypadá to, že ‚x‛ je mezi 0 a 1.
y equals-- when x is equal to 1, this function is equal to 1.
y se rovná… Když je x rovno 1, tak je tahle funkce rovna 1,
When x is equal to 1, this function is also equal to 1.
a když je x rovno 1, tak je i tahle funkce rovna 1.
So this is the point 1 comma 1.
Toto je bod [1; 1]. Tady se protínají.
That's where they intersect. So for this section, this subregion right over here,
V této sekci, v této podoblasti,
y equals square root of x is the upper function the entire time.
je ,y rovno odmocnina z x' celou dobu horní funkcí.
And then we can have a-- we can set up
A pak můžeme...můžeme nastavit různé...
a different-- we can separately tackle figuring out the area of this region.
můžeme odděleně vyřešit plochu této oblasti.
From x is equal to 1 to x is equal to 2,
Od ‚x‛ je rovno 1 do ‚x‛ je rovno 2,
where y equals 2 minus x, is the upper function.
kde y rovno (2 minus x) je horní funkcí.
So let's do it.
Udělejme to.
So let's first think about this first region.
Nejprve se zamysleme nad touto oblastí.
Well, that's going to be the definite integral from x is equal to 0 to x is equal to 1.
Bude to určitý integrál od ‚x‛ rovno 0 do ‚x‛ rovno 1.
And our upper function is square root of x, so square root of x.
A naše horní funkce je odmocnina z x.
And then from that, we want to subtract our lower function--
A od tohoto chceme odečíst naši spodní funkci...
square root of x minus x squared over 4 minus 1.
...(odmocnina z x) minus ((x na druhou lomeno 4) minus 1).
And then of course, we have our dx.
A samozřejmě máme naše ‚dx‛.
So this right over here, this is describing the area in yellow.
Toto zde popisuje tu žlutě vybarvenou plochu.
And you could imagine it, that this part right over here, the difference between these two functions is essentially this height.
A můžete si představit, že tato část zde, čili rozdíl obou funkcí, je výška.
Let me do it in a different color.
Vybarvím to jinou barvou.
And then you multiply it times dx.
A pak to vynásobte krát ‚dx‛.
You get a little rectangle with width dx. And then you do that for each x.
Dostanete malý obdélník široký ‚dx‛. A uděláte to pro každé ‚x‛.
Each x you get a different rectangle. And then you sum them all up.
Pro každé ‚x‛ dostanete jiný obdélník. A pak je všechny sečtete.
And you take the limit as your change in x approaches 0.
A hledáme limitu, kdy se ‚změna x‛ bude blížit 0.
So as you get ultra, ultra thin rectangles,
Když dostanete ultra tenké obdélníky, budete jich mít nekonečně mnoho.
and you have an infinite number of them. And that's our definition, or the Riemann definition of what a definite integral is.
A to je naše, respektive Riemannova, definice toho, co je určitý integrál.
And so this is the area of the left region.
A toto je plocha levé oblasti.
And by the exact same logic, we could figure out
Stejnou logikou můžeme najít plochu oblasti vpravo.
the area of the right region. The right region-- and then we could just
Oblast vpravo... ...a pak můžeme obě oblasti sečíst.
sum the two things together. The right region, we're going from x is equal to 0
Oblast vpravo od ‚x‛ je rovno 0 do ‚x‛...
to x-- sorry, x is equal to 1 to x is equal to 2, 1 to 2.
...promiňte, od ‚x‛ rovno 1 do ‚x‛ rovno 2. Čili od 1 do 2.
The upper function is 2 minus x.
Vrchní funkce je 2 minus x.
And from that, we're going to subtract the lower function, which is x squared over 4 minus 1.
A od tohoto odečteme dolní funkci, což je (x na druhou lomeno 4) minus 1.
And now we just have to evaluate.
A nyní to vypočítáme.
So let's first simplify this right over here.
Nejprve zjednodušíme tohle vpravo.
This is equal to the definite integral from 0 to 1 of square root of x minus x squared over 4 plus 1,
Je to rovno určitému integrálu od 0 do 1 funkce (odmocnina z x) minus ((x na druhou lomeno 4) plus 1) dx...
dx-- I'm going to write it all in one color now--
...nyní to napíši jednou barvou...
plus the definite integral from 1 to 2 of 2 minus x, minus x squared over 4.
...plus určitý integrál od 1 do 2 funkce 2 minus x minus (x na druhou lomeno 4).
Then subtracting a negative is a positive 3-- or a positive 1.
Pak odečteme...minus -1 je kladná 3...teda 1.
We could just add it to this 2. And so this 2 just becomes a 3.
Můžeme to přičíst k této 2. A z této 2 se stane 3.
I said 2 minus negative 1 is 3, dx.
Řekl jsem, že 2 minus -1 je 3…dx.
And now we just have to take the antiderivative and evaluate it at 1 and 0.
A nyní musíme vzít primitivní funkci a vyhodnotit ji pro 1 a 0.
So the antiderivative of this is-- well,
Primitivní funkce tohoto je... ...Tohle je x na 1/2.
this is x to the 1/2.
Přírůstek o 1.
Increment it by 1. Increment the power by 1, you get x to the 3/2,
Mocnina se zvýší o 1, takže dostanete x na 3/2
and then multiply by the reciprocal of the new exponent-- so it's 2/3 x to the 3/2.
a vynásobíte to obrácenou hodnotou nového exponentu...tudíž je to 2/3 x na 3/2.
Minus-- the antiderivative of x squared over 4
Minus...primitivní funkce x na druhou lomeno 4
is x to the third, divided by 3, divided by 4, so divided by 12,
je x na třetí děleno 3, to celé děleno 4, takže děleno 12…plus x.
plus x. That's the antiderivative of 1.
To je primitivní funkce 1.
We're going to evaluate it at 1 and 0.
Vyhodnotíme to v 1 a 0.
And then here the antiderivative is
A zde bude primitivní funkce
going to be 3x minus x squared over 2 minus x to the third over 12.
3x minus (x na druhou lomeno 2) minus (x na třetí lomeno 12).
Once again, evaluate it at-- or not once again.
Znovu to vyhodnotíme...vlastně ne.
Now we're going to evaluate at 2 and 1.
Nyní to vyhodnotíme v 2 a 1.
So over here, you evaluate all of this stuff at 1. You get 2/3 minus 1/12 plus 1.
Zde to všechno vyhodnotíme pro 1. Dostanete 2/3 minus 1/12 plus 1.
And then from that, you subtract this evaluated at 0.
A od tohoto odečtete tuto hodnotu v 0.
But this is just all 0, so you get nothing.
Ale je to jen 0, takže nedostanete nic.
So this is what the yellow stuff simplified to.
Takže to zjednoduší žlutou plochu.
And then this purple stuff, or this magenta stuff, or mauve, or whatever color this is, first you evaluate it at 2.
A tato fialová nebo purpurová nebo lila nebo co to je za barvu... ...nejprve ji vyhodnotíme v 2.
You get 6 minus-- let's see, 2 squared over 2 is 2, minus 8 over 12.
Dostanete 6 minus...podívejte, 2 na druhou lomeno 2 je 2...minus 8 lomeno 12.
And then from that, you're going to subtract
A od tohoto odečtete hodnotu v 1.
this evaluated at 1. So it's going to be 3 times 1-- that's 3-- minus 1/2 minus 1
Takže to bude 3 krát 1...to je 3... ...minus 1/2 minus (1 lomeno 12).
over 12. And now what we're essentially left with
A nyní nám zůstalo několik zlomků.
is adding a bunch of fractions.
Podívejme se, jak to zvládneme.
So let's see if we can do that. It looks like 12 would be the most obvious
Vypadá to, že 12 by mohl být společný jmenovatel.
common denominator. So here you have 8/12 minus 1/12 plus 12/12.
Zde máte 8/12 minus 1/12 plus 12/12.
So this simplifies to-- what's this?
Takže se to zjednodušilo...na co?
This is 19/12, the part that we have in yellow.
Tato žlutá část je 19/12.
And then this business, let me do it in this color.
A pak toto...udělám to jinou barvou.
So 6 minus 2, this is just going to be 4.
6 minus 2, to bude 4.
So we can write this as 48/12-- that's 4-- minus 8/12.
Můžeme to napsat jako 48/12... ...čili 4...minus 8/12.
And then we're going to have to subtract a 3, which is 36/12.
A pak odečteme 3, což je 36/12.
Then we're going to add to 1/2, which is just plus 6/12, and then we're going to add a 1/12.
A pak přičteme 1/2, což je plus 6/12, a pak přidáme 1/12.
So this is all going to simplify to-- let's see, 48 minus 8 is 40, minus 36 is 4, plus 6 is 10, plus 1 is 11.
Toto vše se zjednoduší... ...podívejte...48 minus 8 je 40... minus 36 je 4... ...plus 6 je 10...plus 1 je 11.
So this becomes plus 11/12.
Takže dostaneme 11/12.
Let me make sure I did that right.
Zkontroluji to.
48 minus 8 is 40, minus 36 is 4, 10, 11.
48 minus 8 je 40, minus 36 je 4...10...11.
So that looks right.
Vypadá to správně.
And then we're ready to add these two.
A jsme připravení tyto dva sečíst.
19 plus 11 is equal to 30/12.
19 plus 11 je rovno 30/12.
Or if we want to simplify this a little bit, we can divide the numerator and the denominator by 6.
Pokud to chceme trochu zjednodušit, můžeme vydělit čitatele a jmenovatele 6.
This is equal to 5/2, or 2 and 1/2.
Toto je rovno 5/2 nebo 2 a 1/2.
And we're done.
Jsme hotovi.
We figured out the area of this entire region. It is 2 and 1/2.
Zjistili jsme, že plocha celé oblasti je 2 a 1/2.
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