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Μάθημα 17: Integration by parts- Integration by parts intro
- Integration by parts: ∫x⋅cos(x)dx
- Integration by parts: ∫ln(x)dx
- Integration by parts: ∫x²⋅𝑒ˣdx
- Integration by parts: ∫𝑒ˣ⋅cos(x)dx
- Integration by parts
- Ολοκλήρωση κατά παράγοντες: ορισμένα ολοκληρώματα
- Ολοκλήρωση κατά παράγοντες: ορισμένα ολοκληρώματα
- Integration by parts challenge
- Integration by parts review
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Integration by parts intro
By looking at the product rule for derivatives in reverse, we get a powerful integration tool. Δημιουργήθηκε από τον Σαλ Καν.
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What we're going
to do in this video is review the product
rule that you probably learned a while ago. And from that, we're
going to derive the formula for integration
by parts, which could really be viewed as the inverse product
rule, integration by parts. So let's say that I start
with some function that can be expressed
as the product f of x, can be expressed as a
product of two other functions, f of x times g of x. Now let's take the
derivative of this function, let's apply the derivative
operator right over here. And this, once again, just a
review of the product rule. It's going to be the derivative
of the first function times the second function. So it's going to be
f-- no, I'm going to do that blue color-- it's
going to be f-- that's not blue-- it's going to be f
prime of x times g of x times-- that's not the same
color-- times g of x plus the first function times
the derivative of the second, plus the first function, f
of x, times the derivative of the second. This is all a review
right over here. The derivative of
the first times the second function
plus the first function times the derivative
of the second function. Now, let's take
the antiderivative of both sides of this equation. Well if I take
the antiderivative of what I have here on the
left, I get f of x times g of x. We won't think about
the constant for now. We can ignore that for now. And that's going to
be equal to-- well what's the
antiderivative of this? This is going to be the
antiderivative of f prime of x times g of x dx
plus the antiderivative of f of x g prime of x dx. Now, what I want
to do is I'm going to solve for this
part right over here. And to solve for that, I just
have to subtract this business. I just have to subtract this
business from both sides. And then if I subtract
that from both sides, I'm left with f of x
times g of x minus this, minus the antiderivative
of f prime of x g of x-- let me do that in
a pink color-- g of x dx is equal to what I
wanted to solve for, is equal to the antiderivative
of f of x g prime of x dx. And to make it a
little bit clearer, let me swap sides here. So let me copy and paste this. So let me copy
and then paste it. There you go. And then let me copy and
paste the other side. So let me copy and paste it. So I'm just switching the
sides, just to give it in a form that you might be more used
to seeing in a calculus book. So this is essentially the
formula for integration by parts. I will square it off. You'll often see it squared
off in a traditional textbook. So I will do the same. So this right over here
tells us that if we have an integral or an
antiderivative of the form f of x times the derivative
of some other function, we can apply this
right over here. And you might say, well this
doesn't seem that useful. First I have to identify a
function that's like this. And then still I have
an integral in it. But what we'll see
in the next video is that this can
actually simplify a whole bunch of
things that you're trying to take the
antiderivative of.