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Μάθημα: Ολοκληρωτικός Λογισμός > Ενότητα 1
Μάθημα 10: Κανόνας αντίστροφης δύναμης- Κανόνας αντίστροφης δύναμης
- Κανόνας αντίστροφης δύναμης
- Κανόνας αντίστροφης δύναμης: αρνητικές και κλασματικές δυνάμεις
- Indefinite integrals: sums & multiples
- Κανόνας αντίστροφης δύναμης: αθροίσματα και πολλαπλάσια
- Rewriting before integrating
- Κανόνας αντίστροφης δύναμης: ξαναγράφοντας πριν από την ολοκλήρωση
- Rewriting before integrating: challenge problem
- Reverse power rule review
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Rewriting before integrating: challenge problem
In this example, we find the antiderivative an expression which is not so simple. Δημιουργήθηκε από τον Σαλ Καν.
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So our goal in this video is
to take the antiderivative of this fairly crazy
looking expression. Or another way of saying it is
to find the indefinite integral of this crazy
looking expression. And the key realization
right over here is that this expression is
made up of a bunch of terms. And the indefinite integral
of the entire expression is going to be equal to the
indefinite integral of each of the term. So this is going to be equal
to, we could look at this term right over here, and just
take the indefinite that, 7x to the third dx. And then from that,
we can subtract the indefinite
integral of this thing. So we could say
this is, and then minus the indefinite integral
of 5 times the square root of x dx. And then we can look at
this one right over here. So then we could say plus
the indefinite integral of 18 square root of x. Square roots of x,
over x to the third dx. And then finally, I'm
running out of colors here, finally I need more
colors in my thing. We can take the
antiderivative of this. So plus the antiderivative of x
to the negative 40th power dx. So I've just rewritten this
and color-coded things. So let's take the
antiderivative of each of these. And you'll see that we'll
be able to do it using our whatever we want to call it. The inverse of the power
rule, or the anti-power rule, whatever you might
want to call it. So let's look at the first one. So we have-- what
I'm going to do is, I'm just going to
find the antiderivative without the constant, and just
add the constant at the end. For the sake of this one. Just to make sure we get the
most general antiderivative. So here the exponent is a 3. So we can increase it by 1. So it's going to
be x to the 4th. Let me do that same purple
color, or pink color. It's going to be x
to the 4th, or we're going to divide by x to the 4th. So it's x to the 4th over 4
is the antiderivative of x to the 3rd. And you just had this scaling
quantity, the seven out front. So we can still just
have the seven out front. So we get 7x to the 4th over 4. Fair enough. From that, we're
going to subtract the antiderivative of this. Now at first this
might not be obvious, that you could use our
inverse power rule, or anti-power rule here. But then you just
need to realize that 5 times the
principal root of x is the same thing as 5
times x to the 1/2 power. And so once again, the
exponent here is 1/2. We can increment it by 1. So it's going to
be x to the 3/2. And then divide by the
incremented exponent. So divide by 3/2. And of course we had
this 5 out front, so we still want to
have the 5 out front. Now this next expression
looks even wackier. But once again, we can
simplify a little bit. This is the same thing-- let
me do it right over here-- this is the same thing as
18 times x to the 1/2 times x to the negative 3 power. x to the 3rd in the
denominator is the same thing as x to the negative 3. We have the same base, we
could just add the exponents. So this is going to be equal
to 18 times x to the 2 and 1/2 power. Or another way of
thinking about it, this is the same thing as
18 times x to the 5/2 power. Did I do that right? Yeah. Negative 3. Oh sorry, this is
negative 2 and 1/2. Let me make this very clear. And this is going to be
the negative 5/2 power. x to the negative 3 is the same
thing as x to the negative 6/2. Negative 6/2 plus
1/2 is negative 5/2. So once again, we just have
to increment this exponent. So negative 5/2 plus 1 is
going to be negative 3/2. So you're going to have
x to the negative 3/2. And then you divide
by what your exponent is when you increment it. So divide it by negative 3/2. And then you had
the 18 out front. And we obviously are going
to have to simplify this. And then finally, our
exponent in this term. Let me not use that
purple anymore. The exponent in this term
right over here is negative 40. If we increment it, we get
x to the negative 39 power, all of that over negative 39. And now we can add
our constant, c. And all we need to do is
simplify all of this craziness. So the first one is
fairly simplified. We can write it as
7/4 x to the 4th. Now this term right over
here is essentially negative 5 divided by 3/2. So 5 over 3/2 is
equal to 5 times 2/3, which is equal to 10 over 3. So this thing right over here
simplifies to negative 10/3 x to the 3/2. And then we have all
of this craziness. Now 18 divided by
negative 3/2 is equal to 18 times negative 2/3. Which is equal to, well we can
simplify this a little bit, this is the same thing
as 6 times negative 2. Which is equal to negative 12. So this expression
right over here is negative 12x to
the negative 3/2. And then finally, this
one right over here, we can just rewrite it
as, if we want, we could, well, we could just
write negative 1/39 x to the negative 39 plus c. And we're done. We've found the indefinite
integral of all this craziness. And I encourage you to take
the derivative of this. And you can do it using
really just the power rule, to take the derivative of this. And verify that it does
indeed equal this expression that we took the
antiderivative of.