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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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Μπορείς να διαβάσεις στα Αγγλικά; Κάνε κλικ εδώ για να δείτε περισσότερες συζητήσεις που συμβαίνουν στην αγγλική ιστοσελίδα της Khan Academy.

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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.