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When we found the derivative of a function like: |
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F(X) = (3X2 - 2X + 5)5 |
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| We had an outside function ( stuff ) 5 | |||
| and an inside function 3X 2 - 2X + 5 to deal with. | |||
| We wind up with: | |||
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F'(X) = 5(3X2 - 2X + 5)4(6X - 2) |
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| If we substituted the letter "u" for the inside part, we would have had: | |||
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F(x) = u5 |
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| And the derivative of u 5 is: | |||
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F'(x) = 5u4 du |
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| The du part was created when we found the derivative. | |||
| Since "u" stood for some complicated thing, | |||
| the derivative of u (du) is probably not equal to 1. | |||
| When we do an integral it's like we were given a derivative, | |||
| and we have to find the function that it came from. | |||
| If we are finding the integral of something | |||
| that looks like an outside/inside type function, | |||
| we have to arrange it so the derivative of the inside part (the du) | |||
| is sitting next to the outside part. | |||
| That is, the ( ) n part. | |||
| If we were asked to find the integral of something like: | |||
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(3X - 4)5 |
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| The first thing we do is say: | |||
| "WOW! That looks like one of those outside/inside type things. | |||
| We need a "u," and we need a "du." | |||
| It looks like 3X - 4 should be the "u." | |||
| That makes the expression: | |||
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u5 |
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| And here we hit the snag. | |||
| We've got no du lying around. | |||
| Since we said that 3X - 4 was our "u," then du should be the derivative of this. | |||
| That would be 3dX. | |||
| dX is equal to 1, so that's not a problem. But we don't have a 3 handy. | |||
| Where do we get one? | |||
| The rules say that you can only multiply something by 1 and not change the value. | |||
| So here our fancy name for 1 is going to be 1/3 x 3 x dX. | |||
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| Writing the expression as a function so that we can work with it, | |||
| and multiplying by this fancy 1: | |||
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| So we got the three we wanted, but we also had to take a 1/3 that we didn't want. | |||
| What do we do with the 1/3? | |||
| We put it in a place where it won't be in the way. | |||
| Like to the left of the integral sign. | |||
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| So now we've got what we needed. | |||
| We have the u (3X - 4), | |||
| and we have the du (3dX). | |||
| We have: | |||
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| Working out the integral, we get: | |||
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| Now we put back the (3X - 4) in place of the u ... | |||
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| and simplify. | |||
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| The du or dX or dWhatever, is created when we find the derivative. | |||
| So it is used up, destroyed, absorbed, or whatever when we find the integral. | |||
| In any case, it goes away. | |||
| And as usual, don't forget the "+ C" part! | |||
| We can't always find a du lying around the equation. | |||
| We can change constants around, like we did in the last example. | |||
| But if we need an X or even an extra power of X, we're stopped. | |||
| For now anyway, but stay tuned ... | |||
| So if we have: | |||
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| The u part can be X 4 - 3X 2 + 2, so the du part needs to be 4X 3 - 6X. | |||
| We don't have this term in the equation, | |||
| and we can't get it without changing the value of the equation. | |||
| So we can't do this one with the tricks we have so far. | |||
| Here are some we can solve now. | |||
| These examples will take some studying. | |||
| You may not get all or any of them right off. | |||
| Remember that we can live with extra constants in the problems. | |||
| Example: | |||
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 |
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| So here we tried the obvious first, the 3X. | |||
| That made du equal to 3dx. | |||
| We only had dx, but we can deal with differences in multiplied constants. | |||
| Since 3X probably needs to be u here, what else is there? | |||
| 3dX has to be du. | |||
| It's the first step to really mastering these | |||
| to see that you multiply du = 3dX by 1/3 | |||
| to keep du equal to 3dX and turn the 3dX into what you need | |||
| to substitute into the equation. | |||
| Example: | |||
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 |
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| The tricky part here was realizing that p is just a number, | |||
| so it can be moved to the left and dealt with later. | |||
| From there we have the same "fix the coefficient on the du" | |||
| like the last example. | |||
| Example: | |||
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| Another fix the constants on the du deal. | |||
| The equation was just nastier. | |||
| This fix the constants on the du looks like a common thing, eh? | |||
| Example: | |||
| This is a very tricky one, so watch closely! | |||
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| Most people start this one OK. That is, they set u = X-1. | |||
| The problem is that when they stop there, | |||
| they notice that they are going to have an extra X in the problem. | |||
| The trick is to do the second step on the right. | |||
| That gives you SOMETHING to substitute for that first X. | |||
| Then just multiply everything out and do the integral term by term. | |||
| These problems are really like big puzzles. | |||
| The only way to get good at them is to do a bunch of them | |||
| and learn as many tricks to move the numbers around as you can. | |||
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copyright 2005 Bruce Kirkpatrick |
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