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Unformatted text preview: 1 Antiderivatives & Substitution 1.1 Antiderivatives Reminder about Derivatives First Ill remind you of a few derivatives that well be using a lot: ( x n ) = nx n 1 , (sin x ) = cos x, (cos x ) = sin x, (tan x ) = sec 2 x, (sec x ) = sec x tan x, (sin 1 x ) = 1 1 x 2 , (tan 1 x ) = 1 x 2 + 1 , ( e x ) = e x , (ln  x  ) = 1 x About (ln  x  ) = 1 x The last one deserves some extra attention, since this will show up again and again in integration. You may only remember the derivative (ln x ) = 1 x ; this is of course correct, but is only true for x in the domain of ln x , so for x > 0. For x 0, ln x doesnt exist, and so a derivative wouldnt make sense. We can (and should) extend this differentiation formula to negative x by taking ln  x  instead. Think of the graphs: ln x only has a graph to the right of the yaxis, while ln  x  also has the mirror image of the same graph to the left of the yaxis. We do have to actually check that the derivative is still 1 /x for x < 0, which we do as follows: x <  x  = x (ln  x  ) = (ln( x )) = 1 x ( 1) = 1 x . Antidifferentiation Antidifferentiation (or taking the derivative) is doing differentiation in reverse: given a function f ( x ), find a function F ( x ) such that F ( x ) = f ( x ). This F ( x ) is then called an antiderivative of f ( x ). Here are a few examples of important antiderivatives that you can obtain by reversing the differentiation formulas above: f ( x ) = x n F ( x ) = 1 n + 1 x n +1 ( for n 6 = 1) , f ( x ) = sin x F ( x ) = cos x, f ( x ) = cos x F ( x ) = sin x, f ( x ) = 1 x 2 +1 F ( x ) = tan 1 x, f ( x ) = 1 1 x 2 F ( x ) = sin 1 x f ( x ) = e x F ( x ) = e x , f ( x ) = 1 x F ( x ) = ln  x  Examples Just like with differentiation, we can combine these basic antiderivatives into more complex ones. I will give a few examples first, and then explain some of the steps. Note that an antiderivative is easily checked: just differentiate it and see if you get the orginal function back. But I wont do that in this writeup. f ( x ) = 3 x 7 5 sin( x ) F ( x ) = 3 1 8 x 8 5 ( cos( x )) = 3 8 x 8 + 5 cos( x ) , f ( x ) = e 3 x + 2 1 x 2 + 1 F ( x ) = 1 3 e 3 x + 2 sec( x ) + x, f ( x ) = sin(2 x 3) F ( x ) = 1 2 cos(2 x 3) , f ( x ) = (3 x + 7) 5 + 1 x +1 F ( x ) = 1 6 (3 x + 7) 6 1 3 + ln  x + 1  = 1 18 (3 x + 7) 6 + ln  x + 1  1 Compensating There is one informal trick that I used here several times; let me explain it for sin(2 x 3). We can guess that an antiderivative should involve cos(2 x 3), but if we differentiate that, we would get 2 sin(2 x 3) (with the 2 coming from the chain piece (2 x 3) ), which is not quite what we want. However, we can fix it by compensating for the 2 by putting a 1 2 in front of cos(2 x 3). And indeed, then it works: 1 2 cos(2 x 3) = 1 2 ( sin(2 x 3) 2) = sin(2 x 3) ....
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 Fall '10
 MalabikaPramanik
 Calculus, Antiderivatives, Derivative

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