105-1-AntiderivativesSubstitution

# 105-1-AntiderivativesSubstitution - 1 Antiderivatives&...

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Unformatted text preview: 1 Antiderivatives & Substitution 1.1 Antiderivatives • Reminder about Derivatives First I’ll remind you of a few derivatives that we’ll 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 doesn’t exist, and so a derivative wouldn’t 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 y-axis, while ln | x | also has the mirror image of the same graph to the left of the y-axis. 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 won’t 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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105-1-AntiderivativesSubstitution - 1 Antiderivatives&...

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