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## Calculus

by: Otilia Murray I

16

0

15

# Calculus MAT 021B

Otilia Murray I
UCD
GPA 3.88

Qinglan Xia

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COURSE
PROF.
Qinglan Xia
TYPE
Class Notes
PAGES
15
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 15 page Class Notes was uploaded by Otilia Murray I on Tuesday September 8, 2015. The Class Notes belongs to MAT 021B at University of California - Davis taught by Qinglan Xia in Fall. Since its upload, it has received 16 views. For similar materials see /class/187344/mat-021b-university-of-california-davis in Mathematics (M) at University of California - Davis.

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Date Created: 09/08/15
i 39 3 Derivative problem Given a function how to nd its derivative f c When F is nice enough we have learned many formulas and rules a When F is not that nice we may have to use de nition These are What you have learned from Differential Calculus Mat21A or equivalent courses Antiderivative problem Question If the derivative fX is given can you nd the original Jnction FX such that F x x In other words If the velocity is given can you nd the position If the slope of a curve is given can you gure out the curve Example Can you nd an antideriVatiVe for f 2x AnswerF 2 or 2 1 or 2 7T or x2 C for any arbitrary constant C Are there any other antideriVatiVes for f 2w No If two functions haVe the same derivative then they must differ by a constant So ALL antideriVatiVes of f 2x are in the form of F x2 C Where C is an arbitrary constant What does their graphs look like There are yertical translation of one another In general if F is an antiderivative of f on an interval I then the most general antiderivative of f on I is F C where C is an arbitrary constant a It is a family of functions F C Whose graphs are vertical translations of one another Slopes of all these graphs are equal to f Inde nite Integral This family of antiderivatives F C is very important we denote it by a special symbol f ltccgt den called the inde nite integral of f with respect to x Here a f is the integral a x is the variable of integration o f is an integral sign That is f x dx F x C all antiderivatives of f x Finding a particular antiderivative Example Find an antiderivative of f 3 sin x that satis es F 0 5 The general antiderivative of f is 30sx C Substitute F O 5 an I BCOSOO 3C So C 8 Thus the solution is 30sx 8 Example Find the inde nite integral x2dx This is the same as nd antiderivatives of x The answer is 3 x2dx C 299 Some formulas of antiderivatives YABLE 52 Antideriva ve formulas k a nonzero constant Function General amiderivative Function General amiderivalive u 1 MI In 1 L1 1 x on cnal s e ke 0 2 sinct coslotC 9 lnlxC x 0 3 cos1x ism1x C 10 ismquot kx C V 7 kzx39 I 4 secllcz xanlac 4 1 u k n I kle ktan 1a c 1 5 05620 quotcotor C 1 4 k 12 7 sec lo 4 C kt gt I l kale I 5 1m k1 Io S ks C 13 a l a c 11 gt o a se 1 1 k In a 7 csckxcoth iicsckxiC The formulas in the table are easily proved by differentiating the antideriV atives and verifying that the result agrees with the original function Antiderivative Rules Also derivatives rules also lead to corresponding antiderivative rules TABLE 43 Antiderivative lirhsearii gr rules Functinn General antiderivative 1 Crmsram Multiple Rule k x A39Fix C It a canstam 2 Negative Rule fx Fx C 3 Sum JrDifference Rule x t gm Fix 1 GI C Inde nite Integral Rules That is kfxdx kfxdx fxdx f xdx fltwgtigltxgtdwfltxgtdrrigltrrgtd eg 3 4x2dx4x2dx4x0 3 x x2smxd 2dSi d COSLEC Example Find the inde nite integral 1 2 100 3x 1 42 38 1 2 100 3 1 42 1 8 1 2 5x100dx dx dx 3e3xdx x 1 4x2 5 101 1x01 1n arctan 2x e3m C Initial Value Problem Now we use antiderivative to solve the Initial Value Problem Given a continuous function f nd a function y that satis es the differential equation 2 f the initial condition y 0 340 Note that o f just means that y is an antiderivative of f a The initial condition y x0 340 means that the function y has the value 340 when x x0 As a result solving the above initial value problem is same as Find a particular antiderivative y of f that satis es y x0 yO Example Solve dy 3x2 sin x dx y 0 1 Solution LEB COSIE l C Substitute y 0 1 yields 1 340 O oos0C l l C C 2 Therefore the solution is 3 x oosx2 Example Solve dZy Z 5 240 27310 1 Solution d2y 56 yields dy 56 2 0 dm 2 139 Using 3 O l we have 02 01 So 2 dy x 1 2 Cm 2 andy 0 Thus 3 yxx02 Usingylt0gt 2yields 00CZ y02 022 Therefore the solution is

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