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print barcode image c# With all the needed values &/dt can be evaluated Watch the signs closely 2(54) + 4 in Visual Studio .NET
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Make EAN128 In ObjectiveC Using Barcode creation for iPad Control to generate, create EAN128 image in iPad applications. Decoding EAN / UCC  14 In Visual Basic .NET Using Barcode decoder for VS .NET Control to read, scan read, scan image in .NET framework applications. dP 6 1OO~ dt
dP = ( 2 6 ~ dP 100) dt dt
The population and the rate of increase in population are given in the problem so we have
= [(26)(200) 1001j02thousand per month] = 84 thousand of
pounds per month
This is also an interesting maxmin problem Take
= ( 2 6  100) ~
and set equal to zero to find p = 38
d2T The second derivative = 26 so the point p = 38 is a minimum dP2
The city dump can accommodate the trash from 38 thousand p ople, but at 200 thousand the curve becomes progressively more positive and the tras problem progressively worse INTEGRATION
There are many calculus problems where the derivative of a function is known and the h c t i o n is desired For example, if a mathematical expression for the rate of population growth dP/df is known, is it possible to "work backwards" to find the expression for P, how the population varies over time The process of starting with a derivative and working back to the function is quite naturally called the antiderivative The antiderivative of a h c t i o n is an easy concept but often is operationally difficult There are many integragtion problems where h d i n g the antiderivative will prove a major challenge In some problems the integral can be viewed as the area under the curve of the function being integrated This is often very helpful in getting a physical "feel" for the problem ,and the process of integration This view of the integral will be discussed later in the chapter Some problems in integration require a great deal of imaginative thinking and manipulative ability The simplest first approach to integration is via the antiderivative After that we will move on to using the area under the curve approach and finally to the more difficult integral problems The Antiderivative
Start with a simple function, y = x 2 The derivative of that fimction is written as
CHAPTER^
Keeping this short review of differentiation in mind, suppose we encountered a derivative
du =2v dv
and want to know how U varies with v Keep the differential (of y = x2 ) in fiont of you and just work backwards = 2v can be written as du = 2vdv
Now all we need to do is perform the inverse or anti derivative operation to fmd U in terms of v This being mathematics, no operation can be performed without a symbol For integration we use this elongated s shape, so write Jdu = S2d~ The left side of this equation is the integral of the differential, two inverse operations acting on du is the antiderivative The The d acting on U is the derivative while the result of these inverse operations on U is that the left side of this equation is U The operation is somewhat like squaring a square root The right side is not so easy except that we have the differential example just above us The differential of x 2 is 2x&, so the integral of 2vdv is v 2 The function described by the lfferential statement du = 2vdv is therefore U = v L Conceptually the antiderivative is not difficult Actually finding the antiderivative of a complicated function is often not at all easy Polynomials are the easiest to work with and that is where we will start Find l y 3 d y
Solution: We seek a function that differentiates to y3dy
The differential of y 4 is 4y dy which is very close to what we want Thedifferentialof Y4 is y3dy so the J y dy 4

