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many elements the list will contain This prevents the overhead associated with the allocation of more memory Conversely, if you want to reduce the size of the array that underlies an ArrayList, you can set Capacity to a smaller value However, this value must not be less than Count Recall that Count is a property defined by ICollection that holds the number of objects currently stored in a collection Attempting to set Capacity to a value less than Count causes an ArgumentOutOfRangeException to be generated To obtain an ArrayList that is precisely as large as the number of items that it is currently holding, set Capacity equal to Count You can also call TrimToSize( ) The following program demonstrates ArrayList It creates an ArrayList and then adds characters to it The list is then displayed Some of the elements are removed, and the list is displayed again Next, more elements are added, forcing the capacity of the list to be increased Finally, the contents of elements are changed

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The Fundamental Theorem of Calculus tells us that the left-hand side is u v Thus u v = or u v dx = u v v u dx u v dx + u v dx

// Demonstrate ArrayList using System; using SystemCollections; class ArrayListDemo { static void Main() { // Create an array list ArrayList al = new ArrayList(); ConsoleWriteLine("Initial number of elements: " + alCount); ConsoleWriteLine(); ConsoleWriteLine("Adding 6 elements"); // Add elements to the array list alAdd('C'); alAdd('A'); alAdd('E'); alAdd('B'); alAdd('D'); alAdd('F'); ConsoleWriteLine("Number of elements: " + alCount); // Display the array list using array indexing ConsoleWrite("Current contents: "); for(int i=0; i < alCount; i++) ConsoleWrite(al[i] + " "); ConsoleWriteLine("\n"); ConsoleWriteLine("Removing 2 elements"); // Remove elements from the array list alRemove('F'); alRemove('A');

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// Display a triangle's style public void ShowStyle() { ConsoleWriteLine("Triangle is " + Style); } } class Shapes7 { static void Main() { Triangle t1 = new Triangle("right", 80, 120); // Make a copy of t1 Triangle t2 = new Triangle(t1); ConsoleWriteLine("Info for t1: "); t1ShowStyle(); t1ShowDim(); ConsoleWriteLine("Area is " + t1Area()); ConsoleWriteLine(); ConsoleWriteLine("Info for t2: "); t2ShowStyle(); t2ShowDim(); ConsoleWriteLine("Area is " + t2Area()); } }

In this program, t2 is constructed from t1 and is, thus, identical The output is shown here:

It is traditional to abbreviate u (x) dx = du and v (x) dx = dv Thus the integration by parts formula becomes u dv = uv v du

Info for t1: Triangle is right Width and height are 8 and 12 Area is 48 Info for t2: Triangle is right Width and height are 8 and 12 Area is 48

lection, collecting goods for recycling, including updating guard railings on highways We can demonstrate and show that turnaround time from initial complaint or report receiving to ultimate repair, replacement, or adjusting it has taken 23 percent longer, on average

Pay special attention to this Triangle constructor:

public Triangle(Triangle ob) : base(ob) { Style = obStyle; }

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It receives an object of type Triangle, and it passes that object (through base) to this TwoDShape constructor:

We observe that the integrand is a product Let us use the integration by parts formula by setting u( x) = x and dv = cos x dx Then u( x) = x v( x) = sin x du = u ( x) dx = 1 dx = dx dv = v ( x) dx = cos x dx

public TwoDShape(TwoDShape ob) { Width = obWidth; Height = obHeight; }

Part I:

The key point is that TwoDShape( ) is expecting a TwoDShape object However, Triangle( ) passes it a Triangle object As explained, the reason this works is because a base class reference can refer to a derived class object Thus, it is perfectly acceptable to pass TwoDShape( ) a reference to an object of a class derived from TwoDShape Because the TwoDShape( ) constructor is initializing only those portions of the derived class object that are members of TwoDShape, it doesn t matter that the object might also contain other members added by derived classes

Of course we calculate v by anti-differentiation According to the integration by parts formula, x cos x dx = u dv v du sin x dx

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