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Since b2 4ac < 0, the nal expression in parentheses is positive For simplicity, let = b/2a and let = c b2 /(4a) Then our integral is 1 dx + a (x + )2 Of course we can handle this using II above We nd that ax 2 1 dx = + bx + c 1 dx + a (x + )2 a 1 arctan (x + ) + C = a

// Demonstrate the short-circuit operators using System; class SCops { static void Main() { int n, d; n = 10; d = 2; if(d != 0 && (n % d) == 0) ConsoleWriteLine(d + " is a factor of " + n); d = 0; // now, set d to zero // Since d is zero, the second operand is not evaluated if(d != 0 && (n % d) == 0) ConsoleWriteLine(d + " is a factor of " + n); // Now, try the same thing without short-circuit operator // This will cause a divide-by-zero error if(d != 0 & (n % d) == 0) ConsoleWriteLine(d + " is a factor of " + n); } }

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To prevent a divide-by-zero error, the if statement first checks to see if d is equal to zero If it is, then the short-circuit AND stops at that point and does not perform the modulus

The right-hand side of Eq (747) is readily recognized see Eq (723c) as the projection onto -ni of the matrix second moment of segment Gi, with respect to O, represented here as JO Thus, i

4:

IV Even More on Integrals of Reciprocals of Quadratic Expressions Evaluation of the integral ax 2 1 dx + bx + c

lstAdd('E'); lstAdd('B'); lstAdd('D'); lstAdd('F'); ConsoleWriteLine("Number of elements: " + lstCount); // Display the list using array indexing ConsoleWrite("Current contents: "); for(int i=0; i < lstCount; i++) ConsoleWrite(lst[i] + " "); ConsoleWriteLine("\n"); ConsoleWriteLine("Removing 2 elements"); // Remove elements from the list lstRemove('F'); lstRemove('A'); ConsoleWriteLine("Number of elements: " + lstCount); // Use foreach loop to display the list ConsoleWrite("Contents: "); foreach(char c in lst) ConsoleWrite(c + " "); ConsoleWriteLine("\n"); ConsoleWriteLine("Adding 20 more elements"); // Add enough elements to force lst to grow for(int i=0; i < 20; i++) lstAdd((char)('a' + i)); ConsoleWriteLine("Current capacity: " + lstCapacity); ConsoleWriteLine("Number of elements after adding 20: " + lstCount); ConsoleWrite("Contents: "); foreach(char c in lst) ConsoleWrite(c + " "); ConsoleWriteLine("\n"); // Change contents using array indexing ConsoleWriteLine("Change first three elements"); lst[0] = 'X'; lst[1] = 'Y'; lst[2] = 'Z'; ConsoleWrite("Contents: "); foreach(char c in lst) ConsoleWrite(c + " "); ConsoleWriteLine(); // Because of generic type-safety,

q iO = - n i J iO Furthermore, 3 1 3 1 I iO = r r ( r n i ) - rn iT d Gi = 1( n i v i ) - v i n iT 8 Gi 2 8 2 with vi, a third moment of Gi, de ned as v i = ( r r) rd Gi

24:

// } }

(748)

when the discriminant b2 4ac is 0 will be a consequence of the work we do below with partial fractions We will say no more about it here

// the following line is illegal lstAdd(99); // Error, not a char!

The output, shown here, is the same as that produced by the non-generic version of the program:

(749)

Initial number of elements: 0 Adding 6 elements Number of elements: 6 Current contents: C A E B D F Removing 2 elements Number of elements: 4 Contents: C E B D Adding 20 more elements Current capacity: 32 Number of elements after adding 20: 24 Contents: C E B D a b c d e f g h i j k l m n o p q r s t Change first three elements Contents: X Y Z D a b c d e f g h i j k l m n o p q r s t

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