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The output from the program is shown here:

Unchecked result: 1 SystemOverflowException: Arithmetic operation resulted in an overflow at CheckedDemoMain()

1 1 ln |x + 2| + ln |x 5| + C 7 7 1/2 x/2 + 1/2 1 = hence + 2 + 1) x+1 x2 + 1 (x + 1)(x dx = (x +1)(x 2 +1) 1/2 dx + x +1 x/2 dx + x 2 +1 1/2 dx x 2 +1

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k(1 << k < 20p) This portion of R* is shown in more detail in Fig 1316 It was obtained from an FFT analysis of Vi*(h) which gave Vi*(2pk), k = 0, 1/8, 1/4, 3/8 The complete residual vibration spectrum for the limiting pro le (ie, v*) is shown in Fig 1317 A convenient way to express V1*(h) is in terms of its Fourier series which is as follows vi*(h) = Vi*( 0) + 2 [Re(Vi*(2pm)) cos(2pmh) - Im(Vi*(2pm)) sin(2pmh)]

As is evident, the unchecked expression resulted in a truncation The checked expression caused an exception The preceding program demonstrated the use of checked and unchecked for a single expression The following program shows how to check and uncheck a block of statements

// Using checked and unchecked with statement blocks using System; class CheckedBlocks { static void Main() { byte a, b; byte result; a = 127; b = 127; try { unchecked { a = 127; b = 127; result = unchecked((byte)(a * b)); ConsoleWriteLine("Unchecked result: " + result); a = 125; b = 5; result = unchecked((byte)(a * b)); ConsoleWriteLine("Unchecked result: " + result); }

Part I:

(1376)

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1 1 1 = ln |x +1| ln |x 2 +1|+ Tan 1 x +C 4 2 2 (c) Now x 3 2x 2 5x + 6 = (x 3)(x + 2)(x 1) Then x3 As a result, dx x 3 2x 2 5x +6 = = (d) 1/10 dx + x 3 1/15 dx + x +2 1/6 dx x 1 2x 2 1/10 1/15 1/6 1 = + + x 3 x+2 x 1 5x + 6

checked { a = 2; b = 7; result = checked((byte)(a * b)); // this is OK ConsoleWriteLine("Checked result: " + result); a = 127; b = 127; result = checked((byte)(a * b)); // this causes exception ConsoleWriteLine("Checked result: " + result); // won't execute } } catch (OverflowException exc) { ConsoleWriteLine(exc); } } }

The output from the program is shown here:

The Fourier series coef cients for several of the Vi*(h) are given in Table 134 When V0*(h) is chosen as the velocity curve for simple harmonic motion, then the envelopes of the residual vibration spectra are as shown in Fig 1318 The Fourier series for V2*(h) is given as

Unchecked result: 1 Unchecked result: 113 Checked result: 14 SystemOverflowException: Arithmetic operation resulted in an overflow at CheckedBlocksMain()

Now x 4 1 = (x 2 1)(x 2 + 1) = (x 1)(x + 1)(x 2 + 1) Hence 1/4 x 1/4 x/2 = + + 2 4 1 x 1 x+1 x +1 x We conclude that 1 x dx 1 1 = ln |x 1| + ln |x + 1| ln |x 2 + 1| + C 4+1 4 4 4 x

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