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Returns the index of the first occurrence of any character in anyOf that is found within the invoking string Returns 1 if no match is found Ordinal searching is used Returns the index of the first occurrence of any character in anyOf that is found within the invoking string Searching begins at the index specified by startIndex Returns 1 if no match is found Ordinal searching is used Returns the index of the first occurrence of any character in anyOf that is found within the invoking string Searching begins at the index specified by startIndex and runs for count elements Returns 1 if no match is found Ordinal searching is used Returns the index of the last occurrence of value within the invoking string Returns 1 if value is not found Ordinal searching is used Returns the index of the last occurrence of value within the invoking string Returns 1 if value is not found Culture-sensitive searching is used Returns the index of the last occurrence of value within a range of the invoking string The search proceeds in reverse order, beginning at the index specified by startIndex and stopping at zero Returns 1 if the value is not found Ordinal searching is used Returns the index of the last occurrence of value within a range of the invoking string The search proceeds in reverse order, beginning at the index specified by startIndex and stopping at zero Returns 1 if value is not found Culture-sensitive searching is used Returns the index of the last occurrence of value within the invoking string The search proceeds in reverse order, beginning at the index specified by startIndex and running for count elements Returns 1 if value is not found Ordinal searching is used

> 0,

public int IndexOfAny(char[ ] anyOf, int startIndex)

public int IndexOfAny(char[ ] anyOf, int startIndex, int count) public int LastIndexOf(char value) public int LastIndexOf(string value)

public int LastIndexOf(char value, int startIndex)

public int LastIndexOf(string value, int startIndex)

1 = x2 < 0,

public int LastIndexOf(char value, int startIndex, int count)

You could revise the sentence to end at the conclusion of any one of these phrases For example, you might revise the sentence to read:

TABLE 22-3

The Search Methods Offered by String (continued)

(iii) ln(1) = 0, we know that ln x is an increasing, concave down function whose graph passes through (1, 0) There are no relative maxima or minima (since the derivative is never 0) Certainly ln 09 < 0; the formula ln(09n ) = n ln 09 therefore tells us that ln x is negative without bound as x 0+ Since ln x = ln(1/x), we may also conclude that ln x is positive without bound as x + A sketch of the graph of y = ln x appears in Figure 62 We learned in the last paragraph that the function ln x takes negative values, which are arbitrarily large in absolute value when x is small and positive In particular, the negative y axis is a vertical asymptote Since ln(1/x) = ln x, we then find that ln x takes arbitrarily large positive values when x is large and positive Since we have only defined the function ln x when x > 0, the graph is only sketched in Figure 62 to the right of the y-axis However it certainly makes sense to discuss the function ln |x| when x = 0 (Figure 63): If x = 0 then 1 d (ln |x|) = dx x In other words, 1 dx = ln |x| + C x

22:

Method public int LastIndexOf(string value, int startIndex, int count) public int LastIndexOf(string value, StringComparison comparisonType) public int LastIndexOf(string value, int startIndex, StringComparison comparisonType) public int LastIndexOf(string value, int startIndex, int count, StringComparison comparisonType) public int LastIndexOfAny(char[ ] anyOf) public int LastIndexOfAny(char[ ] anyOf, int startIndex) public int LastIndexOfAny(char[ ] anyOf, int startIndex, int count) public bool StartsWith(string value) public bool StartsWith(string value, StringComparison comparisonType) public bool StartsWith(string value, bool ignoreCase, CultureInfo culture)

TABLE 22-3

For clarity we set ( x) = 3 sin x 4, ( x) = 3( cos x) The integral then has the form 1 3 ( x) 1 dx = ln | ( x) | + C ( x) 3

.