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Without saying so explicitly, we have implicitly assumed in our discussion of area in the last section that our function f is positive, that is, its graph lies about the x-axis But of course many functions do not share that property We nevertheless would like to be able to calculate areas determined by such functions, and to calculate the corresponding integrals This turns out to be simple to do Consider the function y = f (x) shown in Figure 411 It is negative on the interval [a, b] and positive on the interval [b, c] Suppose that we wish to calculate the shaded area as in Figure 412 We can do so by breaking the problem into pieces Of course, because f 0, the area between x = b and x = c is given by c the integral b f (x) dx, just as we have discussed in the last section But our

In the program, pay special attention to this line:

var posNums = numsWhere(n => n > 0)Select(r => r);

This creates a query called posNums that creates a sequence of the positive values in nums It does this by use of the Where( ) method (to filter the values) and Select( ) (to select the

discussions do not apply directly to the area between x = a and x = b What we can do is instead consider the function g = f Its graph is shown in Figure 413 Of course g is a positive function, except at the endpoints a and b; and the area under g ---between x = a and x = b---is just the same as the shaded area between x = a and x = b in Figure 414 (refer also to Figure 412) That area is

Part I:

values) The Where( ) method can be invoked on nums because all arrays implement IEnumerable<T>, which supports the query extension methods Technically, the Select( ) method in the preceding example is not necessary because in this simple case, the sequence returned by Where( ) already contains the result However, you can use more sophisticated selection criteria, just as you did with the query syntax For example, this query returns the positive values in nums increased by an order of magnitude:

var posNums = numsWhere(n => n > 0)Select(r => r * 10);

Hi Team:

What we have learned is this: If f (x) < 0 on the interval under discussion, then the integral of f will be a negative number If we want to calculate positive area then we must interject a minus sign

As you might expect, you can chain together other operations For example, this query selects the positive values, sorts them into descending order, and returns the resulting sequence:

var posNums = numsWhere(n => n > 0)OrderByDescending(j => j);

Here, the expression j => j specifies that the ordering is dependent on the input parameter, which is an element from the sequence obtained from Where( ) Here is an example that demonstrates the GroupBy( ) method It reworks the group example shown earlier

// Demonstrate the GroupBy() query method // This program reworks the earlier version that used // the query syntax using System; using SystemLinq; class GroupByDemo { static void Main() { string[] websites = { "hsNameAcom", "hsNameBnet", "hsNameCnet", "hsNameDcom", "hsNameEorg", "hsNameForg", "hsNameGtv", "hsNameHnet", "hsNameItv" }; // Use query methods to group websites by top-level domain name var webAddrs = websitesWhere(w => wLastIndexOf('') != 1) GroupBy(x => xSubstring(xLastIndexOf(""))); // Execute the query and display the results foreach(var sites in webAddrs) { ConsoleWriteLine("Web sites grouped by " + sitesKey); foreach(var site in sites) ConsoleWriteLine(" " + site); ConsoleWriteLine(); } } }

Calculate the (positive) area, between the graph of f ( x) = x 3 2x 2 11x + 12 and the x-axis, between x = 3 and x = 4

This version produces the same output as the earlier version The only difference is how the query is created In this version, the query methods are used Here is another example Recall the join query used in the JoinDemo example shown earlier:

19:

var inStockList = from item in items join entry in statusList on itemItemNumber equals entryItemNumber select new Temp(itemName, entryInStock);

Consider Figure 416 It was drawn using the technique of Section 31, and it plainly shows that f is positive on [ 3, 1] and negative on [1, 4] From the discussion preceding this example, we know then that