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EXAMPLE 15 The continuous function f(x) = xx has an antiderivative, but it cannot be put into closed form in terms of elementary functions. However, Mathematica can deal with it as a function defined by an integral. Since all antiderivatives of f(x) differ by a constant, we define F(x) to be the antiderivative for which F(0) = 0. Let us plot this antiderivative for 0 x 4.

When it comes to file systems, MS Windows is the 24-flavors ice cream shop of operating systems. Popular flavors include FAT, FAT16, FAT32, OS/2 HPFS, NTFS, and DFS. Then there are the UNIX interoperability flavors: NFS and Samba. Windows provides a great deal of functionality in supporting legacy and contemporary file systems. This capability is especially useful in heterogeneous environments, regardless of whether there are multiple versions of Windows or a mixture of Windows and other operating systems.

log2 8 3 1 4 log1=8 128 ,

f[x_]= x^x; F[x_]= f[t] t;

By making the lower limit 0, we force F(0) = 0.

Plot[F[x], {x, 0, 4}]

(a) 4

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The file allocation table (FAT) file system represents the basic common denominator for file and directory support across the family of Windows platforms. The FAT architecture is essentially an indexed table that references data containers called clusters within a partition of contiguous space on a disk. A cluster represents the smallest unit of storage on a disk. Cluster size ranges from 512 to 32K bytes depending on disk size and format. Because cluster size represents the smallest container for a file or directory, it may be important to tailor the file system and drive format to meet the file size profile for a particular data set. FAT file systems are defined by four control areas. The first or reserve area contains the boot sector, bootstrap program, and partition table. The next area is the FAT index of clusters, followed by the root directory table, which is a directory of files and top level directories. The fourth area is the file area collection of clusters where data are stored. FAT architecture has evolved over the years in keeping with the availability of larger disks and the demand for larger file sizes. Basically, this has meant an increase in the number of bits in the FAT used to index clusters. More index bits means more clusters can be - 50 -

60 50 40 30 20 10

9.14 Let F ( x ) = esin t dt . Find F '( x ) . 1

s 0:00004 25,000 , (c) 0:02 5 0:125 (d) 1/25, (e) 7=144

F[x_]= Exp[Sin[t]] t;

F' [x]

addressed. Originally the FAT architecture used 12 bits as a cluster index. In DOS 4.0 the index size was increased to 16 bits. In the second release of Windows 95, the FAT index grew once again to 32 bits in FAT32.

Sin[x]

(d) 3 2 log3 5 ,

9.15 Sketch, on one set of axes, the graphs of the three antiderivatives of f(x) = e sin x, 0 x 2 , for which F(0) = 0, F(1) = 0, and F(2) = 0.

f[x_]= Exp[Sin[x]]; F1[x_] NIntegrate[f[t], {t, 0, x}] F2[x_] NIntegrate[f[t], {t, 1, x}] F3[x_] NIntegrate[f[t], {t, 2, x}] Plot[{F1[x], F2[x], F3[x]}, {x, 0, 2 o}]

8 6 4 2

(e) 1 4=3 27 2=3 8

Windows NT introduced another player in the suite of Windows file systems, the New Technology File System (NTFS). In keeping with NT's role as a workgroup or domain server, NTFS provides a scalable and recoverable file system environment to better support and secure server data. NTFS again extended the cluster index to 64 bits. This allows NTFS to address large RAID-based data sets that span multiple disk drives. The structural contents and state of NTFS is tracked and maintained in a relational database called the master file table (MFT). MFT information, along with redundant copies of critical file system data and optional file system mirroring, improves NTFS fault tolerance over FAT file systems and enables automated recovery in the event of a system failure. NTFS also provides enhanced data security over FAT. Access controls can be applied at the individual file level. FAT file systems only apply access controls at the directory level.

1 2 4

9.16 Consider the semicircle x2 + y2 = 16, y 0 shown in the figure. Find the height, h, so that the shaded y area is half the area of the semicircle.

We solve for x as a function of y: Solve[x2 + y2 16, x]

(a) 64,

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