Microsoft Mathematics

Ini adalah applikasi untuk memudahkan kita beroperasi di matematika, bukan hanya kalkulator biasa, tetapi semua jenis hitungan baik aljabar, kalkulus, grafik, akar, dan banyak lagi, silahkan coba disini .sya juga menggunakan applikasi ini, karena memudahkan saya untuk menjawab soal, seperti kalkulus I dan II.

Kritikan Hati

Malam tanpa bulan
bukanlah sesuatu hal yang luar biasa
tetapi sudah biasa
apabila bulan sudah pergi
maka sang bintang pun akan datang untuk menyapa

begitu juga dengan hidup kita, jangan memandang sesuatu cuma dari satu sisi,
tetapi pandanglah sisi lainnya,
kelak itu akan mengubah segalanya yang kita anggap merupakan suatu persoalan yang tidak bisa diselesaikan,

setiap orang punya sisi negative dan positive,
begitu juga diri kita, pandanglah diri kita didepan cermin dan ketuk hati kita, siapa aku ini, dan apa aku ini,.???
"ah.....sebenarnya aku ini seorang yg negative atau positive...?????"
renungkanlah diri kita sendiri, sebelum kita merenungkan seseorang....

cuma itu sob yg bisa ku sampaikan," semoga bermamfaat "
wassalam

Folder Protects

Sebuah applikasi untuk menyimpan data dengan aman dan tidak terimpeksi virus , dengan fitur data anda bisa ditampilkan+tidak bisal di delete+dan dicopy, atau Data anda ditampilkan+tidak bisa diakses, atau tidak anda tampilkan sama sekali, bahkan anti virus tidak bisa mendeteksi data anda, bila anda berminat klik disini.

 bila anda memerlukan keynya anda bisa meminta pada saya. saya akan dengan senang hati untuk memberikannya.

Mempercepat Koneksi Internet

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Kumpulan Trik Registry

Beberapa tips dan registry windows dibawah ini cuma berlaku buat Windows XP. Sebelum mengutak-atik Registry, ada baiknya melakukan backup terlebih dahulu. * Klik tombol Start > Run atau tekan tombol Windows+R pada keyboard. * Ketik regedit dan tekan Enter setelah berada didalam jendela Run. * Didalam Registry Editor, pilih menu File > Export. * Setelah Export Registry File muncul, masukkan nama file ke bagian File Name, misalnya backup-registry dan sebagainya. * tekan tombol Save. Proses Di Atas Untuk MengBackUp Registry / Regedit. Hal ini bertujuan jika terjadi kesalahan, kekeliruan, kefatalan atau apalah namanya yang intinya (pengendalian diri) membuat sistem windows sobat tidak normal lagi, maka pergilah ke Windows Explorer dan bukalah file backup dari regedit tadi dan doble klik file tersebut atau klik kanan, kemudian klik perintah paling atas atau “Merge”, maka sistem windows sobat akan kembali normal lagi seperti sebelum diedit registrynya (seperti sediakala). INGAT: BACKUP DULU REGISTRY WINDOWS (REGEDIT) ANDA. Pada pembahasan regedit di sini akan dibagi menjadi dua bagian, yang dikelompokkan atas part1 yang terdiri dari 160 point dan part2 yang terdiri dari 152 point, OK?! Part1 [001] Mengganti Wallpaper HKEY_CURRENT_USER/Control Panel/Desktop Klik ganda pada wallpaper dan masukkan path gambar yang diinginkan pada Value Data. [002] Mengganti Nama Recycle Bin HKEY_CLASSES_ROOT/CLSID/{645FF040-5081-101B-9F08-00AA002F954E} Klik ganda pada option (Default value) dan beri nama baru sesuai yang diinginkan pada Value Data. [003] Memunculkan Rename Pada Recycle Bin HKEY_CLASSES_ROOT/CLSID/{645FF040-5081-101B-9F08-00AA002F954E}/ShellFolder Klik ganda pada Attributes > Edit Binary Value. Pada Value Data, ganti angka tersebut menjadi 0000 50 01 00 20. [004] Menyembunyikan Recycle Bin HKEY_LOCAL_MACHINE/Software/Microsoft/Windows/CurrentVersion/Explorer/Desktop/NameS pace Hapus subkey {645FF040-5081-101B-9F08-00AA002F954E}, kemudian Restart komputer untuk melihat hasilnya. Untuk memunculkannya kembali, buat kembali kombinasi angka {645FF040- 5081-101B-9F08-00AA002F954E}. [005] Menambah Isi Shortcut Menu Pada Recycle Bin HKEY_CLASSES_ROOT/CLSID/{645FF040-5081-101B-9F08-00AA002F954E}/ShellFolder Klik ganda Attributes dan ganti angka di Value Data dengan angka berikut ini: 0000 50 01 00 20 > Rename 0000 60 01 00 20 > Delete 0000 70 01 00 20 > Rename & Delete 0000 41 01 00 20 > Copy 0000 42 01 00 20 > Cut 0000 43 01 00 20 > Copy & Cut 0000 44 01 00 20 > Paste 0000 45 01 00 20 > Copy & Paste 0000 46 01 00 20 > Cut & Paste 0000 47 01 00 20 > Cut, Copy & Paste [006] Menambah Isi Shortcut Menu Pada Recycle Bin Dengan Menu Pilihan HKEY_CLASSES_ROOT/CLSID/{645FF040-5081-101B-9F08-00AA002F954E}/Shell Klik menu Edit > New > Key dan beri nama yang diinginkan (Misalnya: Go To Windows Explorer). Dibawah key yang baru tersebut, tekan lagi menu Edit > New > Key dan buat sebuah key baru bernama Command. Klik ganda option (Default), dan pada bagian Value Data, isi dengan path Windows Explorer ( C:WINDOWSExplorer.exe). [007] Mengembalikan Folder Documents Yang Hilang Di My Computer HKEY_LOCAL_MACHINE/Software/Microsoft/Windows/CurrentVersion/Explorer/DocFolderPaths Pilih menu Edit > New > String Value dan beri nama sesuai dengan username yang digunakan di Windows (contohnya: Chippers). Klik ganda pada value tersebut dan masukkan path tempat dimana Documents anda berada (contohnya D: Documents) [008] Menyingkirkan File Stored Dari My Computer HKEY_LOCAL_MACHINE/SOFTWARE/Microsoft/Windows/CurrentVersion/Explorer/MyComputer/ NameSpace/DelegateFolders Hapus subkey {59031a47-3f72-44a7-89c5-5595fe6b30ee} dengan menekan tombol Del. [009] Menyembunyikan My Recent Documents HKEY_CURRENT_USER/Software/Microsoft/Windows/CurrentVersion/Policies/Explorer Klik menu Edit > New > DWORD Value dan beri nama NoRecentDocsMenu. Kemudian klik ganda pada DWORD Value tersebut, dan berikan angka 1 untuk mengaktifkannya. [010] Menyembunyikan Menu Find HKEY_CURRENT_USER/Software/Microsoft/Windows/CurrentVersion/Policies/Explorer Klik menu Edit > New > DWORD Value dan beri nama NoFind. Kemudian klik ganda pada DWORD Value tersebut dan berikan angka 1 untuk mengaktifkannya. Restart komputer.

Matematika Diskrit : Relasi dan Fungsi

Bagian Satu Relasi Pada pertemuan mengenai himpunan telah dijelaskan himpunan pasangan terurut diperoleh dari perkalian kaertesian antara dua himpunan. A x B = {(a, b)| a �� A dan b} Definisi. Relasi biner R antara A dan B adalah himpunan bagian dari A x B. notasi: R ⊆ (A x B) Himpunan A disebut daerah asal (domain) dari R, dan himpunan B disebut daerah hasil (range atau codomain) dari R. Contoh Misalkan A = {Amir, Budi, Cecep}, dan B = { IF221, IF251, IF342,IF323}. Perkalian kartesian A dan B (AxB) menghasilkan 12 pasangan terurut, yaitu A x B = {(Amir, IF221), (Amir, IF251), (Amir,IF342), … , (cecep, IF323)} Misalkan R adalah relasi yang menyatakan mata kuliah yang diambil oleh mahasiswa pada semester ganjil, yaitu R = {(Amir, IF251), (Amir, IF323), (Budi, IF221), (Cecep, IF342), (Cecep, IF323)} Contoh Misalkan P = { 2, 3, 4} dan Q = {2, 4, 8, 9, 15}. Relasi R dari P ke Q dengan (p, q) �� R jika phabis membagi q Maka diperoleh R = {(2, 2), (2, 4) , (2, 8), (4, 4), (4,8), (3,9), (3,15)}.  Relasi dapat direpresentasi dengan diagram Venn dan Tabel Diagram Venn Amir Budi Cecep IF221 IF251 IF342 IF323 2 3 4 2 4 8 9 15 Tabel A B A B Amir Amir Budi Cecep Cecep IF251 IF323 IF221 IF342 IF323 2 2 2 3 3 2 4 8 9 15 4 4 4 8  Sifat-sifat Relasi 1. Refleksif Definisi, relasi R pada himpunan A disebut refleksif jika (a , a) ∈ R untuk setiap a ∈ A. Contoh: Misalkan A = {1,2,3} dan dengan R = {(1,1), (1,3), (2,1), (2,2), (3,2)} 2. Setangkup Definisi, relasi R pada himpunan A disebut setangkup jika (a, b) ∈ R, maka (b, a) ∈ R, untuk semua a,b ∈ R Contoh Misalkan A = {1,2,3} dan dengan R = {(1,1), (1,2), (2,1), (2,2), (3,2)}  Relasi n-ary Definisi, misalkan A1 , A2, … , An adalah himpunan. Relasi n-ary R pada himpunan-himpunan tersebut adalah himpunan bagian dari A1 x A2x … x An. Contoh Misalkan NIM = {132450, 132452, 132457} Nama = {Dono, Dina, Dita} MatKul = {Matematika, Algoritma, Struktur Data } Nilai = {A, B} Salah satu contoh relasi yang bernama MHS adalah NIM Nama MatKul Nilai 132450 132450 132452 132452 132457 Dono Dono Dina Dina Dita Matematika Struktur Data Algoitma Matematika Algoritma A B B A A Basisdata (database) adalah kumpulan tabel. Salah satu model basisdata adalah model basisdata relasional yang didasari oleh konsep relasi n-ary. Dimana, setiap kolom disebut atribut. Setiap tabel diimplementasikan sebagai sebuah file. Satu baris menyatakan sebuah record, dan setiap atribut menyatakan sebuah field. Operasi yang dilakukan terhadap sebuah basis data dilakukan dengan perintah pertanyaan yang disebut query, contohnya: “tampilkan semua mahasiswa yang mengambil mata kuliah matematika” “tampilkan daftar nilai mahasiswa dengan NIM = 132450” Beberapa operasi yang digunakan dalam query,diantaranya: a. Seleksi Yaitu, memilih baris tertentu dari sebuah tabel. Operator ��. Contoh. Kita ingin menampilkan daftar mahasiswa yang mengambil matakuliah Matematika. Jawab. Operasi seleksi �������������� = “Matematika”(MHS) Menghasilkan (132450, Dono, Matematika) dan (132452, Dina, Matematika) b. Proyeksi Yaitu, memilih kolom tertentu dari sebuah tabel. Operator ��. Contoh Operasi proyeksi ����������,������������ (MHS) Menghasilkan Nama MatKul Dono Dono Dina Dina Dita Matematika Struktur Data Algoitma Matematika Algoritma c. Join Yaitu, menggabungkan dua tabel menjadi satu. Operator ��. Contoh Tabel 1 tabel 2 NIM Nama MatKul NIM Nama JK 132450 132450 132452 132452 132457 Dono Dono Dina Dina Dita Matematika Struktur Data Algoitma Matematika Algoritma 132450 132452 132457 Dono Dina Dita L P P Operasi join ��NIM, Nama(MHS1, MHS2) Menghasilkan NIM Nama MatKul JK 132450 132450 132452 132452 132457 Dono Dono Dina Dina Dita Matematika Struktur Data Algoitma Matematika Algoritma L L P P P Bagian Dua Fungsi Pertama Pengertian Fungsi Misalkan A dan B adalah himpunan. Relasi biner f dari A ke B merupakan suatu fungsi jika setiap elemen di dalam A dihubungkan dengan tepat satu elemen di dalam B. Definisi lengkap fungsi menjadi: set pasangan urut dengan anggota-anggota set X yang dinamakan wilayah, sebagai unsur pertama, dan anggota-anggota set Y, yang dinamakan jangkauan sebagai unsur kedua, yang dihubungkan dengan suatu kaidah yang demikian sehingga tidak ada pasangan urut yang unsur pertamanya sama. Fungsi dapat dispesifikasikan dalam berbagai bentuk, diantaranya: 1. Himpunan pasangan terurut Ingat bahwa fungsi adalah relasi, relasi biasanya dinyatakan sebagai himpunan pasangan terurut. 2. Kata-kata Fungsi dapat dinyatakan secara eksplisit dalam rangkaian kata-kata. Misalnya “f adalah fungsi yang memetakan jumlah bit 1 di dalam suatu string biner”. 3. Kode program (source code) Fungsi dispesifikasikan dalam bentuk kode program komputer. Misalnya dalam Bahasa Pascal, fungsi yang mengembalikan nilai mutlak dari sebuah bilangan bulat x, yaitu |x|, ditulis sbb: Function abs (x : integer) : integer; Begin If x < 0 then Abs := -x Else Abs := x ; End; 4. Formula pengisian nilai (assignment) Seperti y = x2 – 4 yang dapat dituliskan juga f(x) = x2 – 4. Contoh Misalkan f adalah fungsi dari x = {-2, 0, 1} yang didefinisikan oleh y = 2x2 + 3. Tentukan nilai fungsi tersebut Penyelesaian y = 2x2 + 3 artinya f(x) = 2x2 + 3 Untuk x = -2 maka f(-2) = 2(-2)2 + 3 = 11 x = 0 maka f(0) = 2(0)2 + 3 = 3 x = 1 maka f(1) = 2(1)2 + 3 = 5 Bergantung pada bayangan, fungsi dibedakan menjadi: 1. Fungsi satu-ke-satu (one-to-one) atau injektif ; jika tidak ada dua elemen himpunan A yang memiliki bayangan sama. 2. Fungsi pada (onto) atau surjektif ; jika setiap elemen himpunan B merupakan bayangan dari satu atau lebih elemen himpunan A. 3. Fungsi berkoresponden satu-ke-satu atau bijeksi ; jika ia merupakan fungsi satu-ke-satu dan juga fungsi pada. Kedua Fungsi Invers Jika f adalah fungsi berkoresponden satu-ke-satu dari A ke B,maka kita dapat menemukan balikan atau invers dari f . Fungsi invers dari f dilambangkan dengan f –1. Contoh Tentukan invers dari fungsi-fungsi berikut. a. f(x) = 2x – 1 penyelesaian f(x) = 2x – 1 y = 2x – 1 y + 1 = 2x �� + 12 = x f –1(x) = �� + 12 b. f(x) = x2 + 1 penyelesaian f(x) = x2 + 1 y = x2 + 1 y – 1 = x2 ��−1 = x f –1(x) = ��−1 Ketiga Komposisi Fungsi Karena fungsi merupakan bentuk khusus dari relasi, kita dapat melakukan komposisi dari dua buah fungsi. misalkan g adalah fungsi dari A ke B, dan f adalah fungsi dari B ke C. Komposisi f dan g, dinotasikan dengan f o g, adalah fungsi dari A ke C yang di definisikan oleh. �� �� �� (a) = f(g(a)) Diberikan fungsi f(x) = 2x – 1 dan g(x) = x2 + 1 . Tentukan �� �� �� dan �� �� ��. Penyelesaian (i) (�� �� ��)(x) = f(g(x)) = f(x2 + 1) = 2(x2 + 1) – 1 = 2x2 + 2 – 1 = 2x2 + 1 (i) (�� �� ��)(x) = g(f(x)) = g(2x – 1) = (2x – 1)2 + 1 = (4x2 – 4x + 1) + 1 = 4x2 – 4x + 2

Teman adalah Hidupku

Persahabatan sangatlah indah, pertemanan adalah suatu langkah untuk maju kedepan dan menjadi lebih baik, karena dari teman dan bersama temanlah mencari suatu pengalaman akan menjadi menyenangkan dan akan kita ingat seumur hidup, maka janganlah kita menghianati teman kita, jagalah selalu mereka.... "My Friends My Life".

Matematika Diskrit : INDUKSI MATEMATIKA

 Induksi Matematika merupakan suatu teknik yang dikembangkan untuk membuktikan pernyataan  Induksi Matematika digunakan untuk mengecek hasil proses yang terjadi secara berulang sesuai dengan pola tertentu  Induksi Matematika digunakan untuk membuktikan universal statements  n  A S(n) dengan A  N dan N adalah himpunan bilangan positif atau himpunan bilangan asli.  S(n) adalah fungsi propositional TAHAPAN INDUKSI MATEMATIKA  Basis Step : Tunjukkan bahwa S(1) benar  Inductive Step : Asumsikan S(k) benar Akan dibuktikan S(k)  S(k+1) benar  Conclusion : S(n) adalah benar untuk setiap n bilangan integer positif PEMBUKTIAN INDUKSI MATEMATIKA Contoh 1 : Buktikan bahwa : 1 + 2 + 3 + … + n = ½ n(n+1) untuk setiap n bilangan integer positif Jawab :  Basis : Untuk n = 1 akan diperoleh : 1 = ½ 1 . (1+1)  1 = 1  Induksi : misalkan untuk n = k asumsikan 1 + 2 + 3 + …+ k = ½ k (k+1)  adib. Untuk n = k+1 berlaku 1 + 2 + 3 + …+ (k+1) = ½ (k+1) (k+2) Jawab :  1 + 2 + 3 + …+ (k+1) = (k+1) (k+2) / 2 1 + 2 + 3 + …+ k + (k+1) = (k+1) (k+2) / 2 k (k+1) / 2 + (k+1) = (k+1) (k+2) / 2 (k+1) [ k/2 +1 ] = (k+1) (k+2) / 2 (k+1) ½ (k+2) = (k+1) (k+2) / 2 (k+1) (k+2) / 2 = (k+1) (k+2) / 2  Kesimpulan : 1 + 2 + 3 + …+ n = ½ n (n +1) Untuk setiap bilanga bulat positif n Contoh 2 : Buktikan bahwa : 1 + 3 + 5 + … + (2n - 1) = n2 untuk setiap n bilangan bulat positif Jawab :  Basis : Untuk n = 1 akan diperoleh : 1 = 12  1 = 1  Induksi : misalkan untuk n = k asumsikan 1 + 3 + 5 + …+ (2k – 1) = k2  adib. Untuk n = k + 1 berlaku 1 + 3 + 5 + …+ (2 (k + 1) – 1) = (k + 1)2 1 + 3 + 5 + …+ (2k + 1) = (k + 1)2 1 + 3 + 5 + …+ ((2k + 1) – 2) + (2k + 1) = (k + 1)2 1 + 3 + 5 + …+ (2k - 1) + (2k + 1 ) = (k + 1)2 k 2 + (2K + 1) = (k + 1)2 k 2 + 2K + 1 = k 2 + 2K + 1 Kesimpulan : 1 + 3 + 5 + … + n = (2n - 1) = n2 Untuk setiap bilangan bulat positif n Contoh 3 : Buktikan bahwa : n 3 + 2n adalah kelipatan 3 untuk setiap n bilangan bulat positif Jawab :  Basis : Untuk n = 1 akan diperoleh : 1 = 13 + 2(1)  1 = 3 , kelipatan 3  Induksi : misalkan untuk n = k asumsikan k 3 + 2k = 3x  adib. Untuk n = k + 1 berlaku (k + 1)3 + 2(k + 1) adalah kelipatan 3 (k 3 + 3k 2 + 3 k+1) + 2k + 2 (k 3 + 2k) + (3k 2 + 3k + 3) (k 3 + 2k) + 3 (k 2 + k + 1) Induksi 3x + 3 (k 2 + k + 1) 3 (x + k 2 + k + 1) Kesimpulan : n 3 + 2n adalah kelipatan 3 Untuk setiap bilangan bulat positif n

Kalkulus : Definite Integrals

Definite Integrals The formal definition of a definite integral is stated in terms of the limit of a Riemann sum. Riemann sums are covered in the calculus lectures and in the textbook. For simplicity's sake, we will use a more informal definiton for a definite integral. We will introduce the definite integral defined in terms of area. Let f(x) be a continuous function on the interval [a,b]. Consider the area bounded by the curve, the x-axis and the lines x=a and x=b. The area of the region that lies above the x-axis should be treated as a positive (+) value, while the area of the region that lies below the x-axis should be treated as a negative (-) value. The image below illustrates this concept. The positive area, above the x-axis, is shaded green and labelled "+", while the negative area, below the x-axis, is shaded red and labelled "-". The integral of the function f(x) from a to b is equal to the sum of the individual areas bounded by the function, the x-axis and the lines x=a and x=b. This integral is denoted by where f(x) is called the integrand, a is the lower limit and b is the upper limit. This type of integral is called a definite integral. When evaluated, a definite integral results in a real number. It is independent of the choice of sample points (x, f(x)). Properties of Definite Integrals The following properties are helpful when calculating definite integrals. Examples 1 | Evaluate the integral by finding the area beneath the curve ________________________________________ The Fundamental Theorem of Calculus The Fundamental Theorem of Calculus defines the relationship between the processes of differentiation and integration. That relationship is that differentiation and integration are inverse processes. The Fundamental Theorem of Calculus : Part 1 If f is a continuous function on [a,b], then the function denoted by is continuous on [a,b], differentiable on (a,b) and g'(x) = f(x). If f(t) is continuous on [a,b], the function g(x) that's equal to the the area bounded by the u-axis and the function f(u) and the lines u=a and u=x will be continuous on [a,b] and differentiable on (a,b). Most importantly, when we differentiate the function g(x), we will find that it is equal to f(x). The graph to the right illustrates the function f(u) and the area g(x). The Fundamental Theorem of Calculus : Part 2 If f is a continuous function on [a,b], then where F is any antiderivative of f. If f is continuous on [a,b], the definite integral with integrand f(x) and limits a and b is simply equal to the value of the antiderivative F(x) at b minus the value of F at a. This property allows us to easily solve definite integrals, if we can find the antiderivative function of the integrand. Parts one and two of the Fundamental Theorem of Calculus can be combined and simplified into one theorem. The Fundamental Theorem of Calculus Let f be a continuous function on [a,b]. ________________________________________ Indefinite Integrals An indefinite integral has the form When evaluated, an indefinite integral results in a function (or family of functions). An indefinite integral of a function f(x) is also known as the antiderivative of f. A function F is an antiderivative of f on an interval I, if F'(x) = f(x) for all x in I. This is a strong indication that that the processes of integration and differentiation are interconnected. Table of Indefinite Integrals The following tables list the formulas for antidifferentiation. These formulas allow us to determine the function that results from an indefinite integral. Since the formulas are for the most general indefinite integral, we add a constant C to each one. With these formulas and the Fundamental Theorem of Calculus, we can evaluate simple definite integrals. The next table lists indefinite integrals involving trigonometric functions. Note: After finding an indefinite integral, you can always check to see if your answer is correct. Since integration and differentiation are inverse processes, you can simply differentiate the function that results from integration, and see if it is equal to the integrand. Examples 2 | Find the general indefinite integrals 3 | Evaluate the definite integral 4 | Evaluate the definite integral of the absolute value of a function ________________________________________ The Total Change Theorem The total change theorem is an adaptation of the second part of the Fundamental Theorem of Calculus. The Total Change Theorem states: the integral of a rate of change is equal to the total change. If we know that the function f(x) is the derivative of some function F(x), then the definite integral of f(x) from a to b is equal to the change in the function F(x) from a to b. Examples 5 | Given the velocity function, find the displacement during a period of time ________________________________________ The Substitution Rule Suppose that we have an integral such as With our current knowledge of integration, we can't find the general equation of this indefinite integral. There are no antidifferentiation formulas for this type of integral. However, from our knowledge of differentiation, specifically the chain rule, we know that 4x3 is the derivative of the function within the square root, x4 + 7. We must also account for the chain rule when we are performing integration. To do this, we use the substitution rule. The Substitution Rule states: if u = g(x) is a differentiable function and f is continuous on the range of g, then Note: Recall that if u = g(x), then du = g'(x)dx. If we substitute u into the left side of the equation for g(x) and du for g'(x)dx, then we get the integral on the right side of the equation. From our previous example, if we let u = (x4+7), then du = 4x3dx. If we substitutite these values into the integral, we get an integral that can be solved using the antidifferentiation formulas. However, this answer is still in terms of u. We must substitute u = (x4+7) into the resulting function, so that it is a function of x, rather than u. The substitution rule also applies to definite integrals. The Substitution Rule for Definite Integrals states: If f is continuous on the range of u = g(x) and g'(x) is continuous on [a,b], then Examples 6 | Find the general indefinite integrals using the substitution rule 7 | Evaluate the definite integral using the substitution rule ________________________________________ Integrals of Symmetric Functions If f(x) is continuous on [-a, a] and f is an even function, then If f(x) is continuous on [-a, a] and f is an odd function, then These properties of integrals of symmetric functions are very helpful when solving integration problems. Some of the more challenging problems can be solved quite simply by using this property. Examples 8 | Evaluate the definite integral of the symmetric function ________________________________________ Integration By Parts Suppose that we have an integral such as Similar to integrals solved using the substitution method, there are no general equations for this indefinite integral. However there do not appear to be any clear substitutions that could be made to simplify this integral. This brings us to an integration technique known as integration by parts, which will call upon our knowledge of the Product Rule for differentiation. The Product Rule states: If f and g are differentiable functions, then By taking the indefinite integral of both sides of the equation we have: and we can rearrange this equation as To make it easier to remember it is commonly written in the following notation. Let u=f(x) and v=g(x). Then the differentiables are du=f'(x)dx and dv=g'(x)dx, so by the substitution rule, the formula for integration by parts becomes: From our previous example, if we let u=x and dv=cosx, then du=dx and v=sinx. If we substitute these values into the formula we have: Note: By choosing u=x we obtain a simpler integral than we started with. Had we chose u=cosx and dv=x then du=-sinx and v=(1/2)x2 so integration by parts gives: This equation is correct, but the integral is more difficult than the one we started with. When choosing u and dv always try to choose u=f(x) to be a function that becomes simpler when differentiated (or at least not more complicated) and to choose dv=g'(x) to be a function that can be easily integrated to give v. Examples 9 | Find the general indefinite integral by integration by parts 10 | Evaluate the definite integral by integration by parts ________________________________________ Trigonometric Integrals Suppose we have an integral such as The easy mistake is to simply make the substitution u=sinx, but then du=cosxdx. So in order to integrate powers of sine we need an extra cosx factor. Similarily, in order to integrate powers of cosine we need an extra sinx factor. Thus for this example knowing we need an extra sinx factor to integrate powers of cosine we can separate one sine factor and convert the remaining sin4x to an expression involving cosine using the identity sin2x + cos2x = 1. Now by using our knowledge of substitution we can evaluate the integral by letting u=cosx, then du=-sinxdx and Now consider the integral If we were to use the method from the previous example and separate one cosine factor we would be left with a factor of cosine of odd degree which isn't easily converted to sine. We must now consider the half angle formulas Using the half angle formula for cos2x, we have: Strategy for Evaluating (a) If the power of sine is odd (m=2k+1), save one sine factor and use the identity sin2x + cos2x = 1 to convert the remaining factors in terms of cosine. then substitute u=cosx. (b) If the power of cosine is odd (n=2k+1), save one cosine factor and use the identity sin2x + cos2x = 1 to convert the remaining factors in terms of sine. then substitute u=sinx. (c) If the powers of both sine and cosine are even then use the half angle identities. In some cases it may be helpful to use the identity 11 | Find the indefinite trigonometric integral 12 | Using the half angle formulas solve the indefinite trigonometric integral 13 | Find the definite trigonometric integral ________________________________________ Now that we have learned strategies for solving integrals with factors of sine and cosine we can use similar techniques to solve integrals with factors of tangent and secant. Using the identity sec2x = 1 + tan2x we are able to convert even powers of secant to tangent and vice versa. Now we will consider two examples to illustrate two common strategies used to solve integrals of the form Suppose we have an integral such as Observing that (d/dx)tanx=sec2x we can separate a factor of sec2x and still be left with an even power of secant. Using the identity sec2x = 1 + tan2x we can convert the remaining sec2x to an expression involving tangent. Thus we have: Then substitute u=tanx to obtain: Note: Suppose we tried to use the substitution u=secx, then du=secxtanxdx. When we separate out a factor of secxtanx we are left with an odd power of tangent which is not easily converted to secant. Consider the integral Since (d/dx)secx=secxtanx we can separate a factor of secxtanx and still be left with an even power of tangent which we can easily convert to an expression involving secant using the identity sec2x = 1 + tan2x. Thus we have: Then substitute u=secx to obtain: Note: Suppose we tried to use the substitution u=tanx, then du=sec2xdx. When we separate out a factor of sec2x we are left with an odd power of secant which is not easily converted to tangent. Strategy for Evaluating (a) If the power of secant is even (n=2k, k>2) save a factor of sec2x and use the identity sec2x = 1 + tan2x to express the remaining factors in terms of tanx. then substitute u=tanx. (b) If the power of tangent is odd (m=2k+1), save a factor of secxtanx and use the identity sec2x = 1 + tan2x to express the remaining factors in terms of secx. then substitute u=secx. Note: If the power of secant is even and the power of tangent is odd then either method will suffice, although there may be less work involved to use method (a) if the power of secant is smaller, and method (b) if the power of tangent is smaller. 14 | Find the indefinite trigonometric integral 15 | Find the definite trigonometric integral ________________________________________ Integrals of cotangent and cosecant are very similar to those with tangent and secant. it is easy to see that integrals of the form can be solved by nearly identical methods as are integrals of the form . 16 | Find the indefinite trigonometric integral ________________________________________ Unlike integrals with factors of both tangent and secant, integrals that have factors of only tangent, or only secant do not have a general strategy for solving. Use of trig identities, substitution and integration by parts are all commonly used to solve such integrals. For example, If we make the substitution u=secx, then du=secxtanxdx, and we are left with the simple integral Similarily we can use the same technique to solve 17 | Find the definite trigonometric integral 18 | Find the definite trigonometric integral 19 | Find the indefinite trigonometric integral ________________________________________ Another problem that may be encountered when solving trigonometric integrals are integrals of the form Using the product formulas which are deduced from the addition/subtraction rules we have the corresponding identities 20 | Find the indefinite trigonometric integral using the product formulas ________________________________________ Trigonometric Substitution Sometimes trigonometric substitutions are very effective even when at first it may not be so clear why such a substitution be made. For example, when finding the area of a circle or an ellipse you may have to find an integral of the form where a>0. It is difficult to make a substitution where the new variable is a function of the old one, (for example, had we made the substitution u = a2 - x2, then du= -2xdx, and we are unable to cancel out the -2x.) So we must consider a change in variables where the old variable is a function of the new one. This is where trigonometric identities are put to use. Suppose we change the variable from x to by making the substitution x = a sin θ. Then using the trig identity we can simplify the integral by eliminating the root sign. By changing x to a function with a different variable we are essentially using the The Substitution Rule in reverse. If x=g(t) then by restricting the boundaries on g we can assure that g has an inverse function; that is, g is one-to-one. In the example above we would require to assure has an inverse function. If we look at the Substitution Rule and replace u with x and x with t, we obtain This is known as the "inverse substitution". ________________________________________ Integration of Rational Functions By Partial Fractions Integration of rational functions by partial fractions is a fairly simple integrating technique used to simplify one rational function into two or more rational functions which are more easily integrated. Think back to the steps taken when adding or subtracting fractions that do not have the same denominator. First you find the lowest common multiple of the two denominators and then cross multiply with the numerators accordingly. eg. Well the same process applies when dealing with polynomial fractions. eg. Now by reversing this process we can simplify a function such as into two fractions which are more easily integrated. This process is possible when the function is proper; that is the degree of the numerator is less than the degree of the denominator. If the function is improper; that is the degree of the numerator is greater than or equal to the degree of the denominator, then we must first use long division to divide the denominator into the numerator until we obtain a remainder, such that it's degree is less than the denominator. Then if possible the above process is used to simplify the proper function. To complete some of the problems in this section it will be useful to know the table integral In general there are 4 cases to consider to express a rational function as the sum of two or more partial fractions. Case 1 The denominator is a product of distinct linear factors (no factor is repeated or a constant mulptiple of another). For example, Since the degree of the numerator is less than the degree of the denominator we don't need to divide. The denominator can be factored as follows: Since the denominator has distinct linear factors we can write the rational fraction as the sum of two or more partial fractions as follows: By multiplying both sides by we have: From this equation we can match terms of the same degree to determine the coefficients by solving the following system of equations: Case 2 The denominator is a product of linear functions, some of which are repeated. For example, Since the degree of the numerator is greater than the degree of the denominator we must factorize by long division. So we can now factor the denominator to obtain: Since the linear factor (x-2) occurs twice, the partial fraction decomposition is: When we multiply both sides by the least common denominator we get: From this equation we can match terms of the same degree to determine the coefficients by solving the following system of equations: Case 3 The denominator contains irreducible quadratic factors, none of which are repeated. When reducing such functions to partial fractions if there is a term in the denominator of the form ax2 + bx + c, where b2 - 4ac < 0, then the numerator for that partial fraction will be of the form Ax + B. For example, Since the degree of the numerator is less than the degree of the denominator we do not have to divide first. Since x3 + 4x = x(x2 + 4) can't be factored any further we have: multiplying both sides by x(x2 + 4), we have: From this equation we can match terms of the same degree to determine the coefficients by solving the following system of equations: Case 4 The denominator contains a repeated irreducible quadtratic factor. Functions of this form are the same as those in case 3 only there is a term in the denominator that is repeated or is a constant multiple of another. For example, If we were to expand the denominator we would see that its degree is greater than the the degree of the numerator so we do not have to divide first. Since the function cannot be factored any further we have: multiplying both sides by (x + 1)(x2 + 4)2, we have: From this equation we can match terms of the same degree to determine the coefficients by solving the following system of equations: ________________________________________ Improper Integrals In all of the previous tutorials we have dealt with integrals with a continous function f on a finite interval [a,b]. In this section we will consider two types of integrals known as improper integrals. The first type of improper integral are those defined on an infinite interval, and the second are those where the function f has an infinite discontinuity in [a,b]. Type 1: Infinite Intervals Type 2: Discontinous Integrands ________________________________________ For more practice with the concepts covered in this tutorial, visit the Integral Problems page at the link below. The solutions to the problems will be posted after these chapters are covered in your calculus course. To test your knowledge of integration problems, try taking the general integrals test on the iLrn website or the advanced integrals test at the link below. Integral Problems General Integrals Test on iLrn Advanced Integrals Test ________________________________________

Kalkulus : Tangents and Limits

Tangents and Limits A tangent to a curve is a straight line that touches the curve at a single point but does not intersect it at that point. For example, in the figure to the right, the y-axis would not be considered a tangent line because it intersects the curve at the origin. A secant to a curve is a straight line that intersects the curve at two or more points. In the figure to the right, the tangent line intersects the curve at a single point P but does not intersect the curve at P. The secant line intersects the curve at points P and Q. The concept of limits begins with the tangent line problem. We want to find the equation of the tangent line to the curve at the point P. To find this equation, we will need the slope of the tangent line. But how can we find the slope when we only know one point on the line? The answer is to look at the slope of the secant line. It's slope can be determined quite easily since there are two known points P and Q. As you slide the point Q along the curve, towards the point P, the slope of the secant line will become closer to the slope of the tangent line. Eventually, the point Q will be so close to P, that the slopes of the tangent and secant lines will be approximately equal. A limit of a function is written as : We want to find the limit of f(x) as x approaches a. To do this, we try to make the values of f(x) close to the limit L, by taking x values that are close to, but not equal to, a. In short, f(x) approaches L as x approaches a. Examples 1 | Find the equation of the tangent line to the curve at point P ________________________________________ Left and Right Hand Limits The previous example shows that the value a can be approached from both the left and right sides. Each side has its own limit. For example, as x approaches a from the right side we have and as x approaches a from the left we have . The graph to the right shows an example of a function with different right and left hand limits at the point x = 1. As x approaches 1 from the left side, the limit of f(x) approaches 1. As x approaches 1 from the right side however, the limit of f(x) approaches 4. In this case, the limit of f(x) as x approaches 1 does not exist, because the left and right hand limits do not approach the same value. This idea leads to the following theorem: For the limit to exist, the left and right hand limits must approach the same value. In our example, as x approaches 3, the left and right hand limits both approach a value of 4. Since the left and right hand limits are the same, the limit of f(x) as x approaches 3 exists and is equal to 4. Even though the actual value of f(3) is equal to 2, the limit is equal to 4. This gives the following theorem: Examples 2 | Evaluate the limit, if it exists Note: The example above uses basic concepts that are covered in the sections below. ________________________________________ Infinite Limits If a function is defined on either side of a, but the limit as x approaches a is infinity or negative infinity, then the function has an infinite limit. The graph of the function will have a vertical asymptote at a. A curve y=f(x) will have a vertical asymptote at x = a if any of the following conditions hold: Note: When dealing with rational functions, there will always be a vertical asymptote at values of x that make the denominator equal to 0. This is due to the fact that the function is undefined at points where the denominator is 0. However, you must still show that one of the conditions above holds true to prove that there is a vertical asymptote at that point. Examples 3 | Find the vertical asymptotes of the function ________________________________________ Laws of Limits Calculating limits using graphs and tables takes a lot of unnecessary time and work. Using the limit laws listed below, limits can be calculated much more quickly and easily. From the limit laws above, comes the property of direct substitution. This property makes it possible to solve most rational and polynomial functions. The property of direct substitution states: For any rational or polynomial function f, if a is in the domain of f then Often, the method of direct substitution cannot be used because a is not in the domain of f. In these cases, it is sometimes possible to factor the function and eliminate terms so that the function is defined at the point a. For an example of factoring, see example 6 below. Examples 4 | Evaluate the limit using limit laws 5 | Evaluate the limit using the property of direct substitution 6 | Evaluate the limit by factoring f(x) and eliminating terms ________________________________________ The Squeeze Theorem The squeeze theorem is an important concept that will be very helpful in upper year calculus courses. The squeeze theorem states: In simpler terms, the squeeze theorem states that if the graph of g is squeezed between the graphs of f and h when x is near a, and if f and h have the same limit L as x approaches a, then the limit of g as x approaches a is also L. The graph to the right illustrates the squeeze theorem. Note: The function g does not have to be completely contained between f and h. It must only be contained between f and h while x is near a. The graph illustrates where the functions f and g cross each other. Examples 7 | Evaluate the limit using the squeeze theorem ________________________________________ Continuity and Discontinuity A function f is continuous at a number a if: Note: When proving that a function is continuous, you may only show that the limit of f(x) as x approaches a is equal to f(a). This property implies that f(a) is defined and that the limit exists. A function is continuous on an interval if it is continuous at every number that falls within that interval. Continuous functions have the following properties for simple operations. If functions f and g are continuous at a, and c is a constant, then the following functions will also be continuous at a: A function is discontinuous at a if it is defined near a but not continuous at a. To prove that a function is discontinous, we must show which of the requirements of continuous functions that it fails to hold for. There are several different types of discontinuity, which are listed below. Removable discontinuity: A function has a removable discontinuity at a if the limit as x approaches a exists, but either f(a) is different from the limit or f(a) does not exist. It is called removable discontuniuity because the discontinuity can be removed by redefining the function so that it is continuous at a. In example #6 above, the function has a removable discontinuity at x = 3 because if the function is redefined so that f(3) = -4/7, it will be continuous at x = 3. Infinite discontinuity: A function has an infinite discontinuity at a if the limit as x approaches a is infinite. In example #3 above, the function has an infinite discontinuity at every point a = k*pi, since each point has an infinite limit. Jump discontinuity: A function has a jump discontinuity at a if the left- and right-hand limits as x approaches a exist, but are different. It is called jump discontinuity because the function jumps from the left-hand limit to the right-hand limit at each point. In example #2 above, the function has a jump discontinuity at x = 0, since the right and left hand limits approach different values. Note: Polynomial functions are continuous everywhere. Rational, root and trigonometric functions are continuous at every number in their domain. In other words, they are continuous wherever they are defined. Examples 8 | Determine where the function is continuous 9 | Explain why the function is discontinuous at each point ________________________________________ The Intermediate Value Theorem The intermediate value theorem states: If f is a continuous function on the closed interval [a, b] and N is a number between f(a) and f(b), where f(a) does not equal f(b), then there exists a number c in interval (a, b) such that f(c) = N. The intermediate value theorem simply says that if a function is continuous on a closed interval [a, b], and f(a) is different from f(b), then every value on the y-axis between f(a) and f(b) has a corresponding value on the x-axis between a and b. Note: As the graph to the right illustrates, there can be more than one value c in (a, b) for a number N betwen f(a) and f(b). The graph shows that there exist 3 numbers, c0, c1 and c2, such that f(c0)=f(c1)=f(c2)=N. The intermediate value theorem is helpful when proving that a root of a function exists in a certain interval. Examples 10 | Show that the function has a root between a and b ________________________________________ Tangents and Limits (Revisited) As explained at the beginning of this tutorial, a tangent to a curve is a line that touches the curve at a single point, P(a,f(a)). The tangent line T is the line through the point P with the slope : given that this limit exists. The graph to the right illustrates how the slope of the tangent line is derived. The slope of the secant line PQ is given by f(x)-f(a)/x-a. As x approaches a, the slope of PQ becomes closer to the slope of the tangent line T. If we take the limit of the slope of the secant line as x approaches a, it will be equal to the slope of the tangent line T. The slope of the tangent line becomes much easier to calculate if we consider the following conditions. If we let the distance between x and a be h, so that x=a+h, and substitute that equality for x in the slope formula, we get: Note: Either of the limit formulas above can be used to find the slope. You will obtain the same answer using either formula. These formulas have many practical applications. They can be used to find the instantaneous rates of change of variables. For example, if we use the formula above, the instantaneous velocity at time t=a is equal to the limit of f(a+h)-f(a)/h as h approaches 0. Examples 11 | Find the equation of the tangent line to the curve at the point P 12 | Find the instantaneous rate of change ________________________________________ For more practice with the concepts covered in the limits tutorial, visit the Limit Problems page at the link below. The solutions to the problems will be posted after the limits chapter is covered in your calculus course. To test your knowledge of limits, try taking the general limits test on the iLrn website or the advanced limits test at the link below. Limit Problems General Limit Test on iLrn Advanced Limit Test ________________________________________

A Personal Message from Friendster's CEO

Dear fellow Friendster members, As many of you may know, Friendster announced that it is re-launching itself as a social gaming portal and launched a beta version of the new Friendster a couple of weeks ago. The beta version was well received. I am pleased to announce that the new Friendster is going live thereby enabling all our users to login to the new Friendster using your existing Friendster username and password. Friendster has touched the lives of many. Since MOL, the company I founded acquired Friendster in early last year; many people have come up to me to tell me how Friendster has changed their lives. Many have told me that they have found their life partners over Friendster. Just last week, a successful Internet entrepreneur in Singapore told me that her success was triggered by promoting her business on Friendster. Friendster pioneered social networking and ignited the social media industry that has created billion dollar companies such as Facebook and Twitter, companies that may not have existed in their present form if not for Friendster's early innovation. Today, Friendster is in a unique position to take advantage on the growth of social gaming. Through its relationship with MOL, which has a 10 year history in working with gaming companies, Friendster has both the experience and track record to make innovations in this space. Today, as Friendster reinvents itself as a social gaming destination that enables its users to create multiple avatars, play games and enjoy rewards; I hope that all of you will wish us luck and continue to support us in our new reincarnation. The new Friendster is not perfect and we will continue to add new games and features such as localization and rewards over the next few months. Our team is working hard on adding these features and welcomes your suggestions and comments on how we can better serve your needs as a social gaming and entertainment destination. I would like to take this opportunity to thank all of you for your support and hope that all of you will enjoy the new Friendster as Friendster continues to innovate to serve and entertain you better. Yours truly, Ganesh Kumar Bangah Chief Executive Officer ceo@friendster.com