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Örnek soruda mutlak minimumun 0, mutlak maksimumun ise 2 olduğu bulunmuştur."]},"endTime":1603,"title":"Çok Değişkenli Fonksiyonlarda Kapalı Bölgede Mutlak Maksimum ve Minimum Bulma","beginTime":0}],"fullResult":[{"index":0,"title":"Çok Değişkenli Fonksiyonlarda Mutlak Maksimum ve Minimum","list":{"type":"unordered","items":["Çok değişkenli fonksiyonlarda kapalı bir bölge için mutlak maksimum ve minimum değerlerinin bulunması konusu ele alınacaktır.","İki değişkenli fonksiyonlar xy düzlemi üzerinde tanımlanır ve z koordinatı olarak sonuç verirler.","Fonksiyonlar sonuç olarak yüzeyler oluşturur ve mutlak maksimum-minimum değerleri bulmak için bir kapalı aralık tanımlanması gerekir."]},"beginTime":0,"endTime":116,"href":"/video/preview/10254658771907585367?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=0&ask_summarization=1"},{"index":1,"title":"Sınırlanmış Bölgede Maksimum ve Minimum Bulma","list":{"type":"unordered","items":["Girdileri belli bir bölge ile sınırladığımızda, bu bölge dikdörtgensel, üçgensel veya dairesel olabilir.","Sınırlanmış bölgede mutlak maksimum ve minimum değerleri, bölgenin içindeki noktalardan hangisinin en büyük veya en küçük değer verdiğini belirlemektir.","Dairesel bölgelerde mutlak maksimum-minimum soruları Lagrange çarpanı metodu ile çözülür ve bu soru tipi üniversite sınavlarında sıkça sorulur."]},"beginTime":116,"endTime":317,"href":"/video/preview/10254658771907585367?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=116&ask_summarization=1"},{"index":2,"title":"Çözüm Yöntemi","list":{"type":"unordered","items":["İlk adım olarak, bizi sınırlayan bölgeyi çizmek ve tüm sınır fonksiyonlarını belirlemektir.","İkinci adım, mutlak maksimum ve minimum adaylarını belirlemektir ve bunlar üç durumda bulunur:","Köşe koordinatları doğrudan adaydır.","Fonksiyonun kritik noktaları (gradyanı sıfır olan noktalar) içindeki adaylardır.","Sınır fonksiyonları üzerindeki kritik noktalar da adaydır."]},"beginTime":317,"endTime":648,"href":"/video/preview/10254658771907585367?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=317&ask_summarization=1"},{"index":3,"title":"Adayların Tespiti ve Sonuç","list":{"type":"unordered","items":["Sınır fonksiyonlarında kritik noktalar, sınır denklemi fonksiyona konularak tek değişkenli fonksiyon elde edilir, türevi alınır ve sıfıra eşitlenir.","Tüm adaylar fonksiyona konularak çıkan sonuçlardan en büyüğü mutlak maksimum, en küçüğü mutlak minimum değeridir.","Çözüm yöntemi, Calculus II derslerinin vazgeçilmez sorularından biridir ve özellikle üst düzey üniversitelerde sıkça sorulur."]},"beginTime":648,"endTime":796,"href":"/video/preview/10254658771907585367?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=648&ask_summarization=1"},{"index":4,"title":"Bölgenin Çizimi ve Sınırlar","list":{"type":"unordered","items":["Probleme ilk adım olarak bölgeyi çizmek gerekiyor, bu örnekte x=0, x=2, y=0 ve y=2 doğruları ile sınırlanan bir kare bölgesi.","Bölgenin tüm sınırlarının fonksiyonları biliniyor, bu nedenle tekrar bulmaya gerek yok."]},"beginTime":845,"endTime":916,"href":"/video/preview/10254658771907585367?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=845&ask_summarization=1"},{"index":5,"title":"Aday Noktaların Belirlenmesi","list":{"type":"unordered","items":["Köşeler aday noktalar olarak kabul edilir: (0,0), (2,0), (0,2) ve (2,2).","f(x,y) fonksiyonunun kritik noktaları da aday olarak değerlendirilir.","Kritik noktaları bulmak için f fonksiyonunun gradientini sıfıra eşitleriz: f_x = 2 - 2x = 0 ve f_y = 2 - 2y = 0, bu denklemlerden (1,1) noktası elde edilir."]},"beginTime":916,"endTime":1074,"href":"/video/preview/10254658771907585367?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=916&ask_summarization=1"},{"index":6,"title":"Sınır Fonksiyonları Üzerindeki Kritik Noktalar","list":{"type":"unordered","items":["Her sınır fonksiyonu üzerinde kritik noktaları bulmak için y=0, y=2, x=0 ve x=2 değerleri yerine konularak tek değişkenli fonksiyonlar elde edilir.","y=0 sınırında (1,0), y=2 sınırında (1,2), x=0 sınırında (0,1) ve x=2 sınırında (2,1) noktaları kritik noktalar olarak bulunur."]},"beginTime":1074,"endTime":1317,"href":"/video/preview/10254658771907585367?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=1074&ask_summarization=1"},{"index":7,"title":"Aday Noktaları Fonksiyonda Değerlendirme","list":{"type":"unordered","items":["Tüm aday noktalar f(x,y) = 2x + 2y - x² + y² fonksiyonunda değerlendirilir.","Değerlendirmeler sonucunda mutlak minimum 0 değerini (0,0), (2,0), (0,2) ve (2,2) noktalarında alırken, mutlak maksimum 2 değerini (1,1) noktasında alır."]},"beginTime":1317,"endTime":1553,"href":"/video/preview/10254658771907585367?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=1317&ask_summarization=1"},{"index":8,"title":"Çember veya Daire Bölgesi İçin Yaklaşım","list":{"type":"unordered","items":["Çember veya daire şeklinde bir bölge tanımlandığında Lagrange çarpanları metodu kullanılmalıdır.","Sınavlarda dairesel bir bölge tanımlandığında sınır fonksiyonu üzerinde Lagrange çarpanı ile çözüm istenir, fonksiyonsal çözüm istenmez."]},"beginTime":1553,"endTime":1599,"href":"/video/preview/10254658771907585367?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=1553&ask_summarization=1"}],"linkTemplate":"/video/preview/10254658771907585367?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=%%timestamp%%&ask_summarization=1"},"isAdultDoc":false,"relatedParams":{"text":"Calculus-II : Çok Değişkenli Fonksiyonlarda Mutlak Maksimum ve Minimum Bulma (Absolute Max-Min)","related_orig_text":"Calculus Booster","related_porno":false,"related_less_3m_off":true,"client":"d2d","no_cnt":1,"related_src":"serp","related":"{\"porno\":false,\"vfp\":1,\"orig_text\":\"Calculus 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Video, matematiksel kavramları grafikler üzerinden görselleştirerek anlatmakta ve bir sonraki videoda değişken değiştirme metodu ve rasyonel metot gibi konuların işleneceğini belirterek sona ermektedir."]},"endTime":1962,"title":"İntegral Dersi: Belirsiz İntegraller ve İlke Fonksiyonlar","beginTime":0}],"fullResult":[{"index":0,"title":"İntegral Serisi ve Belirsiz İntegral","list":{"type":"unordered","items":["Calculus serisi için integral konusunu anlatacak ve notları kendisi hazırlayarak hızlı bir şekilde ilerleyecek.","Belirsiz integralleri ve integrasyon metotlarını anlatacak, altı başlıkta inceleyecek ve ispatlı gidecek.","Videoda belirsiz integraller, ilkel fonksiyonlar ve türevle ilgili temel kavramlar ele alınacak."]},"beginTime":0,"endTime":52,"href":"/video/preview/8219283975777781596?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=0&ask_summarization=1"},{"index":1,"title":"İlke Fonksiyon 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türevlidir."]},"beginTime":220,"endTime":359,"href":"/video/preview/8219283975777781596?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=220&ask_summarization=1"},{"index":3,"title":"İlke Fonksiyonların Özellikleri","list":{"type":"unordered","items":["İki fonksiyonun aynı fonksiyonun ilkel fonksiyonları olması için, bu iki fonksiyonun farkının sabit bir fonksiyon olması gerekir.","Bir fonksiyonun ilkel fonksiyonuna sabit bir sayı eklendiğinde, elde edilen fonksiyon da aynı fonksiyonun ilkel fonksiyonudur.","F ve G fonksiyonları, f fonksiyonunun ilkeleri ise F(x) - G(x) = C (sabit) olur."]},"beginTime":359,"endTime":548,"href":"/video/preview/8219283975777781596?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=359&ask_summarization=1"},{"index":4,"title":"Belirsiz İntegral Tanımı","list":{"type":"unordered","items":["Belirsiz integral, bir fonksiyonun üzerinde tanımlı tüm ilkel fonksiyonların kümesini ifade eder.","Belirsiz integral, bir ifadenin türevi verilen fonksiyonu bulmayı amaçlar.","İntegral sembolü kullanılırken, integrand (entegre edilen ifade) ve integrasyon değişkeni belirtilir."]},"beginTime":551,"endTime":659,"href":"/video/preview/8219283975777781596?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=551&ask_summarization=1"},{"index":5,"title":"İntegrasyon Sabiti","list":{"type":"unordered","items":["İntegral hesaplarında C olarak gösterilen sabit, integrasyon sabiti olarak adlandırılır.","İntegrasyon sabiti, belirsiz integralin tüm ilkel fonksiyonları kapsamasını sağlar.","Her fonksiyonun belirsiz integralinin var olduğu anlamına gelmez, bu konu ileride incelenecektir."]},"beginTime":659,"endTime":692,"href":"/video/preview/8219283975777781596?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=659&ask_summarization=1"},{"index":6,"title":"Signum Fonksiyonunun İntegrali","list":{"type":"unordered","items":["Signum fonksiyonu, x>0 için 1, x=0 için 0, x\u003c0 için -1 değerlerini verir.","Signum fonksiyonunun eksi sonsuz ile sonsuz aralığındaki ilkel fonksiyonu yoktur çünkü sıfır noktasında türevi yoktur.","Sıfır noktasında sivri nokta olduğu için fonksiyon süreksiz olur ve bu nedenle ilkel fonksiyonu bulunamaz."]},"beginTime":692,"endTime":907,"href":"/video/preview/8219283975777781596?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=692&ask_summarization=1"},{"index":7,"title":"İntegral Özellikleri ve Diferansiyel","list":{"type":"unordered","items":["İntegralin türevi, orijinal fonksiyona eşittir.","Diferansiyel operatörü (d/dx) türevi temsil ederken, d(f(x)) ifadesi diferansiyeli temsil eder.","İntegral dışındaki diferansiyel, d(f(x)) = f'(x)dx şeklinde ifade edilir."]},"beginTime":907,"endTime":985,"href":"/video/preview/8219283975777781596?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=907&ask_summarization=1"},{"index":8,"title":"İntegral Kavramı ve Özellikleri","list":{"type":"unordered","items":["Diferansiyel integrali alınabilir ve diferansiyel efix'in integrali büyük fix artı c olarak bulunur.","İntegral alma sırasında alfa sıfırdan farklıysa, alfa integral dışına çıkabilir.","İntegral alma sırasında alfa artı beta çarpı gx gibi bir ifade varsa, alfa ve beta dışarı çıkarılabilir."]},"beginTime":993,"endTime":1117,"href":"/video/preview/8219283975777781596?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=993&ask_summarization=1"},{"index":9,"title":"İntegral Özellikleri ve Örnekler","list":{"type":"unordered","items":["a ile b arasında büyük eşit küçük ise, integral alınırken a katsayısı varsa bir bölü a katsayısı eklenmelidir.","Sinüs ve kosinüs fonksiyonlarının integrali, türevleri belli olan fonksiyonlarla ilişkilendirilerek bulunabilir.","İntegral sabiti olan c her zaman hesaplanırken dikkate alınmalıdır."]},"beginTime":1117,"endTime":1318,"href":"/video/preview/8219283975777781596?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=1117&ask_summarization=1"},{"index":10,"title":"Temel İntegral Formülleri","list":{"type":"unordered","items":["x üzeri alfa fonksiyonunun integrali, alfa artı bir kuvveti alıp alfa artı bir'e bölmekle bulunur, alfa eksi bir'den farklı olmak zorundadır.","1 bölü x dx integrali ln|x| artı c olarak bulunur.","a üzeri x fonksiyonunun integrali a üzeri x bölü ln(a) artı c olarak bulunur, a bir'den farklı olmak zorundadır."]},"beginTime":1318,"endTime":1471,"href":"/video/preview/8219283975777781596?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=1318&ask_summarization=1"},{"index":11,"title":"Trigonometrik Fonksiyonların İntegrali","list":{"type":"unordered","items":["Sinüs fonksiyonunun integrali eksi kosinüs x artı c olarak bulunur.","Kosinüs fonksiyonunun integrali sinüs x artı c olarak bulunur.","Tanjant fonksiyonunun integrali ln|sekant x| artı c olarak bulunur, kotanjant fonksiyonunun integrali eksi ln|sinüs x| artı c olarak bulunur."]},"beginTime":1471,"endTime":1615,"href":"/video/preview/8219283975777781596?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=1471&ask_summarization=1"},{"index":12,"title":"Hiperbolik Fonksiyonlar","list":{"type":"unordered","items":["Hiperbolik kosinüs fonksiyonu cosh(x) = (e^x + e^-x) / 2 olarak tanımlanır.","Hiperbolik sinüs fonksiyonu sinh(x) = (e^x - e^-x) / 2 olarak tanımlanır.","Tanjant ve kotanjant formülleri trigonometrik formüllerle aynı şekilde geçerlidir."]},"beginTime":1617,"endTime":1653,"href":"/video/preview/8219283975777781596?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=1617&ask_summarization=1"},{"index":13,"title":"Türevler ve İntegral Formülleri","list":{"type":"unordered","items":["Kosinüs x'in türevi sinüs x'tir, sinüs x'in türevi ise kosinüsteir.","Tanjant x'in türevi 1/cos²x veya sekant²x'tir.","Tanjant x'in integrali 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üniversite matematiği derslerinden calculus-2 dersine ait \"Çok Değişkenli Fonksiyonlarda Mutlak Max ve Min Bulma Örnek Soru-1\" videosudur. Hazırlayan: Kemal Duran (Matematik Öğretmeni)...","preview":{"posterSrc":"//avatars.mds.yandex.net/get-vthumb/3829577/bada03e826a2946215c69c560bde6d03/564x318_1","videoSrc":"https://video-preview.s3.yandex.net/L1fEDgEAAAA.mp4","videoType":"video/mp4"},"target":"_self","position":"8","reqid":"1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL","summary":{"isFull":true,"fullTextUrl":"/video/result?ask_summarization=1&numdoc=1&noreask=1&nomisspell=1&parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=videoid:11187584981951555564","teaser":[{"list":{"type":"unordered","items":["Bu video, bir eğitmen tarafından sunulan matematik eğitim içeriğidir. Eğitmen, çok değişkenli fonksiyonlarda mutlak maksimum ve minimum bulma konusunu anlatmaktadır.","Videoda, kapalı bir bölgede (örneğin bir üçgen) çok değişkenli fonksiyonların maksimum ve minimum değerlerinin nasıl bulunacağı adım adım gösterilmektedir. Eğitmen, bu tür soruları çözmek için dört adımlı bir yöntem sunmaktadır: kapalı bölgeyi çizme, köşe koordinatlarını bulma, kritik noktaları tespit etme ve sınır fonksiyonlarının üzerinde maksimum-minimum değerlerini arama.","Video, toplam beş örnek soru üzerinden konuyu pekiştirmeyi amaçlamakta ve özellikle sınır fonksiyonlarının üzerinde maksimum-minimum değerlerini bulma aşamasında karşılaşılan zorluklar ele alınmaktadır."]},"endTime":1205,"title":"Çok Değişkenli Fonksiyonlarda Mutlak Maksimum ve Minimum Bulma","beginTime":0}],"fullResult":[{"index":0,"title":"Çok Değişkenli Fonksiyonlarda Mutlak Maksimum ve Minimum Bulma","list":{"type":"unordered","items":["Çok değişkenli fonksiyonlarda mutlak maksimum ve minimum bulma konusunda örnek soru incelenecek.","Bir fonksiyonun maksimum ve minimum değerleri ancak ve ancak kapalı bir bölge ile tanımlandıysa sorulabilir.","Çok değişkenli fonksiyonlarda maksimum-minimum demek mutlak maksimum-minimumu kast etmektedir."]},"beginTime":1,"endTime":13,"href":"/video/preview/11187584981951555564?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=1&ask_summarization=1"},{"index":1,"title":"Örnek Sorunun Analizi","list":{"type":"unordered","items":["f(x,y) = x² - y² fonksiyonu verilmiş ve köşe koordinatları (0,0), (-1,1), (1,1) olan üçgen üzerindeki f(x,y) fonksiyonunun maksimum ve minimum değerleri bulunması isteniyor.","Maksimum-minimum değerlerini bulmak için önce kapalı bölgeyi çizmek gerekir.","Herhangi bir bölge ile sınırlanılmadığında mutlak maksimum-minimum sorulamaz, ancak yerel maksimum sorulabilir."]},"beginTime":13,"endTime":213,"href":"/video/preview/11187584981951555564?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=13&ask_summarization=1"},{"index":2,"title":"Maksimum ve Minimum Değerlerin Bulunması","list":{"type":"unordered","items":["Maksimum-minimum değerlerini bulmak için üç aday nokta vardır: köşe koordinatları, kritik noktalar ve sınır noktaları.","Köşe koordinatları tartışmasız maksimum-minimum adaylarıdır ve fonksiyonda yerleştirilerek değerleri hesaplanır.","Kritik noktalar, gradient f = 0 olan noktalar olup, bu örnekte (0,0) noktası kritik nokta olarak bulunmuştur."]},"beginTime":213,"endTime":606,"href":"/video/preview/11187584981951555564?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=213&ask_summarization=1"},{"index":3,"title":"Sınır Noktalarının İncelenmesi","list":{"type":"unordered","items":["Kapalı bölgenin sınırlarında da maksimum-minimum değer olabilir ve bu sorunun en zorlu bölümüdür.","Üçgenin sınırları için denklemler bulunur: y = x, y = -x ve y = 1.","Her sınır üzerinde maksimum-minimum adayı olup olmadığını tespit etmek için, sınır denklemleri fonksiyona yerleştirilir ve tek değişkenli fonksiyon olarak ele alınır."]},"beginTime":606,"endTime":821,"href":"/video/preview/11187584981951555564?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=606&ask_summarization=1"},{"index":4,"title":"Sonuçların Değerlendirilmesi","list":{"type":"unordered","items":["(1,1) noktası en küçük değer (-1) veren aday nokta olarak bulunmuştur.","y = x ve y = -x doğruları üzerindeki noktalar, köşe koordinatlarından biriyle aynı sonucu verdiği için yeni aday nokta getirmemiştir.","Tüm aday noktaların sonuçları karşılaştırılarak maksimum ve minimum değerler belirlenir."]},"beginTime":821,"endTime":996,"href":"/video/preview/11187584981951555564?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=821&ask_summarization=1"},{"index":5,"title":"Fonksiyonların Mutlak Maksimum ve Minimum Değerleri","list":{"type":"unordered","items":["Bir fonksiyonun mutlak maksimum ve minimum değeri, kapalı bir bölge ile tanımlandığı zaman sorulabilir.","Kapalı bölge bazen üçgen, dörtgen veya yamuksal bir bölge olabilir.","Kapalı bölgeyi çizdıktan sonra, bölgenin köşeleri mutlak maksimum ve minimum adaylarıdır."]},"beginTime":1034,"endTime":1152,"href":"/video/preview/11187584981951555564?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=1034&ask_summarization=1"},{"index":6,"title":"Maksimum ve Minimum Değerlerin Bulunması","list":{"type":"unordered","items":["Fonksiyonun kritik noktaları da mutlak maksimum ve minimum adaylarıdır.","Sınır bölgelerinin fonksiyonlarını doğru denklemlerini elde edip, türevi sıfır yapan noktalar da aday olur.","Tüm adayları kıyaslayarak en büyük çıkaran mutlak maksimum, en küçük çıkaran da mutlak minimum değeri 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Learn more math at https://TCMathAcademy.com/. Algebra 2 https://tabletclass-academy.teachable... Pre-Calculus https://tabletclass-academy.teachable... 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Video, başlangıç düzeyi seri testleri konusunda örnek soru çözümü yapmak isteyen öğrenciler için hazırlanmıştır."]},"endTime":1575,"title":"Seri Testleri Örnek Soru Çözümleri","beginTime":0}],"fullResult":[{"index":0,"title":"Seri Testleri ve Amacı","list":{"type":"unordered","items":["Video, başlangıç düzeyi seri testlerine ait örnek soru çözümlerini üst düzey olarak işleyecektir.","Seri testleri sekiz türden oluşur: n'inci terim testi, geometrik seri, p serisi, integral testi, kıyaslama testleri, oran testi, kök testi ve arterne seri testi.","Seri testlerinin amacı, verilen bir serinin yakınsak mı yoksa ıraksak mı olduğunu belirlemektir."]},"beginTime":1,"endTime":158,"href":"/video/preview/7721137960984409545?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=1&ask_summarization=1"},{"index":1,"title":"İlk Soru Çözümü","list":{"type":"unordered","items":["İlk soruda, 2'den sonsuza kadar (3n-2)/(2n+5) serisinin karakteri 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Soru Çözümü","list":{"type":"unordered","items":["Üçüncü soru, geometrik seri testi ile çözülür; geometrik serilerde ortak çarpan bulunur.","Geometrik seride ortak çarpan r\u003c1 ise seri yakınsak, r>1 ise seri ıraksaktır.","Sorudaki ortak çarpan 2/3 olduğundan, seri yakınsaktır."]},"beginTime":410,"endTime":524,"href":"/video/preview/7721137960984409545?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=410&ask_summarization=1"},{"index":4,"title":"Dördüncü Soru Çözümü","list":{"type":"unordered","items":["Dördüncü soru, integral testi ile çözülür; integral testi, serinin içindeki ifadenin integralini alabileceğimiz durumlarda uygulanır.","Integral testinde, 1'den sonsuza kadar integral alınır ve genelleştirilmiş integral (improper integral) olarak hesaplanır.","İntegralin sonucu sonsuz çıkarsa seri ıraksak, sayı çıkarsa seri yakınsaktır."]},"beginTime":524,"endTime":881,"href":"/video/preview/7721137960984409545?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=524&ask_summarization=1"},{"index":5,"title":"Beşinci ve Altıncu Soru Çözümü","list":{"type":"unordered","items":["Beşinci soru, p serisi testi ile çözülür; p=1/3 olduğundan seri ıraksaktır.","Altıncı soru, faktöriyel ve üst üstel ifadeler içeren seriler için oran testi ile çözülür.","Oran testinde, limit 1'den küçük çıkarsa seri yakınsak, 1'den büyük çıkarsa seri ıraksaktır, 1'e eşit çıkarsa test sonuç vermez."]},"beginTime":881,"endTime":1018,"href":"/video/preview/7721137960984409545?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=881&ask_summarization=1"},{"index":6,"title":"Oran Testi ve Faktöriyel","list":{"type":"unordered","items":["Oran testi için n+1 faktöriyel ifadesi n+1 çarpı n faktöriyel şeklinde yazılabilir ve n faktöriyeller birbirini götürür.","Limit n sonsuza giderken n+1 üzeri n bölü n üzeri n ifadesinin sonucu belirlenmesi zordur ve bu oran testini iptal etmek gerekir.","Faktöriyel içeren serilerde oran testi çözebilir ancak bu işlemler oldukça zordur."]},"beginTime":1046,"endTime":1116,"href":"/video/preview/7721137960984409545?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=1046&ask_summarization=1"},{"index":7,"title":"N'inci Terim Testi","list":{"type":"unordered","items":["Matematikte en hızlı büyüyen ifadeler n üzeri n'dir, ardından n faktöriyel gelir ve en son üst teller (5³ gibi) gelir.","Payın büyüme hızı faydadan büyük olduğunda sonuç sonsuz olur ve payın derecesi faydadan büyükse sonuç sonsuzdur.","Payın derecesi faydadan büyük olduğu için sonuç sonsuz etti ve bu seri ıraksak oldu."]},"beginTime":1116,"endTime":1209,"href":"/video/preview/7721137960984409545?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=1116&ask_summarization=1"},{"index":8,"title":"Kök Testi","list":{"type":"unordered","items":["Bir ifadenin üzerine en üste konulduysa bu kök testi ile çözülür ve n'inci dereceden kök içine atılır.","Kök testinde limitin sonucu bir'den küçük çıkarsa seri yakınsar, bir'den büyük çıkarsa ıraksaktır, bir'e eşit çıkarsa test işe yaramaz.","N'inci dereceden üs varsa n'inci dereceden kök içine atılır, iki'nci dereceden üs varsa iki'nci dereceden kök içine atılır."]},"beginTime":1209,"endTime":1326,"href":"/video/preview/7721137960984409545?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=1209&ask_summarization=1"},{"index":9,"title":"Alternatif Seri Testi","list":{"type":"unordered","items":["Eksi bir üzeri n içeren seriler alternatif seridir ve alternatif seri testi uygulanır.","Alternatif seri testinin üç şartı vardır: tüm terimler pozitif olmalı, terimler küçülüp küçülmediği kontrol edilmeli ve limit sonsuza giderken sıfır olmalı.","Eksi bir üzeri n içeren bir seri ile karşılaşıldığında eksi bir üzeri n kısmı ve diğer kısmı iki ayrı parça şeklinde ayırmak gerekir, bu üç şartı sağlarsa seri yakınsak olur."]},"beginTime":1326,"endTime":1567,"href":"/video/preview/7721137960984409545?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=1326&ask_summarization=1"}],"linkTemplate":"/video/preview/7721137960984409545?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=%%timestamp%%&ask_summarization=1"},"isAdultDoc":false,"relatedParams":{"text":"Calculus-II : Seri Testleri Örnek Soru Çözümleri (Başlangıç Düzeyi) (www.buders.com)","related_orig_text":"Calculus 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İlk bölümde tanjant ve sekant fonksiyonlarının integral hesaplamaları, ikinci bölümde ise sinüs ve kosinüs fonksiyonlarının integral hesaplamaları detaylı olarak gösterilmektedir. Eğitmen, değişken değiştirme yöntemi kullanarak çeşitli integral sorularını çözmekte ve her çözüm için üçgenler kullanarak trigonometrik ilişkileri açıklamaktadır.","Video, temel integral alma yöntemlerinin birincisi olan değişken değiştirme metodu kapsamında hazırlanmış olup, 9'dan 16'ya kadar olan sorular çözülmektedir. Eğitmen, tanjant dönüşümü, kotanjant dönüşümü ve sekant kare x integralinin çözümünü adım adım göstermektedir."]},"endTime":1149,"title":"İntegral Hesaplama Dersi","beginTime":0}],"fullResult":[{"index":0,"title":"Trigonometrik İntegral Çözümü","list":{"type":"unordered","items":["Bir trigonometrik integral sorusunda a² + x² ifadesi karekök dışına çıkarılabilir.","Tanjant dönüşümü yaparak tanjant u = x/a şeklinde bir üçgen oluşturulur.","İntegral sekant²u du ifadesi, sekantın türevi olan tanjant çarpı sekant kullanılarak çözülür."]},"beginTime":0,"endTime":292,"href":"/video/preview/15684713814669680840?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=0&ask_summarization=1"},{"index":1,"title":"İntegral Sonucu","list":{"type":"unordered","items":["İntegral sonucu e^|sekant u + tanjant u| + C olarak bulunur.","Üçgen kullanılarak sekant u = x/a ve tanjant u = x/a + C olarak ifade edilir.","Sonuç olarak integral x/(a² + x²) dx = e^(x/a + (x²-a²)/a) + C olarak elde edilir."]},"beginTime":292,"endTime":350,"href":"/video/preview/15684713814669680840?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=292&ask_summarization=1"},{"index":2,"title":"İkinci İntegral Sorusu","list":{"type":"unordered","items":["İkinci integral sorusunda payda x² - a² şeklinde bir ifade bulunur.","Sekant dönüşümü yaparak sekant u = x/a şeklinde bir üçgen oluşturulur.","İntegral sonucu e^|sekant u + tanjant u| + C olarak bulunur ve üçgen kullanılarak sekant u = x/a ve tanjant u = (x²-a²)/a olarak ifade edilir."]},"beginTime":350,"endTime":572,"href":"/video/preview/15684713814669680840?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=350&ask_summarization=1"},{"index":3,"title":"Üçüncü İntegral Sorusu","list":{"type":"unordered","items":["Üçüncü integral sorusunda sinx/x ifadesi bulunur.","İntegral, kosinüs ile çarpıp bölerek sekant²x/tanjant x şeklinde dönüştürülür.","Sonuç olarak integral tanjant x + C olarak bulunur ve türevi alınarak doğruluğu kontrol edilir."]},"beginTime":572,"endTime":753,"href":"/video/preview/15684713814669680840?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=572&ask_summarization=1"},{"index":4,"title":"İntegral Alma Yöntemleri","list":{"type":"unordered","items":["Kosinüs ve sinüs fonksiyonları için integral alma yöntemleri ele alınıyor.","Sinüs x = u dönüşümü uygulandığında, kosinüs x dx ifadesi u'nun kareköküne bölünerek integral hesaplanıyor.","Tanjant dönüşümü uygulandığında, tanjant a = u dönüşümü yapılarak integral hesaplanıyor."]},"beginTime":755,"endTime":813,"href":"/video/preview/15684713814669680840?parent-reqid=1773749963183315-6029769923636286936-balancer-l7leveler-kubr-yp-vla-263-BAL&text=Calculus+Booster&t=755&ask_summarization=1"},{"index":5,"title":"Sekant ve Kosekant 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