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Öğretmen, Ergin Hoca olarak da bahsedilen bir kişinin sorularını çözmektedir.","Videoda, belirli integralin tanımı kullanılarak Riemann toplamları ile ilgili sorular çözülmektedir. Öğretmen önce birinci dereceden bir polinom (2x+1) ile başlayıp, ardından ikinci dereceden (1+x+x²) ve üçüncü dereceden (x³-3x+2) polinomlarla örnekler vermektedir. Her soruda limit kavramı kullanılarak toplamlar hesaplanmakta ve sonuçlar integral hesaplamalarıyla kontrol edilmektedir. Video, sınavlarda çıkabilecek soru tiplerini göstermekte ve özellikle toplam sembolleri ile ilgili temel bilgileri içermektedir."]},"endTime":641,"title":"Belirli İntegral ve Riemann Toplamları Dersi","beginTime":0}],"fullResult":[{"index":0,"title":"Riemann Toplamları ve İntegral Tanımı","list":{"type":"unordered","items":["Toplamlarla ilgili sorulara devam ediliyor ve belirli integral tanımı kullanılarak soru sayısı sınırlı olduğu belirtiliyor.","Riemann toplamı formülü: a'dan b'ye f(x) fonksiyonunun limiti, n sonsuza giderken (b-a)/n çarpı toplam k=1'den n'e kadar f(a+k)(b-a) şeklinde yazılır.","Fonksiyonun cinsine göre k-1 yerine k kullanmak işlemlerin kısalığı bakımından daha uygun olur."]},"beginTime":7,"endTime":79,"href":"/video/preview/7947871602571965230?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=7&ask_summarization=1"},{"index":1,"title":"Birinci Örnek Soru Çözümü","list":{"type":"unordered","items":["Örnek 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Eğitmen, Ergin Hoca olarak bahsedilen bir matematik öğretmeninin öğretilerini aktarmaktadır.","Videoda, bir fonksiyonun kritik noktalarının nasıl inceleneceği adım adım anlatılmaktadır. Önce fonksiyonun tanım kümesi belirlenir, ardından kritik noktalar (sıfır yapan ve tanımsız yapan değerler) bulunur. Eğitmen, birinci türev testi ve ikinci türev testi yöntemlerini karşılaştırarak, ikinci türev testinin bazı durumlarda çalışmayabileceğini ve bu durumlarda birinci türev testinin işaret değiştirme yöntemiyle bölgesel maksimum ve minimum noktalarının nasıl bulunabileceğini göstermektedir. Video, bir sonraki videoda son noktada ekstra durumların nasıl inceleneceği anlatılacağı bilgisiyle sonlanmaktadır."]},"endTime":496,"title":"Matematik Dersinde Fonksiyonun Kritik Noktalarının İncelenmesi","beginTime":0}],"fullResult":[{"index":0,"title":"Fonksiyonun Tanım Kümesi","list":{"type":"unordered","items":["Verilen fonksiyon f(x) = 5x²/³ - 2x⁵/³'tür.","Fonksiyonun tanım kümesi tüm reel sayılar (R) çünkü küpkök fonksiyonu negatif ve pozitif değerler için tanımlıdır."]},"beginTime":7,"endTime":76,"href":"/video/preview/18208541705028303073?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=7&ask_summarization=1"},{"index":1,"title":"Birinci Türev ve Kritik Noktalar","list":{"type":"unordered","items":["Fonksiyonun birinci türevi f'(x) = (1/3)(10x⁻¹/³ - 10x²/³) şeklindedir.","Birinci türev sıfır yapan değer x = 1'dir ve tanımsız yapan değer x = 0'dır.","Bu iki değer fonksiyonun kritik noktalarını oluşturur."]},"beginTime":76,"endTime":177,"href":"/video/preview/18208541705028303073?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=76&ask_summarization=1"},{"index":2,"title":"İkinci Türev Testi","list":{"type":"unordered","items":["İkinci türev f''(x) = (-10/9)x⁻⁴/³ - 20x⁻¹/³ olarak hesaplanır.","x = 1 noktasında ikinci türev f''(1) = -10/3 \u003c 0 olduğundan, bu nokta bölgesel maksimumdur.","x = 0 noktasında ikinci türev tanımsız olduğundan, ikinci türev testi bu nokta için çalışmadığı için birinci türev testi kullanılmalıdır."]},"beginTime":177,"endTime":379,"href":"/video/preview/18208541705028303073?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=177&ask_summarization=1"},{"index":3,"title":"Birinci Türev Testi","list":{"type":"unordered","items":["Birinci türevin işaret değişimini inceleyerek kritik noktaların doğasını belirleriz.","x = 0 noktasında fonksiyon solunda azalıp sağında arttığından, bu nokta bölgesel minimumdur.","x = 1 noktasında fonksiyon solunda artıp sağında azaldığından, bu nokta bölgesel maksimumdur."]},"beginTime":379,"endTime":496,"href":"/video/preview/18208541705028303073?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=379&ask_summarization=1"}],"linkTemplate":"/video/preview/18208541705028303073?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=%%timestamp%%&ask_summarization=1"},"isAdultDoc":false,"relatedParams":{"text":"25) Matematik 1(Calculus 1) - 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Hazırlayan: Kemal Duran (Matematik Öğretmeni)...","preview":{"posterSrc":"//avatars.mds.yandex.net/get-vthumb/222140/bada03e826a2946215c69c560bde6d03/564x318_1","videoSrc":"https://video-preview.s3.yandex.net/AEJjAAAAAAA.mp4","videoType":"video/mp4"},"target":"_self","position":"6","reqid":"1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL","summary":{"isFull":true,"fullTextUrl":"/video/result?ask_summarization=1&numdoc=1&noreask=1&nomisspell=1&parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=videoid:12476430028588655811","teaser":[{"list":{"type":"unordered","items":["Bu video, matematik eğitimi formatında bir ders anlatımıdır. Eğitmen, çok değişkenli fonksiyonlarda yerel maksimum, minimum ve eğer noktalarını bulma konusunu örnek üzerinden açıklamaktadır.","Video, önce konu anlatımının yapıldığını belirterek başlıyor ve ardından sınavlarda sorulabilecek kalitede bir örnek üzerinden çözüm sürecini adım adım gösteriyor. Örnek, f(x,y) = 9x³ + 4xy fonksiyonunun yerel maksimum, minimum ve eğer noktalarını bulmayı içeriyor. Çözüm üç adımda ilerliyor: ilk olarak gradient'i sıfıra eşitleyerek kritik noktaları bulma, ikinci olarak d değerini hesaplama ve son olarak bu noktaları sınıflandırma. Video, sınavlarda bu konuyu öğrenmek isteyenler için faydalı bir kaynak niteliğindedir."]},"endTime":618,"title":"Çok Değişkenli Fonksiyonlarda Yerel Maksimum, Minimum ve Eger Noktası Bulma","beginTime":0}],"fullResult":[{"index":0,"title":"Giriş ve Soru Tanıtımı","list":{"type":"unordered","items":["Video, yerel maksimum, yerel minimum ve eğer noktası bulma konusunu ele alıyor.","Konu anlatım videosunu izlememiş kişilerin önce o videoyu izlemesi öneriliyor.","Soru, f(x,y) = 9x³ + y⁴ - 4xy fonksiyonunun yerel maksimum, yerel minimum ve eğer noktalarını bulmayı gerektiriyor."]},"beginTime":1,"endTime":54,"href":"/video/preview/12476430028588655811?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=1&ask_summarization=1"},{"index":1,"title":"Kritik Noktaları Bulma","list":{"type":"unordered","items":["Çok değişkenli fonksiyonlarda yerel maksimum, yerel minimum ve eğer noktalarını bulmak için ilk adım gradient'in sıfıra eşitlenmesidir.","Gradient, f fonksiyonunun x'e ve y'ye göre kısmi türevlerinden oluşur.","Kritik noktalar, fx ve fy türevlerini sıfıra eşitlerken bulunan (x,y) çiftleridir."]},"beginTime":54,"endTime":143,"href":"/video/preview/12476430028588655811?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=54&ask_summarization=1"},{"index":2,"title":"Denklemlerin Çözümü","list":{"type":"unordered","items":["Denklemlerin birinde x veya y'yi yalnız bırakıp diğer denklemde yerine koyarak nokta çiftleri bulunur.","Çözüm sürecinde x = 0 ve x = 4/9 değerleri bulunur.","Bu değerler kullanılarak (0,0) ve (4/9, 4/3) kritik noktaları elde edilir."]},"beginTime":143,"endTime":377,"href":"/video/preview/12476430028588655811?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=143&ask_summarization=1"},{"index":3,"title":"Sınıflama ve Sonuç","list":{"type":"unordered","items":["Kritik 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Eğitmen, önceki videolarda belirsiz integralin tanımını ve özelliklerini hatırlatarak başlayıp, değişken değiştirme yönteminin temel prensiplerini açıklamaktadır. Video, 20 tane kendi hazırladığı soru üzerinden ilerlemekte olup, t² = 2t, arctan(t)¹⁰0, (5t-6)ⁿ⁰⁰, 4x tan(x²) gibi çeşitli fonksiyonların integral hesaplamalarını örneklerle göstermektedir.","Videoda ayrıca tanjant, sekant, kosinüs gibi fonksiyonların integral hesaplamaları, sinüs ve kosinüs dönüşümleri, rasyonel kesirlerin integralini hesaplama teknikleri ve rasyonel kesirleri basit kesirlere ayırma yöntemi gibi konular da ele alınmaktadır. Eğitmen, türev bilgisinin bu metotlarda önemli bir rol oynayacağını vurgulamaktadır."]},"endTime":1695,"title":"Belirsiz İntegral: Değişken Değiştirme Yöntemi","beginTime":0}],"fullResult":[{"index":0,"title":"Belirsiz İntegral ve Değişken Değiştirme Yöntemi","list":{"type":"unordered","items":["Bu video, belirsiz integralin üçüncü videosu olup, ilk iki videoda tanımı ve özellikleri, türevleri hatırlatılmıştır.","Değişken değiştirme metodu ve diğer metotlarda türev bilgisi önemli bir rol oynayacağı için türevlerin iyi hatırlanması gerekmektedir.","Videoda 20 tane kendi hazırladığı soru bulunmaktadır ve yeterli olmazsa ekstra PDF olarak sorular paylaşılacaktır."]},"beginTime":0,"endTime":44,"href":"/video/preview/457903219051232553?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=0&ask_summarization=1"},{"index":1,"title":"Değişken Değiştirme Yönteminin Tanımı","list":{"type":"unordered","items":["Bugün değişken değiştirme yöntemi konuşulacaktır.","Değişken değiştirme yönteminin tanımı, J aralığı üzerinde tanımlı ve sürekli türeve sahip bir f fonksiyonu ile R kümesi üzerinden açıklanmaktadır.","f fonksiyonunun kümesinin f fonksiyonu altındaki görüntüsü belirtilmektedir."]},"beginTime":44,"endTime":90,"href":"/video/preview/457903219051232553?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=44&ask_summarization=1"},{"index":2,"title":"İntegral Özellikleri ve Türev İlişkisi","list":{"type":"unordered","items":["Büyük F fonksiyonu R kümesine gidiyor çünkü kümesi FJ idi ve J R'ye gidiyordu.","Küçük f fonksiyonunun ilkeli üzerinde herhangi bir ilkel fonksiyon olsun, f bileşke f şeklinde yazılırsa yine R'ye gitmek zorundadır.","İntegral e-fix çarpı fx'in integrali büyük F'dir çünkü küçük f'nin ilkeli büyük F'dir."]},"beginTime":92,"endTime":234,"href":"/video/preview/457903219051232553?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=92&ask_summarization=1"},{"index":3,"title":"Değişken Değiştirme Yöntemi","list":{"type":"unordered","items":["İntegralde değişken değiştirme yaparken, içeride türevi olan ifadeye bir değişken dersek ve diferansiyel ile birlikte kullanırız.","Birvirgülon formülünün uygulamalarında çıkan sonucu değişken değiştirmeden önceki şekilde ifade etmek gerektiğinden, integral bulunduktan sonra y=f(x) dönüşümü yardımıyla tekrar x değişkenine dönüştürmemeliyiz.","Değişken değiştirme yapabilmek için f(x) fonksiyonunun tersi olan x=f⁻¹(x) dönüşümünün mevcut olması gerekir."]},"beginTime":234,"endTime":377,"href":"/video/preview/457903219051232553?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=234&ask_summarization=1"},{"index":4,"title":"Değişken Değiştirme Örnekleri","list":{"type":"unordered","items":["İntegralde değişken değiştirme yaparken, türevini içeride görünecek şekilde bir değişken seçilmelidir.","Örnek 1: ∫t² dt integrali için t²=u dönüşümü yapılarak ∫u du = ½u² + C sonucuna ulaşılır.","Örnek 2: ∫(1+t)¹⁰⁰ dt integrali için t=u dönüşümü yapılarak ∫u¹⁰⁰ du = u¹⁰¹/101 + C sonucuna ulaşılır.","Örnek 3: ∫(5t-6)²⁰⁰⁰ dt integrali için 5t-6 = u dönüşümü yapılarak ∫u²⁰⁰⁰ du = (5t-6)²⁰⁰⁰/10000 + C sonucuna ulaşılır.","Örnek 4: ∫4x tanjant x² dx integrali için x²=u dönüşümü yapılarak ∫tanjant u du sonucuna ulaşılır."]},"beginTime":377,"endTime":652,"href":"/video/preview/457903219051232553?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=377&ask_summarization=1"},{"index":5,"title":"Değişken Değiştirme Yöntemi","list":{"type":"unordered","items":["Tanjant fonksiyonunun integrali değişken değiştirme yöntemiyle çözülebilir.","Tanjant x = sinüs x bölü kosinüs x şeklinde yazılabilir ve kosinüs u = k dönüşümü uygulanabilir.","Türev alınarak -sinüs u = du ilişkisi elde edilir ve integral -2 ln |cos u| + C şeklinde çözülür."]},"beginTime":654,"endTime":732,"href":"/video/preview/457903219051232553?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=654&ask_summarization=1"},{"index":6,"title":"Tanjantın İntegrali","list":{"type":"unordered","items":["Tanjant x'in integrali, sinüs x bölü kosinüs x şeklinde yazıldığında -ln |cos x| + C olarak bulunur.","Türev alınarak -cos x'in türevi sinüs x olduğu ve paydasını kosinüs x yazdığımızda tanjant elde edilir.","İntegral çözümlerinde şablon yerine teknik ve mantığı anlamak önemlidir."]},"beginTime":732,"endTime":792,"href":"/video/preview/457903219051232553?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=732&ask_summarization=1"},{"index":7,"title":"Sekant ve Tanjant İntegrali","list":{"type":"unordered","items":["Sekant kare x ve e üzeri tanjant x içeren integralde tanjant x = u dönüşümü uygulanabilir.","Tanjant x'in türevi sekant x dx olduğundan, integral e üzeri tanjant x + C şeklinde çözülür.","Türev alınarak e üzeri tanjant x'in türevi tanjant x çarpı e üzeri x olduğu doğrulanır."]},"beginTime":792,"endTime":838,"href":"/video/preview/457903219051232553?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=792&ask_summarization=1"},{"index":8,"title":"Kosinüs İntegrali","list":{"type":"unordered","items":["Kosinüs beş x'in integrali için cos beş x = u dönüşümü uygulanabilir.","Kosinüs beş x'in türevi -5 sinüs beş x dx olduğundan, integral -1/5 sinüs beş x üzeri altı + C şeklinde çözülür.","Sonuç olarak -1/30 cos altmış beş x + C olarak bulunur."]},"beginTime":838,"endTime":910,"href":"/video/preview/457903219051232553?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=838&ask_summarization=1"},{"index":9,"title":"Karmaşık İntegral Örnekleri","list":{"type":"unordered","items":["Bazı karmaşık integrallerde birden fazla değişken değiştirme gerekebilir.","t üzeri beş bölü üç üzeri on iki artı bir integralinde t üzeri beş = u dönüşümü uygulanabilir.","Sonuç olarak 1/18 tanjant u artı C şeklinde çözülür ve u yerine 3 üzeri beş t üzeri altı artı C yazılır."]},"beginTime":910,"endTime":1077,"href":"/video/preview/457903219051232553?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=910&ask_summarization=1"},{"index":10,"title":"Genel İntegral Formülü","list":{"type":"unordered","items":["t üzeri k bölü a üzeri k integralinde k=1 durumu farklı bir durumdur.","k=1 durumunda integral dt bölü t eksi a artı C şeklinde çözülür.","k>1 durumunda integral eksi bir bölü k eksi bir çarpı a üzeri k eksi bir artı C şeklinde çözülür."]},"beginTime":1077,"endTime":1207,"href":"/video/preview/457903219051232553?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=1077&ask_summarization=1"},{"index":11,"title":"İntegral Hesaplama Yöntemleri","list":{"type":"unordered","items":["İntegral hesaplamalarında dt bölü a kare formülü çok önemlidir ve ileride formüller olarak çıkacaktır.","Sinüs ve kosinüs formüllerini ezberlemek yerine, bilinen formülleri kullanarak hesaplamaları yapmak gerekir.","İntegralde dx bölü bir artı x kare ifadesi tanjant x'e eşittir."]},"beginTime":1210,"endTime":1245,"href":"/video/preview/457903219051232553?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=1210&ask_summarization=1"},{"index":12,"title":"İntegral Hesaplama Örneği","list":{"type":"unordered","items":["İntegral hesaplamasında değişken değiştirme yöntemi kullanılır: t yerine a çarpı u yazılır ve dt yerine au yazılır.","İntegral hesaplaması sonucunda bir bölü a çarpı ark tanjant u artı c ifadesi elde edilir.","İntegral hesaplamasında a'nın mutlak değeri küçüktür bir koşulu önemlidir."]},"beginTime":1245,"endTime":1335,"href":"/video/preview/457903219051232553?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=1245&ask_summarization=1"},{"index":13,"title":"İkinci İntegral Örneği","list":{"type":"unordered","items":["İntegral hesaplamasında dx bölü kök a kare eksi x bölü a kare ifadesi kullanılır.","Değişken değiştirme yöntemi uygulanarak integral hesaplaması yapılır.","İntegral hesaplaması sonucunda aksinüs u artı c ifadesi elde edilir."]},"beginTime":1335,"endTime":1442,"href":"/video/preview/457903219051232553?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=1335&ask_summarization=1"},{"index":14,"title":"Rasyonel Kesirleri Ayırma Yöntemi","list":{"type":"unordered","items":["İntegral hesaplamasında dx bölü a kare eksi x kare ifadesi rasyonel kesirleri ayırma metoduyla çözülür.","İntegral hesaplamasında x yerine a çarpı u dönüşümü uygulanır.","İntegral hesaplaması sonucunda eksi bir bölü a çarpı mutlak değer x eksi bir eksi mutlak değer x artı bir artı c ifadesi elde edilir."]},"beginTime":1442,"endTime":1686,"href":"/video/preview/457903219051232553?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=1442&ask_summarization=1"}],"linkTemplate":"/video/preview/457903219051232553?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=%%timestamp%%&ask_summarization=1"},"isAdultDoc":false,"relatedParams":{"text":"Calculus-1 : Değişken Değiştirme Yöntemi 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Calculus-1 : Metotlar Öncesi En Temel İntegraller -2: • Calculus-1 : Metotlar Öncesi...","preview":{"posterSrc":"//avatars.mds.yandex.net/get-vthumb/1022071/57bb29e83bf51cb5838d73d80e712bd4/564x318_1","videoSrc":"https://video-preview.s3.yandex.net/-It2CgIAAAA.mp4","videoType":"video/mp4"},"target":"_self","position":"12","reqid":"1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL","summary":{"isFull":true,"fullTextUrl":"/video/result?ask_summarization=1&numdoc=1&noreask=1&nomisspell=1&parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=videoid:15684713814669680840","teaser":[{"list":{"type":"unordered","items":["Bu video, bir matematik eğitmeni tarafından sunulan integral hesaplama dersidir. Eğitmen, tahtada veya dijital bir platformda matematiksel işlemler yaparak konuyu anlatmaktadır.","Videoda trigonometrik fonksiyonların integral hesaplamaları ele alınmaktadır. 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=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&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=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&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=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&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=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&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=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=755&ask_summarization=1"},{"index":5,"title":"Sekant ve Kosekant İntegrali","list":{"type":"unordered","items":["Sekant kare a çarpı de bölü sinüs x ifadesi, tanjant çarpı bir artı tanjant kare a şeklinde yazılabilir.","Sekant a bölü tanjant a ifadesi, bir bölü sinüs a veya kosekant a olarak hesaplanıyor.","Kosekant integrali, sekant integrali gibi hesaplanabilir ve kotanjant x çarpı kosekant x şeklinde ifade edilebilir."]},"beginTime":813,"endTime":939,"href":"/video/preview/15684713814669680840?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=813&ask_summarization=1"},{"index":6,"title":"Türev Alma ve İntegral Çözümü","list":{"type":"unordered","items":["Kotanjantın türevi, tanjant çarpı kosinüs bölü sinüs olarak hesaplanıyor.","Kosekantın türevi, eksi kosekant kare x olarak bulunuyor.","İntegral çözümünde, tanjant ve kotanjant değerleri dik üçgen kullanılarak 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isteniyor."]},"beginTime":1113,"endTime":1140,"href":"/video/preview/15684713814669680840?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=1113&ask_summarization=1"}],"linkTemplate":"/video/preview/15684713814669680840?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=%%timestamp%%&ask_summarization=1"},"isAdultDoc":false,"relatedParams":{"text":"Calculus-1 : Değişken Değiştirme Yöntemi 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konu anlatımı videoları için: • Calculus - Kuzey Kampüs ne tür videolar ol... Calculus 1 (Kalkulüs 1) dersi tüm üniversitelerimizin en temel matematik dersi. Benim de en çok sevdiğim...","preview":{"posterSrc":"//avatars.mds.yandex.net/get-vthumb/2882575/b9b37904aec371b7008916ef13ab6df2/564x318_1","videoSrc":"https://video-preview.s3.yandex.net/yGti0wEAAAA.mp4","videoType":"video/mp4"},"target":"_self","position":"13","reqid":"1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL","summary":{"isFull":true,"fullTextUrl":"/video/result?ask_summarization=1&numdoc=1&noreask=1&nomisspell=1&parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=videoid:5400371110304733773","teaser":[{"list":{"type":"unordered","items":["Bu video, Şahin Aksankur tarafından sunulan bir matematik eğitim içeriğidir. Eğitmen, calculus dersinin ilk bölümünde sayı kümelerini ve matematiksel ispatları anlatmaktadır.","Video, doğal sayılar (N), tam sayılar (Z), rasyonel sayılar (Q), irrasyonel sayılar (I) ve reel sayılar (R) kavramlarını detaylı olarak ele almaktadır. İlk bölümde sayı kümelerinin özellikleri, ondalık açılımları ve sayı doğrusu üzerindeki yerleri örneklerle açıklanırken, ikinci bölümde rasyonel ve irrasyonel sayılar arasındaki farklar ispatlanmaktadır.","Videoda ayrıca tekrar eden ondalıklı sayıların rasyonel sayı olduğunu, tekrar etmeyen ondalıklı sayıların ise irrasyonel sayı olduğunu gösteren ispatlar, iki reel sayı arasındaki her zaman başka bir reel sayı olduğunu kanıtlama ve bir sayının karesinin çift olması durumunda o sayının da çift olduğunu kontrapozisyon yöntemiyle ispatlama gibi matematiksel ispatlar da yer almaktadır."]},"endTime":1244,"title":"Calculus Dersi: Sayı Kümeleri ve İspatlar","beginTime":0}],"fullResult":[{"index":0,"title":"Sayı Kümeleri Tanıtımı","list":{"type":"unordered","items":["Calculus dersinin ilk bölümünde doğal sayılar, tam sayılar, rasyonel ve irrasyonel sayılar ile reel sayılar konusu anlatılacaktır.","Doğal sayılar (N) bir'den başlayan ve sonsuza kadar giden sayıları içerir, Türkçe'de genellikle sıfırı da sayarken kalkülüs konusunda bir'den başlar.","Tam sayılar (Z) eksi sonsuzdan başlayarak bir, iki, üç şeklinde sonsuza kadar giden sayıları içerir."]},"beginTime":10,"endTime":117,"href":"/video/preview/5400371110304733773?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=10&ask_summarization=1"},{"index":1,"title":"Rasyonel Sayılar","list":{"type":"unordered","items":["Rasyonel sayılar, iki tam sayının birbirine bölümü şeklinde yazılan sayılardır (a/b şeklinde, b sıfırdan farklı).","Rasyonel sayılar sonlanan veya sonsuz bir dizisi ile son bulan ondalık açılımları olan sayılardır.","Rasyonel sayılar kümesi reel sayıların tüm özellikleri gösterir ancak tamlık özelliği yoktur."]},"beginTime":117,"endTime":437,"href":"/video/preview/5400371110304733773?parent-reqid=1769451656719086-13282591355602346103-balancer-l7leveler-kubr-yp-sas-249-BAL&text=TheCalculusMan&t=117&ask_summarization=1"},{"index":2,"title":"İrrasyonel 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