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ODTÜ midterm soru çözümleri için: • ODTÜ Math 119...","preview":{"posterSrc":"//avatars.mds.yandex.net/get-vthumb/2966453/85ec9800415b759db4d202911b2c57e3/564x318_1"},"target":"_self","position":"0","reqid":"1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL","summary":{"fullTextUrl":"/int_search_summary?data=http%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3DSddYWjvrp18","linkTemplate":"/video/preview/2054721956784963320?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=%%timestamp%%&ask_summarization=1"},"isAdultDoc":false,"relatedParams":{"text":"Calculus: Bölüm 6.1 - Ders 2 | Disk Metodu ile Hacim Bulma (Solids of Revolution - Disk Method)","related_orig_text":"Calculus Methods","related_porno":false,"related_less_3m_off":true,"client":"d2d","no_cnt":1,"related_src":"serp","related":"{\"porno\":false,\"vfp\":1,\"orig_text\":\"Calculus Methods\",\"url\":\"http:\\/\\/www.youtube.com\\/watch?v=SddYWjvrp18\",\"src\":\"serp\",\"rvb\":\"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_wA8u39Aw4E_gEQ8vkB9gD_AOcB9_v4_gEACPsM9P4BAAACCP8C_QAAABD-8Ar__QIAE-r6_AIAAAAWDQAB_gAAAP0H_gv_AQAA_wLy_QP_AAAUCQPz_wAAAAUV8wn-AAAAAQj3-QEAAAAE8Pv8AAAAACAALZzVxjs4E0AJSE5QAipzEAAaYCrxAFYi-NrJ8yYCBdULIMXa-gTMmvP_BfEA4ynL9-xAvsEVF_9Q5-z-nAAAACsTARsaABB6sPu8OB44IMzHuwzvfwMY47T6AOC-1koH9A4VGO0fJQD178nMQJOcByBJ9iAALYEnEDs4E0AJSG9QAiqvBhAMGqAGAADgQQAAcEEAALBBAAAQwQAAgL8AAAAAAAA8QgAAZMIAAHDBAAC4wQAAEEEAAOhBAACIwgAAWMIAADhCAACuwgAAEMEAAGDCAACAwQAAmMEAAJhBAABAwgAA8EEAAOhBAACQQgAAgEAAAITCAAAwwQAAkEEAAFhCAABMwgAAOEIAAIA_AAAAwQAAgMAAAEDBAACQQQAAskIAANjBAAAEQgAA0EEAAAjCAACmQgAAZEIAAABCAAA8wgAAisIAAIA_AABsQgAAMMIAAIDBAABowgAAhsIAAATCAAAkQgAAgEAAAPzCAAAcwgAA0EEAANBBAACUQgAAgEEAAIhBAADKwgAAEMIAAI7CAAAwQQAAEMIAABDCAAAIwgAACEIAAGBCAACawgAAuEEAAEBBAABswgAAsMEAALhBAACWQgAAmMEAABDCAAAwwQAAFEIAAHBBAADowQAAAAAAACRCAABAQgAAwEEAACTCAABAwQAAZEIAAFDBAAD4wQAAQEAAAIDCAAAgwgAAeEIAABhCAACIwQAAAMAAAOBAAACwQQAAEMIAAIbCAADYQQAAPEIAACxCAACgQAAAgD8AAABBAAAwwQAAjMIAACTCAAAgQQAABEIAAIDAAAA8wgAAyMEAANjBAADAwAAAiMEAABjCAADYwQAAQEIAAABCAAAAwQAAEEEAADDCAABgwgAAEEEAAEDAAADIwQAAoMAAAGhCAAAAwQAAIEEAAADBAABwwQAAjMIAAKBBAACAvwAAJMIAAKDAAAA4QgAA6MEAAKDBAACYQgAAVEIAACBCAAAAQgAAIMEAAKDAAAAAwQAAoEAAAPDBAABAQAAAcMEAAEDCAABAQgAAIMIAAFxCAAAAQQAAXMIAAHjCAAAIwgAA-EEAAEBCAACgQQAACMIAAEDAAABIQgAAisIAALLCAAAcwgAABEIAALBBAABUwgAAAEEAAHhCAACwQQAAJEIAAKLCAACgwAAAKEIAAI7CAAAkwgAAAAAAAEBAAADAQQAAiMIAABjCAAAAAAAAEMIAACxCAAAwwgAAGMIAAIBAAADAwQAAEEEgADgTQAlIdVABKo8CEAAagAIAAOg9AAAQPQAARD4AAKA8AAA8PgAAhj4AAIY-AAAPvwAABL4AAOA8AACGvgAAHL4AAI4-AAAsPgAAqL0AAOA8AAAMPgAAUD0AAEA8AAAVPwAAfz8AAES-AAAkPgAAoLwAABS-AAAkPgAAij4AAEC8AAAQPQAAQDwAABQ-AABEvgAAQDwAADw-AAAEPgAAmL0AABS-AABQvQAAur4AAKq-AADYvQAAQLwAAJo-AAAcvgAARL4AAJi9AACmPgAABL4AAFC9AACYvQAAyL0AAEC8AABsPgAAPD4AAHy-AABAvAAAGz8AAEy-AABAPAAAVD4AADC9AABMPgAA-D0AABC9IAA4E0AJSHxQASqPAhABGoACAADgvAAADD4AAJg9AABFvwAAFL4AAAw-AACCPgAAEL0AALi9AABUPgAAmL0AAFS-AAAQvQAATL4AAEw-AACIvQAAyL0AABc_AABwvQAAxj4AAIA7AABAvAAAyL0AABC9AACIvQAATL4AAOA8AAAMPgAANL4AANi9AAAQPQAAuD0AADA9AACCvgAAgLsAADC9AAAkPgAAsj4AAK6-AAAQPQAAqL0AAOA8AACgPAAAML0AAJg9AACYvQAAf78AAMg9AAAsPgAAuD0AADA9AADIvQAAQDwAAKg9AAAcPgAAyD0AAKA8AAAcPgAA4DwAALg9AAAsPgAAND4AAHQ-AAD4PSAAOBNACUh8UAEwCTgBSgBgAGgA\"}","related_url":"http://www.youtube.com/watch?v=SddYWjvrp18","parent-reqid":"1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL","related_vfp":1,"relatedVideo":"yes"},"cwidth":1280,"cheight":720,"cratio":1.77777,"dups":["2054721956784963320"],"episode":0,"season":0,"isEmbedOnly":false,"greenHost":"YouTube","hasTranslation":true,"contentTypeId":null,"censored":false},"2791149328929414095":{"videoId":"2791149328929414095","docid":"34-7-17-Z50D565056045A7CA","description":"Hello, I've been wanting to create a regular playlist for my channel for general mathematics and calculus. 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İlk bölümde x üzeri n ifadelerinin integrali, rasyonel ifadelerde integral alma ve trigonometrik fonksiyonların integrali ele alınırken, ikinci bölümde değişken değiştirme yöntemi ve özellikle kosinüs fonksiyonlarının integrali adım adım gösterilmektedir.","Eğitmen, integral hesaplamalarında kullanılan temel formülleri ve trigonometrik eşitlikleri kullanarak çeşitli soruları çözmekte ve bir sonraki videoda değişken değiştirme yöntemlerinin detaylı olarak anlatılacağını belirtmektedir."]},"endTime":859,"title":"Temel İntegral Dersi","beginTime":0}],"fullResult":[{"index":0,"title":"Temel İntegral Kavramı","list":{"type":"unordered","items":["Videoda temel integral kavramları anlatılacak ve değişken değiştirme yöntemi de ele alınacak.","İntegral, bir ifadenin türevini alarak üzerine olan ifadeyi bulma işlemidir.","Polinom tipteki fonksiyonların integrali, derecenin üstünü bir arttırıp arttırdığımız dereceye bölerek 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cosx/2 şeklinde yazılabilir.","İntegral hesaplamalarında trigonometrik dönüşümler ve temel trigonometrik eşitlikler önemlidir."]},"beginTime":393,"endTime":543,"href":"/video/preview/2791149328929414095?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=393&ask_summarization=1"},{"index":3,"title":"Üstel İntegral Örnekleri","list":{"type":"unordered","items":["Üstel fonksiyonların integrali, türevi alınarak bulunur.","950x² dx integrali, 950x×9/9 + 150x/ln15 + 250x/ln25 + C şeklinde çözülür.","İntegral hesaplamalarında değişken değiştirme yöntemi de kullanılabilir."]},"beginTime":543,"endTime":600,"href":"/video/preview/2791149328929414095?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=543&ask_summarization=1"},{"index":4,"title":"İntegral Hesaplama ve Değişken Değiştirme","list":{"type":"unordered","items":["İntegral hesaplamasında değişken değiştirme yöntemi kullanılabilir, ancak bu yöntem için belirli koşullar gereklidir.","Birinci dereceden bir fonksiyonun türevi sabit olduğundan, bu tür ifadelerde değişken değiştirme yöntemi kullanılabilir.","İkinci dereceden bir fonksiyon (örneğin ax²+bx+c) için değişken değiştirme yöntemi kullanılamaz çünkü türevi sabit değildir."]},"beginTime":601,"endTime":674,"href":"/video/preview/2791149328929414095?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=601&ask_summarization=1"},{"index":5,"title":"İntegral Örneği ve Çözümü","list":{"type":"unordered","items":["İntegral hesaplamasında değişken değiştirme yöntemi uygulandığında, a/a = de/dx şeklinde yazılabilir.","İntegral hesaplandıktan sonra, değişken değiştirme işlemi geri uygulanmalıdır.","İntegral sonucunda elde edilen ifade, değişken değiştirme işlemi geri uygulandığında a sin(ax+b) + c şeklinde 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fonksiyonlarının çarpımının integralini hesaplama yöntemidir.","Formül: ∫u(x)v'(x)dx = u(x)v(x) - ∫v(x)u'(x)dx şeklinde ifade edilir.","Bu formülün amacı, zor integrali daha basit integral haline getirmek ve basit integral daha kolay hesaplanabilir olmasıdır."]},"beginTime":31,"endTime":200,"href":"/video/preview/6880253165433797333?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=31&ask_summarization=1"},{"index":2,"title":"Kısmi İntegrasyon Örneği 1","list":{"type":"unordered","items":["Örnek: ∫x³ln(2x)dx integralini hesaplayalım.","u = ln(2x) ve dv = x³dx olarak seçerek, du = (1/x)dx ve v = x⁴/4 olarak bulunur.","Formülü uygulayarak ∫x³ln(2x)dx = (x⁴/4)ln(2x) - ∫(x⁴/4)(1/x)dx şeklinde yazılır."]},"beginTime":200,"endTime":352,"href":"/video/preview/6880253165433797333?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=200&ask_summarization=1"},{"index":3,"title":"İkinci Kısmi İntegrasyon","list":{"type":"unordered","items":["Kalan integral ∫(x³/4)dx için tekrar kısmi integrasyon uygulanır.","u = x³ ve dv = dx olarak seçerek, du = 3x²dx ve v = x olarak bulunur.","Sonuç: ∫x³ln(2x)dx = (x⁴/4)ln(2x) - (x⁴/16) + C olarak hesaplanır."]},"beginTime":352,"endTime":496,"href":"/video/preview/6880253165433797333?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=352&ask_summarization=1"},{"index":4,"title":"Kısmi İntegrasyon Örneği 2","list":{"type":"unordered","items":["Örnek: ∫x²tan⁻¹(x)dx integralini hesaplayalım.","u = tan⁻¹(x) ve dv = x²dx olarak seçerek, du = (1/(1+x²))dx ve v = x³/3 olarak bulunur.","Formülü uygulayarak ∫x²tan⁻¹(x)dx = (x³/3)tan⁻¹(x) - (1/3)∫x³/(1+x²)dx şeklinde yazılır."]},"beginTime":496,"endTime":554,"href":"/video/preview/6880253165433797333?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=496&ask_summarization=1"},{"index":5,"title":"Polinom Bölmesi ve Sonuç","list":{"type":"unordered","items":["∫x³/(1+x²)dx integralini hesaplamak için polinom bölmesi yapılır.","x³/(1+x²) ifadesi x - (x/(1+x²)) şeklinde yazılır.","Sonuç: ∫x²tan⁻¹(x)dx = (x³/3)tan⁻¹(x) - (1/2)x + (1/2)ln(1+x²) + C olarak hesaplanır."]},"beginTime":554,"endTime":621,"href":"/video/preview/6880253165433797333?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=554&ask_summarization=1"},{"index":6,"title":"Kısmi İntegrasyon Yöntemi","list":{"type":"unordered","items":["Kısmi integrasyon yönteminde, integraldeki ifadeleri u ve dv olarak ayırarak işlem yapılır.","Örneğin, ln(x)sin(x) integrali için u=ln(x) ve dv=sin(x)dx alınabilir.","İntegral işlemi sırasında, aynı integralin iki kez ortaya çıkması durumunda, cebirsel ifadeler gibi düşünülerek bir tarafa toplanıp payda olarak alınabilir."]},"beginTime":628,"endTime":690,"href":"/video/preview/6880253165433797333?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=628&ask_summarization=1"},{"index":7,"title":"Kosinüs ve Üstel Fonksiyonların İntegrali","list":{"type":"unordered","items":["cos(x)e^x integrali, kısmi integrasyon yöntemiyle çözülebilir.","Bu tür integralde, kosinüs fonksiyonu sinüse dönüşür ve başlangıç integraline geri döner.","İntegral işlemi sırasında, aynı integralin iki kez ortaya çıkması durumunda, bir tarafa toplanıp payda olarak alınarak integralin sonucu bulunabilir."]},"beginTime":690,"endTime":846,"href":"/video/preview/6880253165433797333?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=690&ask_summarization=1"},{"index":8,"title":"Ark Kosinüs Fonksiyonunun İntegrali","list":{"type":"unordered","items":["x²arccos(x) integrali için u=arccos(x) ve dv=x²dx alınabilir.","İntegral işlemi sırasında değişken değiştirme yöntemi kullanılarak integral çözülebilir.","Değişken değiştirme yönteminde, t=1-x² dönüşümü yapılarak integral sonucu bulunabilir."]},"beginTime":846,"endTime":1164,"href":"/video/preview/6880253165433797333?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=846&ask_summarization=1"},{"index":9,"title":"İntegral Çözümü","list":{"type":"unordered","items":["İntegral ∫ arcsin(x) / x² dx sorusu çözülüyor.","arcsin(x) = u dönüşümü yapılıyor ve du = dx / √(1-x²) bulunuyor.","İntegral ifadesi ∫ u * (-1/x) dx + ∫ (1/x) * (1/√(1-x²)) dx şeklinde yazılıyor."]},"beginTime":1165,"endTime":1244,"href":"/video/preview/6880253165433797333?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=1165&ask_summarization=1"},{"index":10,"title":"Dönüşüm İşlemleri","list":{"type":"unordered","items":["1-x² = u dönüşümü yapılıyor ve dx = -du/2x bulunuyor.","İntegral ifadesi ∫ (-du/2u) / √u * (1/(1-u)) şeklinde dönüştürülüyor.","√u = m dönüşümü yapılıyor ve du = 2m dm bulunuyor."]},"beginTime":1244,"endTime":1394,"href":"/video/preview/6880253165433797333?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=1244&ask_summarization=1"},{"index":11,"title":"İntegralin Tamamlanması","list":{"type":"unordered","items":["İntegral ifadesi ∫ (1/2) * (1/(m-1) - 1/(m+1)) dm şeklinde yazılıyor.","İntegral sonucu (1/2) * (ln|m-1| - ln|m+1|) + C olarak bulunuyor.","Sonuçta m yerine √(1-x²) yazarak integralin tam çözümü elde ediliyor."]},"beginTime":1394,"endTime":1569,"href":"/video/preview/6880253165433797333?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=1394&ask_summarization=1"},{"index":12,"title":"Yeni İntegral Sorusu","list":{"type":"unordered","items":["∫ √(a²-2x) dx integrali çözülüyor.","Kök içindeki ifade a²-x² şeklinde olursa kolay olacağını belirtiyor.","a²-x² ≠ a²-2x olduğundan kısmi integrasyon yöntemi kullanılması gerektiği söyleniyor."]},"beginTime":1569,"endTime":1604,"href":"/video/preview/6880253165433797333?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=1569&ask_summarization=1"},{"index":13,"title":"İntegral Hesaplama","list":{"type":"unordered","items":["İntegral hesaplamasında a²-x² ifadesi kullanılıyor ve dönüşüm yapılıyor.","Paydadaki ifadeyi düzenlemek için x = a.sin(t) dönüşümü uygulanıyor ve dx = a.cos(t)dt olarak bulunuyor.","Üçgen oluşturarak x/a = sin(t) ilişkisi kuruluyor ve a²-x² ifadesi a.cos(t) olarak hesaplanıyor."]},"beginTime":1605,"endTime":1771,"href":"/video/preview/6880253165433797333?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=1605&ask_summarization=1"},{"index":14,"title":"Dönüşüm ve İntegral Hesaplaması","list":{"type":"unordered","items":["Kosinüs iki katlı açı formülü kullanılarak sin²(t) ifadesi 1/2(1-cos(2t)) olarak yazılıyor.","İntegral hesaplaması sonucunda a²/2(t - 1/2sin(2t)) ifadesi elde ediliyor.","t ifadesi arcsin(x/a) olarak bulunuyor ve integral sonucu a²/2(arcsin(x/a) - x√(a²-x²)/a²) + C olarak hesaplanıyor."]},"beginTime":1771,"endTime":2005,"href":"/video/preview/6880253165433797333?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=1771&ask_summarization=1"},{"index":15,"title":"Ödev Soruları","list":{"type":"unordered","items":["İzleyicilere ödev olarak integral(e^x.cos(x)) ve integral(e^x.sin(x)) soruları veriliyor.","İntegral(sec²(x)dx) sorusu da ödev olarak belirtiliyor.","Videonun devamında rasyonel kesirleri ayırma metodu ele alınacak."]},"beginTime":2005,"endTime":2055,"href":"/video/preview/6880253165433797333?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=2005&ask_summarization=1"}],"linkTemplate":"/video/preview/6880253165433797333?parent-reqid=1766971802292832-7532805961905792780-balancer-l7leveler-kubr-yp-sas-73-BAL&text=Calculus+Methods&t=%%timestamp%%&ask_summarization=1"},"isAdultDoc":false,"relatedParams":{"text":"Calculus-1 : Kısmi İntegrasyon Yöntemi -5","related_orig_text":"Calculus Methods","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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