Explanation Given: ∫ 1 3 ∫ 0 1 y e x y d y d x = I Calculation: Step 1: ∫ 1 3 ∫ 0 1 y e x y d y d x = Compute the indefinite integral ∫ y e x y d y ∫ y e x y d y = y e y x x − e x y 1 x 2 + C Step 2: Compute the boundaries: ∫ 0 1 y e x y d y Formula used: ∫ 0 1 y e x y d y = e x x − e x 1 x 2 − ( − 1 x 2 ) - = ( e x ( x − 1 ) + 1 x 2 ) Step 3: = ∫ 1 3 ( e x ( x − 1 ) + 1 x 2 ) d x Compute the indefinite integral: ∫ ( e x ( x − 1 ) + 1 x 2 ) d x Expand

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 16.1, Problem 46E

To determine

The value of the iterated integral $\int_{1}^{3} \int_{0}^{1} y e^{x y} d y d x$

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Chapter 16.1, Problem 45E

Chapter 16.1, Problem 47E

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الاسم يمنع استخدام الآلة الحاسبة ظلل في ورقة الإجابة الدائرة التي تحتوي على الحرف (ص) للإجابة الصحيحة والحرف (خ) للإجابة الخطأ بحسب رقم الفقرة لكل مما يأتي: درجة لكل فقرة. ( ) نها جا 元 جتا = صفر س ۱ س س -۱ ( ) يمكن إعادة تعريف الدالة د(س) = س قاس لكي تكون متصلة عند س = 7 ( ) إذا كانت د(س) = (٢) س - س ) ؛ فإن د(١) = ٦ ٢ س ص ( ) إذا كانت س + 0= ؛ فإن عند ) - ١ ، - ٦ ) تساوي (٦) ( ) إذا كانت د(س) = س ه ، و (س) = ٣ س ٢ + ٢ س ؛ فإن ( د ) (۱) = ۸ ) ( معادلة ناظم الدالة ص = د(س) عند النقطة ) ( ، د (۲)) هي ص - (د (م) - - د (۲) ( س - م ) ( ) إذا كانت ص = ظتا٢ س ؛ فإن ص = ٢ ص قتا ٢س ) ( إذا كانت د(س) = س ؛ فإن د (T) = جتاس 1- T ( ) إذا كانت د(س) = 1 - جناس جاس ؛ فإن د () = - 1 ( ) إذا كانت الدالة د (س) تحقق شروط مبرهنة القيمة المتوسطة على [ ، ب ] ، فإنه يوجد جـ ] ، ب [ بحيث (جـ) = (P) + (~)- - ب + P 1 2 3 4 5 6 7 8 9 10 11 ( ) للدالة د(س) = لو ( س ) + (٣) نقطة حرجة عند س = . ( ) إذا كان س = - ٢ مقارباً رأسياً للدالة د(س) 12 10 13 14 15 16 17 س = لو|س | + ث - = ۲ س + ٣ ب س + ٤ ، فإن معادلة…

2. Symmetry Evaluate the following integrals using symmetry argu- ments. Let R = {(x, y): -a ≤ x ≤ a, −b ≤ y ≤ b}, where a and b are positive real numbers. a. SS Sf xye xye¯(x² + y²) dA R b. C sin (x − y) - dA x² + y² + 1 R

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CALCULUS (CLOTH)

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