Explanation Let there be a continuous function f ( x , y ) on R = [ a , b ] × [ c , d ] , then the below integrals are referred to as iterated integrals. ∬ R f ( x , y ) d A = ∫ a b ∫ c d f ( x , y ) d y d x = ∫ c d ∫ a b f ( x , y ) d x d y Given: An iterated integral as ∫ − 1 1 ∫ 0 π x 2 s i n y d y d x Formula Used: Integration of the below functions: 1. Constant Function: ∫ a d x = a x + c 15. Variable Function: ∫ x d x = x 2 2 + c 30. Reciprocal Function: ∫ 1 x d x = l n | x | + c 45. Exponential Function: ∫ e x d x = e x + c 59. Trigonometric Function: a. ∫ s i n ( x ) d x = - c o s ( x ) + c q. ∫ c o s ( x ) d x = s i n ( x ) + c 60. Sum Rule of Derivative: ∫ ( f + g ) d x = ∫ f d x + ∫ g d x 78. Difference Rule of Derivative: ∫ ( f - g ) d x = ∫ f d x - ∫ g d x Calculation: The given problem is ∫ − 1 1 ∫ 0 π x 2 s i n y d y d x Let us first compute the indefinite integral ∫ 0 π x 2 s i n y d y ∫ 0 π x 2 s i n y d y = x 2 ∫ 0 π s i n y d y Applying the trigonometric rule of integration, we get, ∫ 0 π x 2 s i n y d y = x 2 ( − c o s y ) | 0 π Compute the value, by substituting y = ...

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 16.1, Problem 23E

To determine

The value of the iterated integral $\int_{- 1}^{1} \int_{0}^{π} x^{2} s i n y d y d x$

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Chapter 16.1, Problem 22E

Chapter 16.1, Problem 24E

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Chapter 16 Solutions

CALCULUS (CLOTH)

Ch. 16.1 - Prob. 1PQ Ch. 16.1 - Prob. 2PQ Ch. 16.1 - Prob. 3PQ Ch. 16.1 - Prob. 4PQ Ch. 16.1 - Prob. 5PQ Ch. 16.1 - Prob. 6PQ Ch. 16.1 - Prob. 1E Ch. 16.1 - Prob. 2E Ch. 16.1 - Prob. 3E Ch. 16.1 - Prob. 4E

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