Explanation We know that, area under a curve y = f ( x ) region bounded by the given function, vertical lines x = a & x = b and the x axis is given by A r e a = ∫ b a f ( x ) d x Given: Consider ∫ − 2 2 2 2 ( 1 − x 2 − | x | ) d x --------------------- (i) To sketch the region whose area is represented by (i) which mean to find the region of integration. i.e., the curve by y = f ( x ) = 1 − x 2 − | x | From the integral the limits are boundaries. i.e., x = − 2 2 and x = 2 2 are the boundaries. Draw the graph of y = 1 − x 2 − | x | ; x = − 2 2 and x = 2 2 with the scale X Maximum 5 X Minimum -5 X Scale 1 Y Maximum 3 Y Minimum -3 Y Scale 1 Graph is given by The shaded region gives the required area of integration. ∫ − 2 2 2 2 ( 1 − x 2 − | x | ) d x = Area of the shaded region O A B C O ----------- (ii) By Geometry Area of the region O A B C O = Area of O A B O + Area of O B C O --------- (iii) = 2 ( Area of O B C O ) [ O A B O & O B C O are two equal parts] Thus, Area of the region bounded by the curve O B C O is ∫ 0 2 2 y d x = ∫ 0 2 2 1 − x 2 − | x | d x = ∫ 0 2 2 1 − x 2 − ∫ 0 2 2 | x | d x -------------- (iv) Formula used: By definition of | x | , | x | = x , x ≥ 0 = − x , x ≤ 0 ∫ a 2 − x 2 d x = x a 2 − x 2 2 + a 2 2 S i n − 1 ( x a ) ∫ x n d x = x n + 1 n + 1 Find the integrals in (iv) separately ∫ 0 2 2 1 − x 2 d x = [ x 1 − x 2 2 + 1 2 S i n − 1 x ] 0 2 2 Substituting the limits

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ISBN: 9781319434540

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 6.1, Problem 50E

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Chapter 6.1, Problem 49E

Chapter 6.1, Problem 51E

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Chapter 6 Solutions

ACHIEVE STANDALONE ACCESS F/ CALC 4E

Ch. 6.1 - Prob. 1PQ Ch. 6.1 - Prob. 2PQ Ch. 6.1 - Prob. 3PQ Ch. 6.1 - Prob. 4PQ Ch. 6.1 - Prob. 5PQ Ch. 6.1 - Prob. 6PQ Ch. 6.1 - Prob. 1E Ch. 6.1 - Prob. 2E Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E

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The region of integration and evaluate using geometry.

Solution Summary: The author explains that the region of integration is the shaded region whose area is represented by (i).

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Chapter 6 Solutions