Explanation Given information: The two curves has a function of y = x ( x 2 + 1 ) 2 and y = x 5 − x . Calculation: Procedure to sketch the region bounded by the two curves is explained below: Draw the graph for the function y = x ( x 2 + 1 ) 2 by substituting different values for x . Similarly in the same graph plot for the function y = x 5 − x by substituting different values for x . The region enclosed by the curves y = x ( x 2 + 1 ) 2 and y = x 5 − x is shown in Figure 1. Refer to Figure 1. The curves intersects at x = 0 and x = 1.052 . The curves are bounded by the top and bottom curve. Hence, the integration can be done with respect to x . Therefore, the point of intersection of the curves are at x = 0 _ and x = 1.052 _ . Find the area of the region bounded by the curves using the relation: A = ∫ a b ( f ( x ) − g ( x ) ) d x (1) Here, the top curve is f ( x ) , the bottom curve is g ( x ) , the lower limit is a , and the upper limit is b . Substitute x ( x 2 + 1 ) 2 for f ( x ) , x 5 − x for g ( x ) , 0 for a , and 1.052 for b in Equation (1). A = ∫ 0 1.052 [ ( x ( x 2 + 1 ) 2 ) − ( x 5 − x ) ] d x = ∫ 0 1.052 [ x ( x 2 + 1 ) 2 ] d x − ∫ 0 1.052 [ x 5 − x ] d x (2) Consider u = x 2 + 1 (3) Differentiate Equation (3) with respect to x . d u = 2 x x = d u 2 Find the value of the integral ∫ 0 1.052 [ x ( x 2 + 1 ) 2 ] d x : Substitute u for x 2 + 1 and d u 2 for x in the integral function ∫ 0 1.052 [ x ( x 2 + 1 ) 2 ] d x

9th Edition

ISBN: 9781337613927

Author: Stewart

Publisher: CENGAGE L

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Question

Chapter 6.1, Problem 46E

To determine

To draw: The region enclosed by the given curves.

The point of intersection of the curves.

The area of the region enclosed by the curves.

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Chapter 6.1, Problem 45E

Chapter 6.1, Problem 47E

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To draw: The region enclosed by the given curves. The point of intersection of the curves. The area of the region enclosed by the curves.

Solution Summary: The procedure to sketch the region bounded by the curves is explained below.