Explanation Given : x = y 2 + 4 y − 22 → ( an equation of a parabola ) x = 3 y + 8 → ( an equation of a straight line ) Formula used : The area of the region lying to the right of x = y 2 + 4 y − 22 and to the left of x = 3 y + 8 is given by, A = ∫ a b ( r i g h t f u n c t i o n ) − ( l e f t f u n c t i o n ) d y a ≤ y ≤ b A-area of the region a,b-lower and upper limits Calculation : Equation of curves are x = y 2 + 4 y − 22 x = 3 y + 8 First, we need to find the points of intersection of two curvesby solving the given equations, x = y 2 + 4 y − 22 → ( 1 ) x = 3 y + 8 → ( 2 ) y 2 + 4 y − 22 = 3 y + 8 → ( setting ( 1 ) and ( 2 ) equal ) y 2 + 4 y − 3 y − 22 − 8 = 0 → ( taking all terms to one side ) y 2 + y − 22 − 8 = 0 → ( simplifying y terms ) y 2 + y − 30 = 0 → ( simplifying constants ) y 2 + 6 y − 5 y − 30 = 0 y ( y + 6 ) − 5 ( y + 6 ) = 0 ( y + 6 ) ( y − 5 ) = 0 } ( factorisation ) ( y + 6 ) = 0 and ( y − 5 ) = 0 → ( setting factors equal to 0 ) y = − 6 or y = 5 Therefore, the intersection points are y = − 6 and y = 5 For y = − 6 and y = 5 , we get the value of x as x = 3 ( − 6 ) + 8 = − 10 x = 3 ( 5 ) + 8 = 23 Coordinates of points of intersection of curves are (–10, –6), (23, 5) as shown in the figure: Now, we integrate the ( r i g h t f u n c t i o n ) − ( l e f t f u n c t i o n ) to find the area of the region...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 6.1, Problem 23E

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The area of the region lying to the right of parabolic curve and to the left of straight line.

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Chapter 6.1, Problem 22E

Chapter 6.1, Problem 24E

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The area of the region lying to the right of parabolic curve and to the left of straight line.

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The area of the region lying to the right of parabolic curve and to the left of straight line.

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Chapter 6 Solutions