 Calculus - XP Math - Forums Sign Up FREE! | Sign In | Classroom Setup | Common Core Alignment  XP Math - Forums Calculus 07-24-2008 #1 rousse101 Guest   Posts: n/a Calculus The area of the region:y = x^1/2 and y=-x+6 enclosed by the two graphs and the x-axis is given by: 2 ∫− x +6 − x^1/2 dx .....a) 0 2 ∫− x +6 − x^1/2 dx .....b) -3 2 ∫ 6− y − y^2 dy ....c) 0 2 ∫y2- (6-y) dy .....d) -3  07-28-2008 #2 MyriamK Guest   Posts: n/a Looks like a Calculus question, but ... I did study some calculus once, and know what that integral symbol means. The square root function does not exist for x<0, so b) is not an option. The two functions exist for x=0 and x>0. The functions intersect at (4,2). They intersect the x-axis at (0,0) and (6,0). The region is bound by: the x-axis from (0,0) to (6,0), the line y=-x+6 (which we could call x=-y+6) between (4,2) and (6,0), and the parabola y=x^1/2 (or x=y^2) between (0,0) and (4,2) The "horizontal" distance between the two curves (difference in x values for the same y) is -y+6-y^2. The users of this forum would approximate that area by adding up the areas of thin rectangles of base -y+6-y^2 and small height d, inscribed in that region between y=0 and y=2. The limit of those approximations as d tends to zero is the integral, and that's all the Calculus you need to know for this problem. Integrating the function -y+6-y^2 between y=0 and y=2 is the easiest way to find the area (choice c). The other choices don't make much sense. Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
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