Date: 03/09/2001 at 07:53:16 From: Michelle Subject: Integration Hi there. I'm currently doing integration, and blimey, I personally think it's really hard. Anyway I'm stuck on a couple of questions and haven't a clue how to do them. 1. Calculate the area of the finite region bounded by the curves y = x^2 and y = sqrt(x) 2. The region R is bounded by the curve y = x^2+2, the x and y axes, and the line joining point (2,6) to point (26,0). Find the area of R. Also show that the line joining (2,6) to (26,0) is the normal to the curve at (2,6). 3. Find the area of the finite region between the curve x = y^2 + 3y and the line x = 4y. I think this site is great. Keep it up :) Thanks, Michelle.
Date: 03/09/2001 at 09:00:27 From: Doctor Jaffee Subject: Re: Integration Hi Michelle, The first thing I always do when I try to solve problems like the ones you presented is to draw a graph. In problem 1 you should see that the region is a football-shaped figure that intersects at (0,0) and (1,1). (I'm talking about American football, by the way, not what we Yanks call Soccer.) If you were to draw a thin vertical strip anywhere in the region, you see that the height of the strip is the y number on the curve y = sqrt(x) minus the y number on the curve y = x^2. The thickness of the strip we can call dx. So the area of the region is the integral from 0 to 1 of (sqrt(x) - x^2)dx. In problem 2 you should see that the region can be partitioned into two regions, the first, the region under the curve y = x^2 + 2 bounded by the axes and the vertical line that goes through (2,6). Find the area of that region using the same method as in problem number 1. The second region is a triangle whose vertices are at (2,0), (2,6), and (26,0). Since it is a right triangle, it should be easy to use geometry to find the area of the triangle. Add these two areas together and you will have the area of the entire region. A normal line is the line perpendicular to the tangent line. Since f'(x) = 2x, the slope of the tangent line at (2,6) will be f'(2).You should be able to find the slope of the line that connects (2,6) and (26,0). If the product of these two slopes is -1, you know that they are perpendicular, which means that the line under consideration is the normal line. Do problem 3 the same as problem 1. Hint: the region will be very thin. Give these problems a try and if you want to write back to check your answers, or if you need more help understanding what to do, I'll be on the lookout and I'll try to help you some more. Good luck. - Doctor Jaffee, The Math Forum http://mathforum.org/dr.math/
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