Integration of y = e^(-x)Date: 06/01/99 at 20:46:31 From: Rachel Meyers Subject: Difficult Integration Problem Dear Dr.Math, This is a problem that I cannot solve. I hope you can help me out. A diagram shows a sketch of the graph of y = e^(-x). The points A and B have coordinates (n,0) and (n+1,0). Points C and D are on the curve, such that AD and BC are parallel to the y-axis. 1. Show that B lies on the tangent to the curve at D. 2. Find the area of the region ABCD under the curve and show that the line BD divides this region into 2 parts whose areas are in the ratio e:(e-2). Thanks, Rachel. P.S. I am finding the last part, about the ratios, especially hard to understand. Date: 06/02/99 at 07:02:27 From: Doctor Jerry Subject: Re: Difficult Integration Problem Hi Rachel, Thanks for your question. You described it very clearly. The derivative of y = e^(-x) is y' = -e^(-x), so that at x = n, the slope of the tangent line is -e^(-n). The equation of the tangent line at D is y-e^(-n) = -e^(-n)(x-n), right? So, we ask if when we set x = n+1 in this equation whether y turns out to be 0 or not. You can see that this is so. For part 2, you can find the area beneath line DC. This is just a triangle. Also, because you already have the equation of the tangent line, you can find the area above DC. This would be int(x = n, x = n+1, [e^(-x) - e^(-n) + e^(-n)(x-n)]*dx). Not too bad, because many parts of this are just constants. I feel certain you can finish this. If not, please write back. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
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