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### Integration of y = e^(-x)

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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.
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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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Associated Topics:
College Calculus

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