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Samuelb88

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## Homework Statement

Let C be the arc of the curve y = f(x) between points P(p,f(p)), Q(q,f(q)), and let R be the region bounded by C, the line y = mx + b, and the perpendiculars by the line from P and Q.

http://img15.imageshack.us/img15/6827/pic1ds.jpg

Show that the area R is

[tex]\frac{1}{1+m^2}\right) \int_q^p (f(x) - mx - b)(1+m\frac{df}{dx}\right) ) dx[/tex]

Hint: This formula can be verified by subtracting areas, but it will be helpful to derive region R using rectangles perpendicular to the line, shown in the figure below.

http://img163.imageshack.us/img163/1365/pic2c.jpg

## Homework Equations

## The Attempt at a Solution

Tangent to C @ (x_i,f(x_i)): [tex]y=f'(x_i)(x-x_i) + f(x_i)[/tex]

I let L equal the line segment "?" perpendicular to the line y=mx+b and used the distance equation to determine its value.

[tex] L = ((x_i-x)^2+(f'(x_i)(x-x_i) + f(x_i) - mx - b)^2)^(^1^/^2^)[/tex]

And the area A

A = L(change in u)

So the region R should equal the limit of all the approximating rectangles.

Pretty confused as of how to proceed. Any suggestions would be greatly appreciated. :)

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