Area of Region R Bounded by Line and Curve

[Samuelb88]

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

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

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)): $$y=f'(x_i)(x-x_i) + f(x_i)$$

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

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

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. :)

 

[mighty2000]

Thank you for your post. It seems like you have made some good progress in understanding the problem and determining the necessary equations. However, I would suggest taking a step back and looking at the problem from a more general perspective.

Firstly, it is important to understand what the question is asking for. The problem states that we want to find the area of the region R, which is bounded by the curve y = f(x), the line y = mx + b, and the perpendiculars from points P and Q. This means that we are looking for the integral of the function (f(x) - mx - b) over the interval [q,p]. This is because the area of a region can be calculated by finding the integral of the function that represents the boundary of the region.

Now, the hint given in the problem suggests using rectangles perpendicular to the line y = mx + b to approximate the region R. This means that we can divide the region R into smaller rectangles, each with a height of (f(x) - mx - b) and a width of (dx). The sum of the areas of these rectangles will be an approximation of the total area of R. To get a more accurate approximation, we can increase the number of rectangles, making their width (dx) smaller and smaller.

Using this approach, we can write the area of R as the limit of the sum of the areas of these rectangles as the width (dx) approaches 0. This is the same as taking the integral of the function (f(x) - mx - b) over the interval [q,p].

Therefore, the area of R can be expressed as:

A = ∫q^p (f(x) - mx - b) dx

However, this is not the final answer as the problem asks us to show that the area of R is equal to:

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

To prove this, we need to use the relationship between the slope of the curve (df/dx) and the slope of the line (m) to show that the above expression is equivalent to the one we derived earlier. I would suggest starting by writing the equation of the tangent line to the curve y = f(x) at any point (x_i, f(x_i)) as:

y =