Question

Let A denote the area between the graph of \(\mathrm{f}(\mathrm{x})=1 / \mathrm{x}\) and the interval \([1,2]\), and let \(\mathrm{B}\) denote the area between the graph of \(f\) and the interval \(\left[\frac{1}{2}, 1\right]\). Explain geometrically why \(\mathrm{A}=\mathrm{B}\).

Short answer

Question: Explain geometrically why the area A between the graph of the function f(x) = 1/x and the interval [1,2] is equal to the area B between the graph and the interval [1/2,1]. Answer: The areas A and B are equal because there exists a one-to-one correspondence between the infinitesimally small rectangles of areas A and B, and each pair of corresponding rectangles have the same area. This can be seen by observing that the area of the rectangle at x in interval [1,2] and its corresponding rectangle at 1/x in interval [1/2,1] have the same height times width (f(x)dx = f(1/x)dx/x^2), demonstrating that A = B geometrically.

Step-by-step solution

Visualize the graph of f(x)

First, we need to visualize the graph of the function f(x) = 1/x, to get the idea of the shape of the graph. The graph of 1/x is a hyperbola with the center at the origin (0,0), for x > 0, our focus in this problem.

Analyze the intervals [1,2] and [1/2,1]

Now, we need to analyze the intervals [1,2] and [1/2,1]. In the interval [1,2], the function f(x) takes the values of f(1)=1 and f(2)=1/2. Similarly, for the interval [1/2,1], the function takes the values of f(1)=1 and f(1/2)=2.

Visualize the geometric areas A and B

Plot the function f(x) = 1/x on a coordinate plane. For interval [1,2], mark its end points on the x-axis, and draw the corresponding lines along the y-axis until they intersect with f(x) at points P(1,1) and Q(2,1/2). For interval [1/2,1], mark its end points on the x-axis, and draw the corresponding lines on the y-axis until they intersect with f(x) at points R(0.5,2) and S(1,1). Now consider the areas A and B. - Area A is the area enclosed by the curve of f(x) in the interval [1,2], i.e., the graph of f(x) from P(1,1) to Q(2,1/2) and the corresponding segments between y-axis and the curve. - Area B is the area enclosed by the curve of f(x) in the interval [1/2,1], i.e., the graph of f(x) from R(0.5,2) to S(1,1) and the corresponding segments between y-axis and the curve.

Geometrically show A = B

Divide the area A into infinitesimally small vertical rectangles of width dx and height f(x). Now, consider the rectangle at x in interval [1,2]. Its corresponding rectangle in interval [1/2,1] is at the point 1/x, and it has width dx/x^2 and height f(1/x) = x. Notice that the area of the rectangle at x is equal to the area of the rectangle at 1/x in interval [1/2,1] because their height times width is the same. This can be shown by f(x)dx = f(1/x)dx/x^2 => (1/x)dx = x(dx/x^2) which is true. Therefore, their corresponding rectangles have the same area. Hence, as there exists a one-to-one correspondence between the infinitesimally small rectangles of areas A and B, and each pair of corresponding rectangles have the same area, we can geometrically say that A = B.