prove that 1/3 is less than or equal to the integral of 1/x dx on [4,6] and prove that this integral is smaller than or equal to 1/2

Answer 1

The function f(x) = 1/x is positive and decreasing on the interval [4,6].

So, the value of the integral ∫_{(from 4 to 6)}(1/x)dx, is the area, A, of the region below the graph of y = 1/x and above the x-axis between x = 4 and x = 6.

Since f(x) is decreasing, A is smaller than or equal to the area of the rectangle with height f(4) and width 2, and is larger than or equal to the area of the rectangle with height f(6) and width 2, where the base of each rectangle is the line segment lying on the x-axis with endpoints 4 and 6.

So, f(6)(6-4) ≤ A ≤ f(4)(6-4)

Therefore, (1/6)(2) ≤ A ≤ (1/4)(6-4)

1/3 ≤ A ≤ 1/2

Answer 2

Evaluate the definite integral and verify that it falls in the stated range.

∫₄⁶ dx/x = ln(6/4) = ln(1.5) ≅ 0.405465

This value is between 1/3 and 1/2.