Question

If \(f\) is a continuous function such that \(0 \leq f(x, y) \leq 1\) over a a region \(R\) of area \(1,\) prove that \(0 \leq \int_{R} \int f(x, y) d A \leq 1 .\)

Short answer

The proof is based on the facts that the integral of a non-negative continuous function is non-negative, and the integral of a continuous function over a particular region is less than or equal to the maximum value of the function times the area of the region. Hence, the double integral of the function \(f(x, y)\) over the region \(R\) is non-negative and not more than 1.

Step-by-step solution

Understanding double integral's physical meaning

A double integral of a function over a region is the volume under the function and above the region in the xy-plane. In this case, their values are limited between 0 and 1. So the volume can't be more than the area of the region which is 1.

Proving the lower limit

The function \(f(x,y) \geq 0\), therefore, \( \int_{R} \int f(x, y) d A \geq 0\). Because the integral of a non-negative function is non-negative.

Proving the upper limit

The function \(f(x,y) \leq 1\), and the area of \(R\) is 1, therefore \( \int_{R} \int f(x, y) d A \leq 1 * 1 = 1\). Because the integral of a function over a region is less than or equal to the maximum value of the function times the area of the region.