Question

Integrals of Nonpositive Functions Show that if \(f\) is integrable then See page \(293 .\) \(f ( x ) \leq 0\) on \([ a , b ] \Rightarrow \int _ { 0 } ^ { b } f ( x ) d x \leq 0\)

Short answer

Because a nonpositive integrable function may at most equal zero within any given interval, the result of its integration is always equal to or less than zero, as required: \(\int_{a}^{b} f(x)\, dx \leq 0\). The proof is based on the properties of Riemann sums and the nature of the integral as the limit of these sums.

Step-by-step solution

Understanding the Riemann Sum

A function's integral over a region can be approximated by a Riemann sum. The function must be integrable for a Riemann sum approximation to be useful. To apply it, let's partition the interval [a, b] into n subintervals. Each subinterval will have a length of \(\Delta x\). Then a Riemann sum of the function f on this partition is the sum of the product of the function's value at some point \(x_{i}^*\) in the \(i^{th}\) subinterval and the width of this subinterval, which is \(\Delta x\). The Riemann sum is then given as \(S(f, P) = \sum_{i=1}^n f(x_{i}^*)\Delta x\).

Applying the Nonpositive Property of \(f\)

Since \(f(x)\) is given to be nonpositive for all \(x\) in \([a, b]\), this means that \(f(x) \leq 0\) for all \(x\) in this interval. Applying this inequality to the Riemann sum expression, the sum \(S(f, P)\) becomes less than or equal to zero. This can be written as \(S(f, P) \leq 0\). This is because each term \(f(x_{i}^*)\Delta x\) in the sum is nonpositive as \(f(x_{i}^*) \leq 0\) and \(\Delta x\) is always positive.

Formulating the Result

The integral of function \(f\) from \(a\) to \(b\) is the limit of the Riemann sums as the subintervals' width \(\Delta x\) goes to zero. As \(f(x)\) is integrable, the integral \(\int_{a}^{b} f(x)\, dx\) exists and it is the limit of these Riemann sums. Since each of these Riemann sums is less than or equal to zero, then the limit is less than or equal to zero as well. This can be written as \(\int_{a}^{b} f(x)\, dx \leq 0\). This completes the proof.