Question

True or False If \(\int_{a}^{b} f(x) d x>0,\) then \(f(x)\) is positive for all \(x\) in \([a, b] .\) Justify your answer.

Short answer

False. Just because the definite integral of a function over an interval is positive, it does not mean that the function itself is positive throughout the interval. The integral being positive only means that the overall area above the x-axis is greater than the area below the x-axis within the interval, but the function may still take on negative values within that interval.

Step-by-step solution

Understanding the Definite Integral

The definite integral of a function \(f(x)\) from \(a\) to \(b\) represents the net area enclosed between the function and the \(x\)-axis within the bounds of \(a\) and \(b\). The net area is calculated as the difference between the area above the \(x\)-axis (considered positive) and the area below the \(x\)-axis (considered negative). If an integral is positive, it means the total area above the \(x\)-axis is greater than the total area below the \(x\)-axis in the interval from \(a\) to \(b\).

Evaluate the Statement

Given that the integral from \(a\) to \(b\) of \(f(x)\) is greater than 0. This means that the overall enclosed area is positive, indicating that there is more area above the \(x\)-axis than below. However, this does not mean that \(f(x)\) is positive for all \(x\) in the interval. The function could as well cross the \(x\)-axis one or more times, having portions where it is negative, but in such a way that the positive area exceeds the negative.

Verification through example

Let's consider, for example, the function \(f(x) = x^3 -x\) on the interval \([-1, 1]\). This function crosses the x-axis at -1, 0 and 1 and is clearly negative at some points, but its integral over the interval \([-1, 1]\) is 0, which is positive.