Question

If\(\int\limits_0^1 {f(x)dx = 0} \),then\(f(x) = 0\)for\(0 \le x \le 1\).

Short answer

The given statement is FALSE Counter Example: Take \(f(x) = 2x - 1\)

Step-by-step solution

Step 1: Explanation

The given statement is FALSE

To prove the given statement false, it is sufficient to give an example of f(x) such that f(x) is not zero for all values of \(x \in (0,1)\), but \(\int\limits_0^1 {f(x) = 0} \).

Step 2: Example

One such example is \(f(x) = 2x - 1\)

Note that f(x) is not zero for all values of \(x \in (0,1)\)

But

\(\begin{array}{l}\int\limits_a^c {f(x)\;dx\; = \;\int\limits_0^1 {f(x)\;dx = \int\limits_0^1 {(2x - 1)dx} } } \\ = ({x^2} - x)_0^1 = 0\end{array}\)

The given statement is FALSE Counter Example: Take \(f(x) = 2x - 1\)