Problem 10 - Calculus Early Transcendentals

Question

If \(f\) is continuous on \([a, b]\) and \(\int_{a}^{b}|f(x)| d x=0,\) what can you conclude about \(f ?\)

Step-by-Step Solution

Answer

Answer: If the integral of the absolute value of a continuous function \(f\) on the interval \([a, b]\) is 0, it implies that the function \(f\) must be the zero function on the interval \([a, b]\).

Step 1: Recall the definition of the absolute value of a function

For any function \(f(x)\), its absolute value \(|f(x)|\) is defined as: $$ |f(x)| = \begin{cases} f(x) & \text{if } f(x) \geq 0, \\ -f(x) & \text{if } f(x) < 0. \end{cases} $$

Step 2: Note the properties of the absolute value function

Remember that the absolute value function is always non-negative, i.e., for any \(x\) in its domain, \(|f(x)|\geq0\). This is important because it sets a lower bound to the value that the integral of \(|f(x)|\) can take.

Step 3: Draw a conclusion about the integral given

Since \(|f(x)| \geq 0\) for all \(x\) in \([a, b]\) and the integral \(\int_{a}^{b}|f(x)| dx = 0\), this implies that the absolute value of the function \(f\) must be identically zero on the interval \([a, b]\). So, for all \(x\) in \([a, b]\), \(|f(x)|=0\).

Step 4: Relate the absolute value to the original function

Since \(|f(x)|=0\) for all \(x\) in \([a, b]\), and by the definition of the absolute value, this implies that \(f(x)=0\) for all \(x\) in \([a, b]\). Therefore, we can conclude that \(f\) must be the zero function on the interval \([a, b]\).