Question

Assume that \(\mathrm{f}\) is integrable and nonnegative on \([a, b] .\) If \(\int_{a}^{b} f(x) d x=0\), prove that \(f(c)=0\) at each point of continuity of \(f\).

Short answer

Question: Given that a function f is integrable and nonnegative on an interval [a, b], and the integral of f(x) from a to b is equal to 0, prove that for any point c, where the function f is continuous, the value of f(c) is equal to 0. Answer: Since the integral of f(x) from a to b is 0 and f is nonnegative, we can conclude that the integral of f(x) is also 0 for any continuous subinterval of [a, b]. If f is continuous at a point c, then the limit of f(x) as x approaches c exists and is equal to f(c). We used the properties of definite integrals to first show that the integral of f(x) is 0 for subintervals around point c. We then utilized the continuity of f at point c and the integral property to deduce that f(c) = 0. Therefore, for any point c where the function f is continuous, it follows that f(c) = 0.

Step-by-step solution

Using the properties of the definite integral

First, we need to establish that the integral of f(x) is equal to 0 for any continuous subinterval of [a, b]. We know that \(\int_{a}^{b} f(x) \: dx = 0\). In general, a property of definite integrals states that if \(\int_{a}^{b} f(x) \: dx = 0\) and the function f is nonnegative, then for any subinterval [c, d] in [a, b], \(\int_{c}^{d} f(x) \: dx = 0\). Thus, we have \(\int_{c}^{d} f(x) \: dx = 0\) for any subinterval [c, d] in [a, b] where f(x) is continuous.

Utilize the continuity of f at point c

Now, let's consider f continuous at point c, where c is in [a, b]. By definition, if a function is continuous at a point, then the limit of the function as x approaches that point exists and is equal to the value of the function at the point, i.e. \(\lim_{x \to c} f(x) = f(c)\).

Prove that f(c) = 0 using the integral property

Since f is continuous at point c, we want to find a small neighborhood around c such that f is continuous on that neighborhood. Let's consider an open interval (c - h, c + h) with h > 0. We know from Step 1 that \(\int_{c - h}^{c + h} f(x) \: dx = 0\) since it is a subinterval of [a, b]. Now, because the limit exists for a continuous function, we can say that for a sufficiently small h > 0, \(|f(x) - f(c)| < \epsilon\) for all x in the interval (c - h, c + h). This implies that for a subinterval [c - \delta, c + \delta] of (c - h, c + h) where 0 < \delta < h, f(x) is continuous and nonnegative. We now want to prove that \(\int_{c - \delta}^{c + \delta} f(x) \: dx = 0\). By the properties of definite integrals, the integral over [c - \delta, c] and the integral over [c, c + \delta] should also be zero, i.e. \(\int_{c - \delta}^{c} f(x) \: dx = 0\) and \(\int_{c}^{c + \delta} f(x) \: dx = 0\). Now, using the additivity property of definite integrals, we can write: \(\int_{c - \delta}^{c + \delta} f(x) \: dx = \int_{c - \delta}^{c} f(x) \: dx + \int_{c}^{c + \delta} f(x) \: dx = 0 + 0 = 0\) From this, it is evident that \(\int_{c - \delta}^{c + \delta} f(x) \: dx = 0\), and since f(x) is nonnegative, we conclude that \(f(c) = 0\). Therefore, for any point c where the function f is continuous, it follows that \(f(c) = 0\).