We are tasked with proving that if $\int_a^b f(x) \, dx = 0$, where $f : [a, b] \to [0, \infty)$ is a nonnegative continuous function, then $f(x) = 0$ for all $x \in [a, b]$. Let's proceed step by step.

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Proof (By Contradiction)

1. Assume the contrary: Suppose $f(x) \neq 0$ for some $c \in [a, b]$. That is, there exists a point $c$ such that $f(c) > 0$.

2. Continuity and positivity: Since $f$ is continuous on $[a, b]$, there exists a small interval $[c - \epsilon, c + \epsilon] \subset [a, b]$ (for some $\epsilon > 0$) such that $f(x) > 0$ for all $x \in [c - \epsilon, c + \epsilon]$. This is a result of the definition of continuity, where $f(x)$ remains close to $f(c) > 0$ in a small neighborhood around $c$.

3. Positive contribution to the integral: On the interval $[c - \epsilon, c + \epsilon]$, since $f(x) > 0$ and $f$ is continuous, we can say: $\int_{c - \epsilon}^{c + \epsilon} f(x) \, dx > 0$ because the integral of a strictly positive function over a nonzero interval is positive.

4. Contradiction: Now consider the integral over the whole interval $[a, b]$. Using the additivity of the integral, we have: $\int_a^b f(x) \, dx = \int_a^{c-\epsilon} f(x) \, dx + \int_{c-\epsilon}^{c+\epsilon} f(x) \, dx + \int_{c+\epsilon}^b f(x) \, dx.$ Since $f(x) \geq 0$ on $[a, b]$ and $f(x) > 0$ on $[c - \epsilon, c + \epsilon]$, the term $\int_{c-\epsilon}^{c+\epsilon} f(x) \, dx > 0$, making the total integral $\int_a^b f(x) \, dx > 0$.

   However, this contradicts the given condition that $\int_a^b f(x) \, dx = 0$.

5. Conclusion: The assumption that $f(x) \neq 0$ for some $c \in [a, b]$ leads to a contradiction. Therefore, $f(x) = 0$ for all $x \in [a, b]$.

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Key Takeaways:

• The proof relies on the continuity of $f$, which ensures the existence of an interval where $f(x) > 0$ around any point $c$ where $f(c) > 0$.
• The nonnegativity of $f(x)$ ensures that the integral of $f$ over any subinterval is nonnegative.
• If $\int_a^b f(x) \, dx = 0$, the only way this can hold is if $f(x) = 0$ everywhere on $[a, b]$.

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Do you want more details or have any questions?
Here are 5 related questions to deepen understanding:

1. Why is continuity critical for this proof to work?
2. What would change if $f(x)$ were not nonnegative?
3. How does this proof rely on the property of integrals of nonnegative functions?
4. Can this proof be extended to functions defined on $\mathbb{R}$ (instead of $[a, b]$)?
5. What happens if $f(x)$ is piecewise continuous instead of continuous?

Tip: When dealing with integrals and continuity, always explore the local behavior of the function to identify intervals of positivity or negativity.