Let [a, b] be a bounded closed interval, f ∈ R[a, b] and M > 0. Suppose that |f (x)| ≤ Mfor all x ∈ [a, b]. Prove thatZ baf ≤ M (b − a)
Question
Let [a, b] be a bounded closed interval, f ∈ R[a, b] and M > 0. Suppose that |f (x)| ≤ M for all x ∈ [a, b]. Prove that
Solution
The statement you're asking to prove is that the integral of a function f over a closed interval [a, b] is less than or equal to M times the length of the interval, given that the absolute value of f(x) is less than or equal to M for all x in [a, b]. Here's a step-by-step proof:
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Start with the given inequality: |f(x)| ≤ M for all x in [a, b]. This means that -M ≤ f(x) ≤ M.
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Integrate the inequality over the interval [a, b]. Remember that the integral of an inequality is itself an inequality, provided the functions involved are integrable, which is given in this case. So, we have -M(b - a) ≤ ∫f(x) dx from a to b ≤ M(b - a).
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The left side of this inequality, -M(b - a), is the negative of the quantity we're trying to prove. The right side, M(b - a), is the quantity we're trying to prove.
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Therefore, we've shown that the integral of f over [a, b] is indeed less than or equal to M times the length of the interval.
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