Definite Integration
Mean Value Theorem for Integrals
Grade None
Question:
<p>Let \(f\) be continuous on \([a,b]\). Which are correct about the Mean Value Theorem for integrals?</p>
\existsc\in (a,b): f(c) = \intₐ^b f(x)dx (without dividing by (b-a))
\existsc\in (a,b): f(c)(b-a) = \intₐ^b f(x)dx
The average value of f on [a,b] is (1/(b-a))\intₐ^b f(x)dx
If f \geq 0 on [a,b], then \intₐ^b f(x)dx \geq 0
Step-by-Step Solution
Key Concept: MVT for integrals: \existsc: f(c)(b-a) = \intf. Option A omits the (b-a) factor — wrong.
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<p><strong>A:</strong> ✗ Missing the factor $(b-a)$. The correct statement has $f(c)(b-a)=\int_a^b f$.</p>
<p><strong>B:</strong> ✓ This is exactly the MVT for integrals.</p>
<p><strong>C:</strong> ✓ Definition of average value.</p>
<p><strong>D:</strong> ✓ If $f(x)\ge 0$ everywhere on $[a,b]$, then the integral is non-negative (monotonicity of integration).</p>
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Correct Answer: ['B', 'C', 'D']