Mean Value Theorem for Integrals, Visual Proof, Examples, Practice Problems - Calculus

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The Mean Value Theorem for Integrals states that for a continuous function \(f(x)\) on a closed interval \([a,b]\), there is at least one point \(c\) within the interval where the function's value \(f(c)\) equals the function's average value over the interval, given by \(\frac{1}{b-a}\int _{a}^{b}f(x)\,dx\). Mathematically, this is expressed as \(f(c)=\frac{1}{b-a}\int _{a}^{b}f(x)\,dx\), or equivalently, \(\int _{a}^{b}f(x)\,dx=f(c)(b-a)\). This means a rectangle with width \((b-a)\) and height \(f(c)\) will have the same area as the definite integral of \(f(x)\) over the interval [a, b].

💡Key Aspects of the Theorem
• Continuity is Required: The function \(f(x)\) must be continuous on the closed interval \([a,b]\).
• Existence of 'c': The theorem guarantees that at least one such point \(c\) exists within the interval, but it does not provide a method to find \(c\). • Average Value: The value \(f(c)\) represents the average height of the function over the interval.
• Geometric Interpretation: Geometrically, the theorem states that for a continuous function, there exists a rectangle with height \(f(c)\) and width \((b-a)\) that has the same area as the region under the curve of \(f(x)\) from \(a\) to \(b\).

💡Mathematical Statement
• If \(f(x)\) is continuous on the closed interval \([a,b]\), then there exists a number \(c\in [a,b]\) such that:
\(f(c)=\frac{1}{b-a}\int _{a}^{b}f(x)\,dx\)
• Or, this can be rearranged to:
\(\int _{a}^{b}f(x)\,dx=f(c)(b-a)\)

💡Worksheets are provided in PDF format to further improve your understanding:
• Questions Worksheet: https://drive.google.com/file/d/1GQtYzPXeZpvnm7EdpOAPfEyr-VSltAse/view?usp=drive_link
• Answers: https://drive.google.com/file/d/1urkbDfPaDPsymE8T58TZZJHgeUw-P-qx/view?usp=drive_link

💡Chapters:
00:00 Mean value theorem for integrals
01:00 Worked example

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