anonymous
  • anonymous
what is the exact difference between mean value theorem and average value when it comes to integration?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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shadowfiend
  • shadowfiend
The average value of a derivative is the mean value of its underlying function. So if you have \(f(x)\) and its derivative \(f'(x)\), the average value of \(f'(x)\) is the mean value of \(f(x)\). In essence, this means that \(f'(x)\) between two points \(a) and \(b\) must at some point equal its own average value between those two points.
anonymous
  • anonymous
The average value formula, 1/(b-a) * Integral f(x)dx from a to b, gives the average y-value of function f between a and b. The mean value theorem (MVT) says take any smooth curve. Select two points on that curve and connect them with a straight line. MVT guarantees that somewhere between the two points, there will be a tangent line that is parallel to the line connecting the two points.
anonymous
  • anonymous
thanks guys, i finally get it!

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anonymous
  • anonymous
Actually MVT of integration says that on (a, b) there exist a c such that f(c) = 1/(b-a) * Integral f(x)dx, and the proof is a lot easier than the derivative one

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