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Find the average value of the function f(x)=e^{{2x}} on the interval [0,4].

An integral gives us the total value of a function over an interval. If we then take this and divide it by the length of the interval, then we attain the average value of the function on the interval. Thus the average value of a function on an interval [a,b] is given by

{\frac  {1}{b-a}}\int _{{a}}^{{b}}f(x)\,dx\,

For this specific function, we have

{\frac  {1}{4-0}}\int _{{0}}^{{4}}e^{{2x}}\,dx={\frac  {1}{4}}\left[{\frac  {1}{2}}e^{{2x}}\right]_{{0}}^{{4}}={\frac  {1}{8}}\left[e^{8}-1\right]\,

Thus, the average value of f(x) on the interval [0,4] is {\frac  {1}{8}}\left[e^{8}-1\right]. Though the function increases rapidly and thus is different on every interval, its average value per 1 unit, on this interval, is about 372.49.

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