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If $ f $ is continuous and $ \displaystyle \int_{1}^3 f(x) dx = 8 $, show that $ f $ takes on the value 4 at least once on the interval $ [1, 3] $.

there exists a $c$ such that $f(c)=4$

Applications of Integration

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we know we're gonna be using the mean value theorem, which essentially means that because of his continuous there exists a numbers see between the bounds of 123 such that the average value won over B minus A from A to B of our function after vax d X equals F f c. So we know this means that half of sea is 1/2 times ate half of eight is four. Therefore, there does exist to see such the F of CS four.