# How do you find the average value of: #f(x) = 8 sin x + 6 cos x# on the interval #[0,(10pi)/6 ]#?

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To find the average value of ( f(x) = 8 \sin(x) + 6 \cos(x) ) on the interval ([0, \frac{10\pi}{6}]), you need to use the formula for the average value of a function on a closed interval:

[ \text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) , dx ]

where ( a = 0 ) and ( b = \frac{10\pi}{6} ) in this case. So,

[ \text{Average value} = \frac{1}{\frac{10\pi}{6} - 0} \int_{0}^{\frac{10\pi}{6}} (8 \sin(x) + 6 \cos(x)) , dx ]

[ = \frac{6}{10\pi} \left[ -8 \cos(x) + 6 \sin(x) \right]_{0}^{\frac{10\pi}{6}} ]

[ = \frac{6}{10\pi} \left[ -8 \cos\left(\frac{10\pi}{6}\right) + 6 \sin\left(\frac{10\pi}{6}\right) - (-8 \cos(0) + 6 \sin(0)) \right] ]

[ = \frac{6}{10\pi} \left[ -8 \cdot \frac{1}{2} + 6 \cdot \frac{\sqrt{3}}{2} - (-8 \cdot 1 + 6 \cdot 0) \right] ]

[ = \frac{6}{10\pi} \left[ -4 + 3\sqrt{3} - (-8) \right] ]

[ = \frac{6}{10\pi} \left[ 4 + 3\sqrt{3} \right] ]

[ = \frac{12 + 9\sqrt{3}}{5\pi} ]

[ \approx 1.346 ]

Therefore, the average value of ( f(x) = 8 \sin(x) + 6 \cos(x) ) on the interval ([0, \frac{10\pi}{6}]) is approximately ( 1.346 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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