What is the average value for #f(x)=8x-3+5e^(x-2)# over the interval [0,2]?
Try this to find the average value
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To find the average value of a function ( f(x) ) over an interval ([a, b]), you use the formula:
[ \text{Average Value} = \frac{1}{b-a} \int_{a}^{b} f(x) , dx ]
For the function ( f(x) = 8x - 3 + 5e^{x-2} ) over the interval ([0, 2]), the average value is:
[ \text{Average Value} = \frac{1}{2-0} \int_{0}^{2} (8x - 3 + 5e^{x-2}) , dx ]
[ = \frac{1}{2} \left[ 4x^2 - 3x + 5e^{x-2} \right]_{0}^{2} ]
[ = \frac{1}{2} \left[ (4(2)^2 - 3(2) + 5e^{2-2}) - (4(0)^2 - 3(0) + 5e^{0-2}) \right] ]
[ = \frac{1}{2} \left[ (16 - 6 + 5e^{0}) - (0 - 0 + 5e^{-2}) \right] ]
[ = \frac{1}{2} \left[ 10 - 5e^{-2} \right] ]
[ = 5 - \frac{5}{e^2} ]
[ \approx 5 - \frac{5}{7.389} ]
[ \approx 5 - 0.676 ]
[ \approx 4.324 ]
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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.
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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