What is the surface area produced by rotating #f(x)=3x^3+6x^2-2x+3, x in [-3,2]# around the x-axis?
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The surface area produced by rotating the function f(x) = 3x^3 + 6x^2 - 2x + 3, where x is in the interval [-3, 2], around the x-axis can be calculated using the formula for surface area of revolution, which is given by:
Surface Area = ∫[a, b] 2πf(x) √(1 + (f'(x))^2) dx
After finding f'(x) and plugging it into the formula, integrate it over the interval [-3, 2].
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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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