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What is the surface area of the solid created by revolving #f(x) = e^(2x) , x in [2,7]# around the x axis?

Answer 1

Samantha Adkison

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Answer 2

Scarlett Allison

To find the surface area of the solid created by revolving the function f(x) = e^(2x) on the interval [2, 7] around the x-axis, you can use the formula for the surface area of revolution:

[S = \int_{a}^{b} 2\pi f(x) \sqrt{1 + [f'(x)]^2} dx]

Where:

(a) and (b) are the bounds of integration.
(f(x)) is the function being revolved.
(f'(x)) is the derivative of the function with respect to (x).

For the function (f(x) = e^(2x)):

(f'(x) = 2e^(2x)).

Plugging in these values and integrating from (x = 2) to (x = 7):

[S = \int_{2}^{7} 2\pi e^(2x) \sqrt{1 + [2e^(2x)]^2} dx]

You would then solve this integral to find the surface area.

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Answer from HIX Tutor

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.