# What is the volume of the solid produced by revolving #f(x)=x^3-2x+3, x in [0,1] #around the x-axis?

Here's a TI screenshot of the answer:

As a decimal expansion, we have :

When you integrate area, you get volume. In other words, if you go backwards from area, you'll get volume.

So you just integrate with the standard disc method :

Here,

Here,

That's the actual radius of your disc., varying as a function of x.

Here is a pretty pic:

I did this in Maple...and yes, Maple freakin' rules.

Finally, you need to "get your hands dirty" and fill in all the algebraic details as far as arriving at the antiderivative and then crunching the bounded definite integral.

That's your job.

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To find the volume of the solid produced by revolving the function ( f(x) = x^3 - 2x + 3 ) on the interval ( [0, 1] ) around the x-axis, you would use the method of cylindrical shells. The formula for the volume of revolution using cylindrical shells is:

[ V = 2\pi \int_{a}^{b} x \cdot f(x) , dx ]

For the given function ( f(x) = x^3 - 2x + 3 ) on the interval ( [0, 1] ), the volume becomes:

[ V = 2\pi \int_{0}^{1} x \cdot (x^3 - 2x + 3) , dx ]

Solving this integral will give you the volume of the solid of revolution.

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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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