# How do you test the improper integral #int x^3 dx# from #(-oo, oo)# and evaluate if possible?

The integral is not convergent as:

The two limits should be finite separately and they are not.

and then:

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To test the improper integral ( \int_{-\infty}^{\infty} x^3 , dx ) for convergence or divergence, you can split it into two separate integrals, one from ( -\infty ) to ( 0 ) and the other from ( 0 ) to ( \infty ). Then, evaluate each integral individually.

- For the integral from ( -\infty ) to ( 0 ):

[ \int_{-\infty}^{0} x^3 , dx ]

Apply the antiderivative:

[ \left[ \frac{x^4}{4} \right]_{-\infty}^{0} ]

Since ( x^4 ) grows without bound as ( x ) approaches negative infinity, this integral diverges to negative infinity.

- For the integral from ( 0 ) to ( \infty ):

[ \int_{0}^{\infty} x^3 , dx ]

Apply the antiderivative:

[ \left[ \frac{x^4}{4} \right]_{0}^{\infty} ]

As ( x ) approaches infinity, ( x^4 ) also grows without bound, so this integral diverges to positive infinity.

Since both parts of the improper integral diverge, the entire improper integral ( \int_{-\infty}^{\infty} x^3 , dx ) diverges.

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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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