Determine whether the improper integral diverges or converges. integral_1^infinity 1/x^3 dx converges diverges Evaluate the integral if it converges. (If the quantity diverges, enter DIVERGES.

It can be evaluated using the limit comparison test or by integrating 1/[tex]x^3[/tex] directly to get -1/2[tex]x^2[/tex] evaluated from 1 to infinity, Therefore, the integral converges to 1/2.

The integral can be written as:

∫₁^∞ 1/x³ dx

To determine whether the integral converges or diverges, we can use the p-test for integrals. The p-test states that:

If p > 1, then the integral ∫₁^∞ 1/xᵖ dx converges.

If p ≤ 1, then the integral ∫₁^∞ 1/xᵖ dx diverges.

In this case, p = 3, which is greater than 1. Therefore, the integral converges.

To evaluate the integral, we can use the formula for the integral of xⁿ:

∫ xⁿ dx = x (n+1)/(n+1) + C

Using this formula, we get:

∫₁^∞ 1/x³ dx = lim┬(t→∞)⁡(∫₁^t 1/x³ dx)

             = lim┬(t→∞)⁡[ -1/(2x²) ] from 1 to t

             = lim┬(t→∞)⁡( -1/(2t²) + 1/2 )

             = 1/2

Therefore, the integral converges to 1/2.

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To determine if this integral converges or diverges, we can use the p-test. According to the p-test, if the integral of the form ∫1∞ 1/x^p dx is less than 1, then the integral converges. If the integral is equal to or greater than 1, then the integral diverges.

In this case, p=3, so we have ∫1∞ 1/x^3 dx = lim t→∞ ∫1t 1/x^3 dx.

Evaluating the integral, we get ∫1t 1/x^3 dx = [-1/(2x^2)]1t = -1/(2t^2) + 1/2.

Taking the limit as t approaches infinity, we get lim t→∞ [-1/(2t^2) + 1/2] = 1/2.

Since 1/2 is less than 1, we can conclude that the given improper integral converges.

Therefore, the value of the integral is ∫1∞ 1/x^3 dx = 1/2.
To determine whether the improper integral converges or diverges, we need to evaluate the integral and see if it results in a finite value. Here's the given integral:

∫(1 to ∞) (1/x^3) dx

1. First, let's set the limit to evaluate the improper integral:

lim (b→∞) ∫(1 to b) (1/x^3) dx

2. Next, find the antiderivative of 1/x^3:

The antiderivative of 1/x^3 is -1/2x^2.

3. Evaluate the antiderivative at the limits of integration:

[-1/2x^2] (1 to b)

4. Substitute the limits:

(-1/2b^2) - (-1/2(1)^2) = -1/2b^2 + 1/2

5. Evaluate the limit as b approaches infinity:

lim (b→∞) (-1/2b^2 + 1/2)

As b approaches infinity, the term -1/2b^2 approaches 0, since the denominator grows without bound. Therefore, the limit is:

0 + 1/2 = 1/2

Since the limit is a finite value (1/2), the improper integral converges. Thus, the integral evaluates to:

∫(1 to ∞) (1/x^3) dx = 1/2

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