Evaluate the integral or state that it diverges. ∫ − ∞ − 1 2 x 3 d x \int_{-\infty}^{-1}\frac{2}{x^3}dx

we're asked to evaluate the integral or state whether or not it diverges. Now notice here we have the definite integral that goes from negative infinity up to negative one. So how do we deal with an integral like this? Well, because I see we have an infinity somewhere in the bounds, this is what's known as an improper integral, and we need to rewrite improper integrals in this form. So we're going to have the limit as t approaches this negative infinity. So that's going to leave us with the integral from t all the way up to negative one. Now from here, what I can do is try to evaluate this integral that we have right here, two over x cubed. Now I can write this in a little bit more of a simple way by pulling that two, which is a constant, outside of the integral. So I have from t to negative one of one over x cubed, which is the same thing as x to the negative third power integrated with respect to x. Now how do I integrate x to the negative third power? Well, I just need to do the power rule here, so add one to the exponent and then divide by that new exponent. So we'll keep this limit as t approaches negative infinity, and that's going to be two times, in this case, is going to be negative or x to the negative three plus one, which is x to the negative two over negative two, and all of this is going to be bounded from t to negative one. At this point, I can rewrite what we have inside those brackets. So keep this limit as t approaches negative infinity, it's going to be two times, and this is the same thing as negative one over two times x squared from t to negative one. Now at this point, I can go ahead and apply my bounds. So I'm gonna start with my bound of negative one. So what we're gonna have is the same limit right here as t approaches negative infinity. You wanna make sure you keep this limit on as you go through each case until you actually apply the limit. So we'll keep this limit, and it's going to be two times, and then we'll plug in one. Now plugging in one here, it's gonna be negative one over two times negative one squared, and that's why I'm actually pulling in negative one. And then what I'm going to do is subtract off this same piece with t plugged in. I can cancel this negative sign, just give us plus one over two, and I'm going to plug in t, so it's going to be one over two times t squared. So this is what happens. Now at this point, we know how to evaluate negative one over two times negative one squared. Negative one squared is positive one, so we know that. And we know that in this case, the positive one times two will give us just two, so we're just gonna end up with negative one half right about here. But what happens to this term? Well, notice we have t squared in the denominator. T is getting infinitely big in the negative direction, so it's technically getting infinitely small. But because this is going to be squared, it's actually gonna get infinitely big because whenever you square a negative number, it becomes positive. So we're gonna get a larger and larger and larger positive number in the denominator multiplied by two, which is going to cause this whole thing to just go to zero. If you have a denominator that's infinitely growing, it's going to overtake the top setting the whole thing to zero. So, basically, this whole thing will just be zero when we apply that limit. So since we applied the limit, we actually don't even need to write this limit out anymore in this case. So I can clear the limit here, and I can just focus on what we evaluated. So we have two times this quantity, which is the same thing as just two times negative one half, which I can cancel these twos, giving us just negative one. So negative one is the result that we get, and because we did converge to a value that is not just blowing up to infinity or negative infinity, I can see that this whole thing converges. So converges to negative one is the solution to this problem. Hope you found this helpful.

12. Techniques of Integration

Improper Integrals

Calculus

12. Techniques of Integration

Improper Integrals

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Multiple Choice

Evaluate the integral or state that it diverges.
$\int_{- \infty}^{- 1} \frac{2}{x^{3}} d x$

The integral diverges.

$negative 2$ ; converges.

$negative 1$ ; converges.

$1$ ; converges.

Verified step by step guidance

Rewrite the given improper integral as a limit to handle the infinite bound: $ \int_{-\infty}^{-1} \frac{2}{x^3} \, dx = \lim_{a \to -\infty} \int_{a}^{-1} \frac{2}{x^3} \, dx $.

Evaluate the integral $ \int \frac{2}{x^3} \, dx $ by finding its antiderivative. Recall that $ \int x^n \, dx = \frac{x^{n+1}}{n+1} $ for $ n \neq -1 $. Here, $ n = -3 $, so the antiderivative is $ -\frac{1}{x^2} $.

Substitute the limits of integration into the antiderivative: $ \int_{a}^{-1} \frac{2}{x^3} \, dx = \left[ -\frac{1}{x^2} \right]_{a}^{-1} = -\frac{1}{(-1)^2} - \left(-\frac{1}{a^2}\right) $.

Simplify the expression: $ -\frac{1}{(-1)^2} = -1 $ and $ -\left(-\frac{1}{a^2}\right) = \frac{1}{a^2} $. Thus, the integral becomes $ -1 + \frac{1}{a^2} $.

Take the limit as $ a \to -\infty $: Since $ \frac{1}{a^2} \to 0 $ as $ a \to -\infty $, the result of the integral is $ -1 $. Therefore, the integral converges to $ -1 $.