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# What is wrong with the equation?$\displaystyle \int^2_{-1} \frac{4}{x^3} \, dx = -\frac{2}{x^2} \Bigg]^2_{-1} = -\frac{3}{2}$

## Integrand is not continuous

Integrals

Integration

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### Video Transcript

Okay, so let's go through one of thieve Eri common mistakes that a lot of people make when we do into So you see a rational function right here and then we're going to try to take the integral and evaluated from negative 1 to 2. So if you remember the power rule, the integral basically says if I take the anti derivative with respect to X, you will have X to the n plus one, all divided by M plus one. So, using that idea, you can consider execute as X to the negative three. So the expression basically boils down to negative two over X squared, and then you're going to evaluate it from negative 1 to 2. If you plug in the numbers, the first number is going to be negative to over four. You're going to subtract negative two divided by one. So it simplifies into negative three over to looks good, right? But it turns out that this is actually not a proper integral. It's not one of those things where we are allowed to do these calculations. Okay, Why? It will make sense when you actually graph it. The graph off four over X cubed. As you can see, you have a variable on the denominator. So when you have things like that, by the time you're doing calculates, most of most people are aware of this. But you have to be very mindful that if x zero there's going to be a division by zero, so many weird things could happen, right? So the integration limit is from negative 1 to 2, which actually includes zero. So if I take a look at it, the graph roughly roughly looks something like this. And then what we're trying to do is to try to evaluate the values from here, too. There, quote unquote under the curb, which goes all the way down to infinity, comes all the way up from infinity, and then we're trying to calculate that area. And as you can see, it blows up very quickly. So it doesn't make sense to say that it's going to be negative three over to, especially because we know that this portion looks like it's going to be bigger than at least this portion, right? So things doesn't make sense when you don't do the proper, Um, when you don't use when you don't follow the proper rules. In this case, this is not a continuous function between the interval negative 1 to 2, and that that makes this an improper interval. So you have to take steps to make sure that you're doing it the correct way.

University of California, Berkeley

Integrals

Integration

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