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# Express the limit as a definite integral on the given interval. $\displaystyle \lim_{n \to \infty} \sum_{i = 1}^n [5(x_i^*)^3 - 4x_i^*] \, \Delta x$, [2, 7]

## $$\int_{2}^{7}\left(5 x^{3}-4 x\right) d x$$

Integrals

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### Video Transcript

All right, let's actually move onto the next question. Hopefully, you understood what I explained right here. And you can refer back to this problem any time you want. Okay? Now the question is going to be He's black limit as an approach tested Infinity off the song from my goes toe Want end off when you copy it down correctly. Five x i Q t s cubed minus four x. I squared the entire thing times Delta X, where X is from 2 to 7. And it's the same kind of question, right? This as the form of the definite integral. So let's recap the important parts, the limit and the some. You know, the sigma notation is basically the Greek large s so it makes sense that when you're taking the limit, it becomes this fancy looking s okay, the integration is going to be from A to B. We need to identify what those are. We need to figure out what the function is, the height off that point, and we need to figure out what the excess. And of course, we already know that the Delta X is going to be this guy. Okay, so let's figure out what? My effort axis. Very simple. It's this guy. So we will have the expression five x Q minus four X squared. I already mentioned that the Delta X becomes your DX. I usually don't write it, but some people prefer that you use the parentheses so that it looks better where you know exactly where the integration starts at the Trop. It's quite clear if you put this thing, but some people actually like it better that way. A and B are simply the starting point and the endpoint from 2 to 7. So here is your answer. So, in summary, the Riemann some can be represented as the integral from 2 to 7 of five. X cubed minus four X squared TX and this is how you answer this question?

University of California, Berkeley

Integrals

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