Problem

If $ F(x) = \displaystyle \int^x_2 f(t) \, dt $, …

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Problem 51 Medium Difficulty

For the function $ f $ whose graph is shown, list the following quantities in increasing order, from smallest to largest, and explain your reasoning.

(A) $ \displaystyle \int^8_0 f(x) \, dx $
(B) $ \displaystyle \int^3_0 f(x) \, dx $
(C) $ \displaystyle \int^8_3 f(x) \, dx $
(D) $ \displaystyle \int^8_4 f(x) \, dx $
(E) $ f'(1) $

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February 25, 2021

There is no graph here lol!

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Video Transcript

the problem says for a function F Who's graph is shown last day following quantities in increasing order from the smallest to largest. And explain your reasoning. Right? So here is a graph given to us. Right? So, let me right here of the numbering of the segments like 123 and this is for this is five. This is 678. Alright. We can write in this point. So what do we observe you? We observe that Here. We can see that from 0-3. The area covered is negative here. Right? From 0-3. Can we say this? I'm sorry, I should try it with. And that there's been hell. Let me. Right. So this is Lady citizens. Right? So this area is having uh area covered is negative. Right? So far, 3-28. We can see that the area covered will be positive here quoting. So now moving forward near two, x equals to one. There is minimum point of function. Right? And and X equals to one, slope is negative because it is coming down. Right? So it is giving us a negative slope. Right? So this is actually having a negative slope. All right. So, now we can say that the smallest area is given by The area covered by 0-3 here. Right, So, the 0-3. And the integration of fX the eggs here. Alright, so that's why this has the smallest area. Okay, now slope at X equals to one is bit higher than the uh than that, but still negative. Yes. So what happens is when we integrated. Right. So at X equals to one. Right? The you can say the slope is actually a little higher than Before. This between between 0-1. It will have less slope. Right? But at X equals to one. If we see here this is one. So this slope is slightly higher than before. Uh slope which is which was between 0-1. All right. So, now F dash X. Uh to FDR X after that. The area bounded by the integral ah integration of F X dx limit zero to aid is higher as it is a combination of positive area and negative area. All right. So, this is the combination of positive area and negative. Mhm. Mhm. All right. So then, the area bounded by of F X D X from limit 4 to 8 is greater than because it has only positive area only. Right. So, this this becomes very easy because we have this positive part, right? 4 to 8. Right? So, this will have only positive area. So, that's where it is coming after this. Okay, know what we can say again. Uh Right. And finally, area bounded by the integral 3 to 8. Integration of F X E X. Right. F X D X From 3- eight. That is the biggest as it contains the whole positive area. Right? It contains whole positive area. Okay, So, this is how we number it. So they be option B will be less least area will be having the least area, then we have E. That is f. and we have after that A. D. C. So the order becomes the highest. Uh, the least area will be having the baby. E. 10 A D N D D N C. C. Has the highest covering area in the club. So this is how we solve this problem. I hope you understood the concept. Thank you for watching.

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James Stewart

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Kristen Karbon

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Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Kristen Karbon

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Michael Jacobsen

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James StewartCalculus: Early Transcendentals

Kristen Karbon

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Michael Jacobsen

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James Stewart

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