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Use the properties of integrals to verify the ine…

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Problem 56 Easy Difficulty

Use the properties of integrals to verify the inequality without evaluating the integrals.

$ \displaystyle \int^1_0 \sqrt{1 + x^2} \,dx \le \int^1_0 \sqrt{1 + x} \,dx $

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Video Transcript

all right, we're going to try to calculate or prove this inequality that you see right here. What's important is that the book specifically asks you to do it without evaluating the integral. So how do we approach this? We use one of the properties of Integral Is that says if you have the same limit of integration and if you have an expression that's less then the other function than the integration is also going to be less than it. So long story short, we're going to prove that this portion is larger than that portion and that doesn't involve the evaluation of thes integral. So we're going to be able to do that. Okay, so let's compare the square root off one plus X squared versus squared of one plus X. We know that square root functions are 1 to 1. So we know that the input is what really matters here to decide which ones larger. So I'm gonna look at X squared versus X. I am going to ignore the one because one is just added to both of them. So if I subtract it from one, you can subtract it for the other, and then the relationship between which ones larger and which one smaller doesn't change. Okay, now X squared versus X. Just way looking at it. You might think that X squared could be larger than X. But that's not necessarily true, because in this case, we're looking at the region 0 to 1. Okay, graphically, what's happening is this X squared looks like this. Well, X looks like that between the region 0 to 1, so eventually X squared is going to be larger than X. But when X is a number between 0 to 1, we know that X is actually larger than X square. So what's the relationship that we see right here? X squared is less than X and going going back, we can show that the square root off one plus x squared iss less than the square root off one plus X and according to the properties of Integral, this is also going to be too. Now, what's important is the fact that it's from 0 to 1. Okay, Without this, we're not able to prove this portion, so it wouldn't work. So, keeping that in mind, we were able to prove this situation

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Calculus: Early Transcendentals

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