if-f-is-continuous-on-mathbbr-prove-that-int_ab-fxc-d-xint_acbc-fx-d-x-for-the-case-where-fx-geqslan

Name: if f is continuous on mathbbr prove that int_ab fxc d xint_acbc fx d x for the case where fx geqslan
Uploaded: 2019-07-04
Duration: 36 s
Channel: Numerade
Description: If $f$ is continuous on $\mathbb{R},$ prove that $$\int_{a}^{b} f(x+c) d x=\int_{a+c}^{b+c} f(x) d x$$ For the case where $f(x) \geqslant 0,$ draw a diagram to interpret this equation geometrically as an equality of areas.

Question

If $f$ is continuous on $\mathbb{R},$ prove that $$\int_{a}^{b} f(x+c) d x=\int_{a+c}^{b+c} f(x) d x$$ For the case where $f(x) \geqslant 0,$ draw a diagram to interpret this equation geometrically as an equality of areas.

Fuzail Shakir · Accepted Answer

Okay, so it says if efforts continues on our proof that the into go from a to B f f f x plus e T x is equal to the issue of a plus C to be fussy of f of x t act. So to prove this, we&#x27;re gonna have f of x equals X and X plus c We&#x27;re going to find season with one, So we have X plus one. So therefore, you can quite clearly see that a one is equal to eight years. So even if you just, like, take a mental snapshot of this area here and then bring it over this area here, you&#x27;re gonna see that they are in fact equal.

Amrita Bhasin · Answer

we know that for this question, withdrawing the diagram The first thing you could do a substitute you is expose. See, this indicates that D axe is do you? Because it&#x27;s just one as the derivative of X. Therefore, in of the limits of integration, change to a pussy on the bottom and people see on the top after vax d axe Remember wise half of Expo See you had shifted by C units in the left direction horizontally after vax on the right and affects was seeing left.

Amrita Bhasin · Answer

okay, We know that we&#x27;re gonna have our limits of integration from negative beat too Negative A Because, remember, from A to B f of x, T X is equivalent to negative from B to a FX DX. It&#x27;s the same thing. So we know now that we can draw our diagram for this question, which is what they asked us to. D&#x27;oh! We know this is f of AKs and this is off of negative X recall the fact that half of negative X is a reflection off Michael&#x27;s FX and the Y axis.

Vikash Ranjan · Answer

in the problem. Okay. Be if our A minus six did on my six U minus six people on it is equal to minus. In addition, it will be if off minus six on it Is he minus DX So it is represented as my lesson minus B my s. If you do know. Since we know to be effects he did people miners, we do a defense. The eggs Therefore it is united. Honestly, you you you way You know that tradition? It will be if you do, you is equal to integration into being FX did it? They&#x27;re just given at my b tu minus e from x dx Hennis No, we have to remember that one waas ef minus six is the with election off while Waas X in the You want Texas. So in the graph below, you can see that the do ers are equal. Yeah, You know what? They&#x27;re photographed. I don&#x27;t like this. So here it is like this it is It doesn&#x27;t be granted. Graf won once forfeits and here just don&#x27;t like these. So it is my B minus a day he felt minus six. This is the answer to gurgle in problem

Yiming Zhang · Answer

So now we have inside the region are the integral of after backs. Why d a is always zero, and we&#x27;re going to prove that at a map of X Y zero at every point in our So I&#x27;m gonna prove this by contradictory. So we we said we&#x27;ll assume that. So there&#x27;s a region omega, which is a subset of our such that So for, um, see, there exist some region which is a subset of are such that effort alphabets. What is positive, Um, in this region. Okay, so let&#x27;s see, um, the air integral of every backs way the A, which is zero given by the condition. Also, this is the home to grow because efforts wise, positive, it&#x27;s always known active. So we have this integrates greater than the girl over the separation because the only reason we have some positive points and uh, that&#x27;s gonna cost the integral on the separation is greater than zero. So let&#x27;s see. We have now zero cool and zero zero greater than zero. So which is impossible leading to a contradictory, which means over assumption is false. This is force. Okay, So that implies, for every points in the region are we must have after that&#x27;s way equals zero. And that finished proof

Frank Lin · Answer

hey will try to prove this person equality and ah, he&#x27;LL notice one thing. This absolute value ofthe f works. It&#x27;s glittery. Who&#x27;s wherefore? Fax is also greater or equal to next the Fairfax, but by the defendant. Absolute value is a bigger while, these two. So this world tells us that it this integral, it&#x27;s going to be greater or equal to this integral. And this means this the same into group. It&#x27;s going to be quitting or equal to this integral for them by linear poverty off the integral. Your factor. All of that, the negative side. So what does it mean? It means thiss quantity is greater or equal to post this one and negative off this one. Let me it&#x27;s going to go to this one and also the connective off the same quantity. If, ah, if something is greater or equal to this and negative all the same thing, then it&#x27;s quite a record to the absolute value, because ever so very with the bigger one off these two. But ice greater because of both. So this proof of the first part there&#x27;s second part fof excited to works by the first part. It&#x27;s great very for two. Ah the integral. By putting the absolute value inside and this equals and because this&#x27;s less or equal to one, no sign of anything is between one and one. And if you put the absolute values between zero and the one, that means if we drop it, it&#x27;s going to make the an integral, bigger and therefore the integral value picker, and that&#x27;s the second part we need to prove.

Problem

If $a$ and $b$ are positive numbers, show that $$…

Problem 66 Medium Difficulty

If $f$ is continuous on $\mathbb{R},$ prove that $$\int_{a}^{b} f(x+c) d x=\int_{a+c}^{b+c} f(x) d x$$ For the case where $f(x) \geqslant 0,$ draw a diagram to interpret this equation geometrically as an equality of areas.

Answer

$\int_{a}^{b} f(x+c) d x=\int_{a+c}^{b+c} f(x) d x$

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$\int_{a}^{b} f(x+c) d x=\int_{a+c}^{b+c} f(x) d x$

Biocalculus Calculus for the Life Sciences

Heather Z.

Kristen K.

Samuel H.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

James StewartBiocalculus Calculus for the Life Sciences

Heather Z.

Kristen K.

Samuel H.

Joseph L.

James Stewart

Biocalculus Calculus for the Life Sciences