suppose-int_14-fx-d-x8-and-int_16-fx-d-x5-evaluate-the-following-integrals-a-int_14-3-fx-d-x-mathbfb

Question

Suppose $\int_{1}^{4} f(x) d x=8$ and $\int_{1}^{6} f(x) d x=5 .$ Evaluate the following integrals.
a. $\int_{1}^{4}(-3 f(x)) d x$
$\mathbf{b} . \int_{1}^{4} 3 f(x) d x$
c. $\int_{6}^{4} 12 f(x) d x$
$\mathbf{d .} \int_{4}^{6} 3 f(x) d x$

Foster Wisusik · Accepted Answer

Okay. This question gives us some integral and it wants us to use the linearity properties to find different modified forms. So part A, once the integral from 1 to 4 of negative, three times f of x, the X and we know that we can just factor out Constance. So this is just negative. Three times are integral from 1 to 4 of f of X, and we know that the value of this integral is just eight. So this integral turns out to just be 24 then for part B. It wants the integral from 1 to 4 of three times f of X, and this is the exact same as the last one. Except now are constant is a positive three. So we can just pull it out front. And we know that this integral eight. So it&#x27;s just eight times three, which is 24. Done part C once the integral from 6 to 4 of 12 times after vax d x. So we don&#x27;t really like our limits going this way so we could make it negative and pull out the 12 at the same time. So let&#x27;s get negative 12 times the integral from 4 to 6 of F of x DX. Because again we just pull Constance out in front and to get our limits facing the direction we want, we just pull a negative sign. And now we actually don&#x27;t have an expression for this integral. But we can figure out its value because we know that this whole integral has to be equal to the entire area minus the area from 1 to 4. So this just turns out speak negative 12 times five minus eight, which is just 12 times negative three or positive 36. And again, we found that into girl just by taking the total area from 1 to 6 and subtracting the area we already had at X equals four to get the area between. So then, lastly, Part D wants the integral from 4 to 6 of three f of x dx, and that&#x27;s just equal to three times the integral from 4 to 6 of f of X. And we actually figured out that integral, but we can show it again. The integral from 4 to 6 again is just the integral going all the way from 1 to 6 minus the area we have accumulated up until X equals four and I&#x27;ll change that limit cause I miss wrote it. So it&#x27;s just three times negative three, which is negative and nine.

Foster Wisusik · Answer

Okay. This question wants us to use these integral values to calculate the values of some other integral using properties. So part A wants the integral from 0 to 3 of five f of x dx. And since five is just a constant, we can pull that outside of the integral to get five times the integral from 0 to 3 of f of X. And we know that this integral right here has a value of two. So the end result is just 10. And let&#x27;s just make that three look more like a three. There we go. Then. Part B wants the integral from 3 to 6 of negative three g of x dx. So we&#x27;re going to do the same thing and just pull out our constant of negative three. And we know that this integral right here is equal to one. So it&#x27;s just negative three times one or negative three. Then for part C, we want the integral from 3 to 6 of three f of X minus G of X, and from here we can first split it up with the subtraction signs. So we have the integral from 3 to 6 of three f of X minus the integral from 3 to 6 g of x, the X And then we can just pull out that constant of three in front of the integral. And now we just have the difference of two integral, each of which we know the value of because we have three times this integral, which is negative five and then we&#x27;re subtracting one. So negative 16 is the value of this integral. And then lastly, parte de wants the integral from 6 to 3 of f of X plus two g of x, and we&#x27;re going to do the same exact thing. But first, we&#x27;re just gonna flip the limits by adding a negative sign in front. And then we can just split this up like we have done for We have negative into girl from 3 to 6 of F of X minus two times the ends of girl from 3 to 6 of G of x dx. And now this is just when we plug in. This is negative negative five minus two times one or five minus two, which is three

Jake Dixon · Answer

So in this question, were given to intervals the integral of F of X between 14 and the integral of FX between one and six. And we are asked to evaluate these four into girls using those t that we&#x27;ve been given not be able to do this. We&#x27;re gonna need a few key results which is gonna help us. So the first is that the integral between A and C of a function is equal to the integral between a and B of the same function, plus integral between B and C of the same function where B is some number between A and C that says that you condone either go one way and that way and split the integral into two sub into girls or going the other way. You can add to into girls which join together a to point B to make one integral of the whole thing between A and C. Second result is that the integral between a and B of a function is equal to negative, the integral between being a so you can swap around the limits by changing the sign of the integral. We then have another one similar to the first is that the integral between B and C of a function f of X can be split into the integral between a and C of a function f of X minus integral between a and B of the same function effects. And I should say there were a less than be lesson see, se and you couldn&#x27;t again, um, split this integral going this way into two other interval integral that this time ones with a bigger interval. So you find a point smaller than what you&#x27;ve already been integrating integrate over a larger range and it could still happen. This value a is much nicer to work with your integral and the same going the other way if you if you want to find a difference of two end girls between, um and you&#x27;ve got CNN being a, you can find the difference between them, which would be between B and C. And you can verify that by looking at graphs of functions. And the 4th 1 is that eight times the integral of the integral of a times F of X is equal to eight times the integral of effects so long as a is some arbitrary constant. So Teoh, start this off will start with the 1st 1 So, using this result of the bottom number four, we can pull out the minus two recent smile history of some artery constant or some constant, I should say so you got minus three times into go between one and four of F of X dx, which is minus three times the result. We&#x27;ve already been given up here, which is eight. That makes that minus 24. The next one is similar. So again we can pull out the three since it&#x27;s just a constant on the front and it&#x27;s not any terms in X. If there any times in X because you&#x27;re differentiating with respect to X, you&#x27;d have to keep them in. So f of X is in terms of X, so we have to keep in. Since the three is unrelated, we can bring it out when we get the integral three times the integral between 14 of FX DX, which again we&#x27;ve been given at the top is eight. So that&#x27;s three times eight, which is 24 now. We can start this integral the same by saying by putting out 12 and saying that his equals 12 times into go between six and four f of X dx now, because our bottom integral is bigger than our top one. It&#x27;s probably going to be easy if we stop this round, because most in schools you normally have the top limit is going to be bigger than your bottom limit. So if we make that negative, then we can swap around the intervals we have minus 12 times integral between four and six of FX, TX. Now we can reviews that second result. Now we can use the third years old, which is that the integral between B and C is equal to the integral between A and C minus, integral between A and B. So since we know the intercourse between 14 and one and six, it might be a good idea to set a equals one. And then we know that this integral is equal to minus 12 lots off the integral between one and six F of X dx, minus the integral between 14 If Becks DX, which is equal to minus 12 lots off five minus eight, she&#x27;s minus three, which makes the whole integral 36 and then we can look at the final one. And again we could pull the three hour in fronts. We have the integral between four and six f x dx. This time we don&#x27;t need toe flip their limits and make it negative. We&#x27;ve got them in the right way around the way we want them. So we have three times intercourse 316 f of X DX minus intercourse between 14 FX DX, which is three times five minus eight, which is three times minus three minus nine. So that&#x27;s how foreign girls

Fuzail Shakir · Answer

Okay, So off the property of definite into all states that get me in to change the interval off integral in its value becomes negative. So this means that the integral from A to B off F off X is equal to negative terrace Ian to grow from B to a off off X So when you have the integral from 4 to 5 off f off X t x, this is equal to negative times integral from 5 to 4 off off X dx, which is equal to 36. So therefore of senior 3.6. So therefore, since this is equal to 3.6, this is that equal to negative 3.6. So if we add both the integral Sze wee enough with the integral from 1 to 5 off f off X t X plus the integral from 5 to 4 off f of X d. X, which is equal to 12 minus 3.6, which is equal to eight 0.4

Katelyn Chen · Answer

were given a phone information. The integral from 1 to 5 of effects. D X is equal to 12 integral from 4 to 5 of affects the X equal to 3.6. It grow from 1 to 4 of after vax d of access unknown that you were attempting to figure out. So start off with We know that&#x27;s integral 5 to 4 of F of X DX plus the integral from 4 to 1 with a Fairfax DX is equal to the integral from 5 to 1 of f of x dx. Now, while you notice well, you can notice Over here the lower limit is the same as the upper limit for this integral. So you can kind of think of them as canceling else to leave you with the integral from 5 to 1. So we know that because I know that this is true. We also know that if we use our rules of algebra no, that&#x27;s integral from 4 to 1 of after backs, T X is equal to the integral or 5 to 1 frx d X minus the integral from 5 to 4 of max dx. Now we&#x27;re given these into girls you know, that&#x27;s integral of from what&#x27;s it? Er start from 1 to 5 of vax D X single to 12 and integral from 4 to 5 of next e XT equals 23.6. Then we know that this is equal to 8.4. And so we know that the integral from 1 to 4 about 60 x is equal to 8.4.

Foster Wisusik · Answer

Okay. This question wants us to use thes integral to evaluate various expressions. So part A, once the integral from 1 to 4 of three f of x so we can just use our properties of integration pull that constant out in front. So now we have three times the integral from one before of f of X dx. But it doesn&#x27;t tell us this integral, but we can find it from the other ones it gives because this is just three times the total area from 1 to 6 of f of X, minus the area already. We&#x27;re sorry. Minus the area that&#x27;s been accumulated from 4 to 6 and I should have a d X over here too. But this just three times 10 minus five, 4 15 for party, then for B. It wants the integral from 1 to 6 of f of x minus G of X. So we&#x27;re just gonna split this up using our linearity property. So this is just the integral of F minus. The integral G and the integral of F is 10 in the integral of gs five. So it&#x27;s 10 minus five or five, then for part C, we want the integral from one before of F minus g. So same thing. We&#x27;re just going to split this into two into girls. We have the integral from 1 to 4 of f of X plus You&#x27;re sorry minus. I&#x27;ll just pull that nine. A sign out minus the integral from 1 to 4 of g of x. So tells us the g of X in and grow. And we already found the f of X in a growing part A So we said this one was five and this one is giving us three SAR assaults too. And again, we found that 1 to 4 integral by subtracting the 1 to 6 and a girl from the 4 to 6. All right, other way around. But then Part D wants the integral from 4 to 6 of g of x minus f of x. And we do the same thing we&#x27;ve been doing where we split this up and clean this up a little bit. But the integral from 4 to 6 of g of X is not something they gave us, but we can find it. We can just do the total integral and then subtract the integral from one before and then subtract the integral of f. And just just because I was going to run off the board here, I&#x27;m not gonna bright the DX. But now we know everything so you can plug in. We have, ah, five here. Then we have a two. And then we&#x27;re subtracting the integral from 4 to 6, which it gave us as five star result is negative two, then Part E, once the integral of eight g of X from 4 to 6. And that&#x27;s just eight times the integral from 4 to 6 of G of x. And we just found that integral we found that was just equal, the three in the last part. So our results is 24 for that one. Then finally, Part F wants the integral from 4 to 1 of two f of x. So to get our limits to match up Oh, pull a negative sign now to flip the order and lower at it, I&#x27;ll pull out a to So now it&#x27;s just negative. Two times the integral from 1 to 4 of f of x. And we already found that integral in part A. So it&#x27;s just negative two times that Inter girl, which he said was equal to five. So our answer is negative 10 and now we&#x27;re done.

Problem

Suppose $\int_{0}^{3} f(x) d x=2, \int_{3}^{6} f(…

Problem 42 Hard Difficulty

Suppose $\int_{1}^{4} f(x) d x=8$ and $\int_{1}^{6} f(x) d x=5 .$ Evaluate the following integrals.
a. $\int_{1}^{4}(-3 f(x)) d x$
$\mathbf{b} . \int_{1}^{4} 3 f(x) d x$
c. $\int_{6}^{4} 12 f(x) d x$
$\mathbf{d .} \int_{4}^{6} 3 f(x) d x$

Answer

a. -24
b. 24
c. 36
d. -9

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William Briggs, Lyle Cochran, Bernard Gillett

Calculus for Scientists and Engineers: Early Transcendental

Grace H.

Catherine R.

Heather Z.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

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a. -24b. 24c. 36d. -9

Calculus for Scientists and Engineers: Early Transcendental

Grace H.

Catherine R.

Heather Z.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

William Briggs, Lyle Cochran, Bernard Gillett Calculus for Scientists and Engineers: Early Transcendental

Grace H.

Catherine R.

Heather Z.

Michael J.

a. -24
b. 24
c. 36
d. -9

William Briggs, Lyle Cochran, Bernard Gillett

Calculus for Scientists and Engineers: Early Transcendental