Which represents all antiderivatives for h(t) = 10t* 2t + 17 (A) H(t) = 2t5 ~t2 + 17t + (B) H(t) = 2t5 t2 + 17t + € (C) H(t) = 10t5 2t2 + 17t (D) H(t) = 40t3 ...

Find each antiderivative. Then use the antiderivative to evaluate the definite integral. (A) $\int(4 x+6 y+5) d y$ (B) $\int_{1}^{4}(4 x+6 y+5) d y$

Yeah. We want to find the anti derivative for the indefinite integral in part A. And then we want to use that anti derivative to help us solve the definite integral given in part B. The anti derivative or rather the integral part we want to evaluate is integral or expose six Y plus five G. X. Since our differential is D. X. That means we're going to integrate with respect to X. Any terms with why are going to be treated as a constant. So we've anti derivative two X squared plus six Y X plus five X. We went term by term so four X becomes two X square six Y becomes six Y. X. Five becomes five X. Now let's use this to solve our definite integral part B. So for part B we have the same anti derivative and we're evaluating it from X equals negative 22 X equals three. Thus plugging in for three and then contracting for X equals negative two. We can obtain find the solution 35 plus 30 Y.

In the institution we have to find the quadratic equation and we have you and that Integration 02 x. F D D T Will be equals two X plus integration X to one T into F D D. D. Now I'm going to differentiate both sides so differentiating both side. So we get this is fx is equals to this. Bill comes out to be one minus x fx. From here we get the value of perfect, that is one upon one plus X. Now we have given the route of the equation that is F -1 x two is equal to two And f who is one x 3. So we can say that this will comes out to be X squared minus two plus 1 x three in two X plus two by three X equals to zero. From here we get the equation that is three X squared minus seven X plus two equals to zero. So this is the required quadratic equation and this is given an option seat. So she is the direct answer for Thank you.

We want to find an anti derivative for each of these functions here. So let's just go ahead and apply our anti derivative to the 1st 1 And it may help if we rewrite this in the form that we have right above it for the power rule for integration. So it be thanks to the negative second DX. So what that power rule says is a song is in this article one, we're gonna add one to the power. So it's negative too, plus one, and that we divide by its new power. That would just be negative one, and we would add some constant sees. What we're gonna do is just let c equals zero. So that just drops off, since they just want a solution as opposed to the general solution and simplifying that down, though, would you have us negative one over X? So this is a anti derivative. And if you add any number to this, it will give you another solution. And that's because taking the derivative of a concert just zero Now, let's go ahead and apply this to the 2nd 1 So by the Lady Garrity of this, we can pull the five out and rewrite this SOS five times the integral of one of her X squared DX. And we just found out that one of her X squared DX integrates to negative one over X for one of its possible solutions. So we can replace that with negative one of her x b negative five over X as a possible and tied route. And just like before, we could add any constant to this that we want, and it will also be a solution. But in this case, we're just choosing our constant to be zero for simplicity. Now, entering this well, we're going to distribute it by linearity the interval. So we end up with this here. So to get a constant, that means we had just an ex was gonna be two x than minus. Well, we just solved that five over X squared integrates to negative five over X. Or at least one of the anti derivatives is this, I should say, and we would add our constant C but little nippy zero, and so we could just go ahead at those together and actually we just get rid of that. Plus there this would be one R and tie derivatives. So if you were to take the derivatives of all of these, it would end up being the original function that we started with.

They want us to find the anti derivatives of each of these functions here. So let's just go out and start so we can go ahead and apply the anti derivative to this 1st 1 And one of the properties of anti derivatives is that we can factor out constants so we can factor of this 1/3 and just have one rex DX. And now this follows directly from one of the anti derivatives that they give us in this chart in this chapter. And that's just going to be the natural log of the absolute value of X times 1/3 plus a constancy. Now, since agents a gift and solution as opposed to all possible. First, let's see equal to zero. So that just goes away and that goes away there. So if you just want to get any different solution, we would just let CB any number. But just for simplicity, we can go ahead and let C B A zero so we don't have to write as much now for the next one. We apply the anti derivative again so we can go ahead and factor out our 2/5 which is a constant and again. That's just going to be the natural log of the upside value of X plus the constancy. But we go ahead and just chop that off because it's zero. That would be one possible anti directive. And again add any number you want to this and will give another ballot anti derivative. Now for C, we're gonna have to apply the other property that a linear operator has, which is, you know, to distribute across plus and minus signs so we could rewrite. This is actually the some indifference of three intervals. So one d X plus 4/3 into a row of one over X, the X and then minus. And I'm gonna rewrite one over X, where as X to the negative second DX. Now, these 1st 2 So if we take the derivative of X, that should give us well, one. So the anti derivative one is X, and then this is going to be 4/3 the natural log of the opposite value of X. And now this one here were actually going to use the power room to integrate it. So remember, power rule is the 1st 1 on that chart that they provide so it says add one to the power and then divide by the new power. So that would be negative one. And again we would have our constant plus C. But we're just going to ignore it because it's zero. In this case, at least we're saying it is. And let's go ahead and rewrite this to make it look a little bit cleaner, T X plus or thirds natural log of the upside value of X. So these negatives here cancel. There's gonna be a plus and then it would be X to the negative first power, which is really just one over X. So this here is and anti derivative or C And like I was saying, for each of these three solutions at any number that you want and that will give you another ballot anti Drood