317/7 POINTSSCALCET8 5.5.044.MI.SANAWThis question has several parts that must be completed sequentially. If you skip any points for the skipped part of part; and y...

$49-52$ Evaluate the indefinite integral. Illustrate and check that your answer is reasonable by graphing both the function and its antiderivative (take $C=0 )$ .
$$\int x\left(x^{2}-1\right)^{3} d x$$

Again this question. We have to solve integration X into the bracket. Access square minus one whole cube DX. Okay, so first of all, we will let U equals two x squared minus one here that will give us the U equals to two x dx. Or we can say, Do you divided by two Will be X dx OK. And X DX is presenting our question and the integration X squared minus one whole cube X DX will be integration. This will be you. Then it will be you. Cube and X dx will be. Do you divided by two? Okay, so it will be won by two integration, you cube, do you? So it will be one way to an integration of you. Cube will be used to be powerful, divided by four plus C. Or we can say it will be one by eight. You raised to the power four plus C. Now we will back Substitute the value of you that is X squared minus one. And it will be X squared minus one raised to give our four divided by earth plus c. Okay, so this will be our final answer. And now we have to check that our answer is reasonable. Uh, we have to check by graphing. We have to graph both the function and its derivative. Okay, so we will use the graphic utility hair, okay. And the graph of this will be like this. Okay, this is the graph. Okay? And you can see the red color graph. Okay. The red color graph is for the anti derivative effects. That is our answer. And that is X squared minus one. Race to the powerful divided by eight. Okay, we have taken C equals to zero here, okay? And the function in the graph in the blue color. Okay, glue color is small effects. That is our function. And it is X X squared minus one whole cube. Okay, this is in the blue color, and we can say we will note that the capital effects that is anti derivative has its minimum at points where that is this point. Okay. At this point, where the over function changes sign from negative to positive. Okay. From this point, this function goes to this side that is negative to positive and this point also negative to positive and the enter derivative FX Hazard's maximum head point. Where over, uh, function changes Sign from positive to negative here. Okay, so this proves that we have what we have as our anti derivative is correct. Thank you. Mhm.

Again this question. We have to solve interrogation one minus two X rays to give our nine DX OK, so here we will do the substitution here and we will substitute U equals to one minus two x. Okay, that will give us do u equals two minus two DX. OK, it means the value of DX will be. Do you divided by two? Negative. Okay. Minus one by two d you So now we will Can we can say one minus to express to the power 90 x will be one minus two x will be you. Then it will be you raised to the power nine and the u. D X will be minus. Do you divided by two? Okay. Or we can say it will be minus one by two. Interrogations you raised to the power nine and do you okay? And integration of it will be minus one by two. An integration of you raised to the power nine will be you raised to the power 10 divided by 10 plus c Okay. Or we can say it will be minus one by 20. You raised to the power 10 plus c and now we will back. Substitute the value of you here. That is one minus two x. Okay. And it will be minus one by 20. And you will be one minus two x. Okay, one minus two X raised to the power 10 plus c and that will be our final answer. Thank you.

So for this integral we're gonna use a U. D. U. Substitution. And we're gonna let U equal four plus three X. And do you then would be equal to three times dx. And what this substitution is gonna do is it's going to make this integral um a lot simpler and a lot more straightforward. And so if we substitute in you for four plus three X. Gonna have you to the one half power. And if we substitute D. U. N. Well we can see d'you divided by three is equal to dx if we just divide both sides by three. So we're gonna plug that in for dx here. So we're gonna do you divided by three. And the first thing I'm going to do is take this one third out of this integral. So we have one third times the integral from 0 to 7 of you to the one half power times D. U. And whenever we have an integral of just a variable you races some power and that power isn't um the negative first power. What we're gonna do is we're just gonna add one to that power in divide by the resulting power. So here we're gonna have one third and then times U. To the three have power I added one and then I'm gonna divide by that power which is the same as multiplying by the reciprocal. And so now I'm gonna go ahead and plug in U. Is equal to four plus three X. So we're looking at 1/3 times two times four plus three X. 23 halves power Divided by three. And we're looking from 0 to 7. And so when X. Is equal to seven we're gonna get four plus 21 23 hours power or 25 to the three halfs power and 25 to 3 hours Power is the square root of 25 cubes in the square of 25 is five. So we're gonna have five cubed which is 1 25 times two is 250. So we're going to have and here 250 divided by three and then minus when X is equal to zero we're gonna have four to the three of us power which is two cubed. So it's eight times two is 16 divided by three. So we're gonna have to 50 divided by three minus 16 divided by three and then multiplied by one third. So that's going to be 234 divided by nine.

That's what I'm gonna let you be the L. N. X. And TVB X. To the three House Dx. Yeah so do you is one of her X. Dx. M. B. Is X to the five house Over five House or Times 2/5. So the integral is to 5th x. to the 5/2. Ellen X minus to fifth dinner girl X. To the three house dx. So that's 2/5 X to the five halves L. N. X. When it's 2 5th x to the five house over five house pussy to fifth X. To the five halves. L. N. X minus four 25th X to the five house plus he.