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Bonus: (15 points)Use the form of the definition of the integral;[fkr)tt = limZ f(r)Ar to evaluate the integral,[( +J...

Question

Bonus: (15 points)Use the form of the definition of the integral;[fkr)tt = limZ f(r)Ar to evaluate the integral,[( +J

Bonus: (15 points) Use the form of the definition of the integral; [fkr)tt = limZ f(r)Ar to evaluate the integral, [( +J



Answers

$21-25$ Use the form of the definition of the integral given in
Theorem 4 to evaluate the integral.
$\int_{0}^{5}\left(1+2 x^{3}\right) d x$

Okay, so in order to do this integral, what we can do is we can use properties of integral that were outlined in this chapter, One being that we have the integral of one term plus another. It's going to be equal to the integral of that first term which was just one. So it's the same, it's just the integral of D X. And then plus um the integral of the second term which was three X. And I'm actually gonna take the constant out which is another property of integral. So that we're going to be utilizing here, we're gonna take the three out and multiply it by the derivative or sorry, the integral of just X in times dx. And so the integral of just D. X. That's always equal to X. And the integral of just X is a power rule integral. And so what we do is we add one to the power that this x is being raised too. So here we have X the first, so now we would have X squared. And then we just divide by that power. So we divide by whatever the resulting power is and then we still have this three being multiplied by this integral up front as well. And we're looking from -1, 25. And we can go ahead and just combine these 200 girls into one again. And so when X is equal to five, we get five plus three times five is 75 divided by two are three times five squared, sorry five squared is 25. So three times 25 is 75. And then we have minus um when X is equal to negative one we're gonna get negative one here. So that's actually gonna be plus one and then when X. Is equal to negative one we're gonna get plus three have so that's gonna be actually minus three halves. And so we're going to get 1 -3/2 is negative 1/2 75/2 -1 half would be 74/2 which is equal to 37. So we're going to have five plus 37 here which is equal to 42.

So the same. We need to apply the definition. So we got integral of the function. Oh just he calls to the limit of that with an approach to infinity data. X times sigma. I also want you own of X. I So remember excite here dress equals two a plus items. 30 x. And also in this function in this question we got there to ax equals two B minus A over N. So B equals to two echoes to one. We got there two extra Seiko to one hour. That means our function. Yeah. Yeah just echoes to the limit times 30 X. Times so F. X. I guess equals two. A plus I. Time sir to X. That is I times one over. End to the power of three. That is a function exude power of ST so now we extend us from here got one hour n times sigma. You can see by the function of axa Powell Street. We all got one plus I to the power of three. Yeah so we know the some of the let me just equals to the meat of the some so we can separate that first we got one hour runtime sigma of one. So we got into powerful here. Times out to power of ST cast three hour unscrew wire here. Time sigma don't forgot the sigma here. Bye. And the last time we got three over into the power of three. Okay so to solve this function we can substitute this by several term here so we can see Sikma one from my co 21 to earn. So this term here we just got one plus one plus plus one So we got 1 1 here you so we can see the total number of the sigma dress equals two. And so we all got our Just equals to one. So the limited approach three in vanity after term just equals one here. And then we can see the second term here. We know sigma of I two power of three. Yeah those one just equals two into the power of two times and plus one to power up to our four. So we can substitute the term here that is one hour and two powerful and to power two and plus one to power to our four. The start here we can see the sigma of I that is one plus two plus and And this time dress equals two on times and last one or 2. Yeah So this term three over and Squire times and times and plots 1/2. Yeah and the last one we got the stigma of I Squire which is equal to one times And plus one times to an plus 1/6. So we got three to the power over and to the power of three In times and plus one two n plus 1/6. Mhm. So we can see that is the the meat of all those terms. So one plus we can counsel and Squire here so we all got one plus one to the problem to four and Squire. And for dessert from here we can also consult on here. So we are gas 3/2 and times and plus what And for this one we can counsel the under. We got also we can cancel the street here we got to so one plus 121 plus one. Power to inspire. Yeah so first we can see the term here that is Um last one hour to one squat red. We just put those in the same power princesses. So for this one the same mega 3/2 put them in one for instance is and also for those one we can put on here and then we time another term in the princesses. Yeah I think so if we divided the denominator and numerator by and at the same time you got one plus one our own or were too the same. You got one last one over and over one. So we got one plus one over and over one two plus one hour and over one so we can see even undergo approach to infinity one over and Jessica approach to zero. So we have got our final answer that is one plus one hour full cause we govern our N. S. Zero. So we only got one hour to to the power of to which as equals to one hour four the same. We can see it is firmly in the process is just echoes for what? Same here, we got one hour two times one, and those terms appear equals two, so we got 15 Hour four as a final answer.

Okay, we're gonna find that this center girl by using the definition. So the first thing we need is the width of the sub intervals until two x, which will be minus a over in what, minus zero over in when? Over in next, we need a point inthe e I threat tangle. Let's call it X. I star a plus I Delta x So zero plus I times one of her in or I over in next. We need to find the height of that rectangle So we'll plug that X I star into the function circuit I over in to the third minus three I over in squared So I to third over into the third minus three I squared over in squared So we have the height of the rectangle the width of the rectangle We confined the area of the rectangle Well, just multiply those two things together This times This so I get I cubed over into the fourth minus three. I squared over in Cute. Okay, so we have the area of one rectangle. So let's sum them all up and find the area of the first in rectangles. Okay. My area has two pieces. Someone right that has two separate sums. One over into the fourth. Some I equals one the end of cubed minus three Over in cubed. Some ID equals one day end. I squared. Now put in the identities for those sums whenever into the fourth times in squared and plus one squared over four minus three over in cubed and M plus one to m plus one over six. At the end of this step, you should always have the same number of ends on the bottom as you have on the top. If you don't, then you forgot to multiply by Delta X probably. Okay, so separate them the numbers 1/4. But this end square, take two of those. So in squared over in squared. With this in plus one square, Put the other two plus one squared over in square minus 3/6 times. Okay, put one of each of these ends with each one of these and we're in in plus one over in to M plus one over in. Okay, so now we have 1/4. Uh oh. In, squared over in squared. That's gonna be one times can rewrite this as a one square and put the in over in So you get in tips in over in which is one plus one over in whole thing squared minus one half times, one times one plus one of we're in times two plus one over in. Alright, last step. Now we'll have the answer to what is the integral. After we make the number of of rectangles go to infinity So we'll take the limit as in goes to infinity of the some of the A eyes. So it's limit as it goes to infinity 1/4 times one plus one over end squared my ass one half times one plus one over in times two plus one over in as in goes to infinity one over and goes to zero one over in goes to 01 over n goes to zero So you're left with 1/4 times one minus one half times one times two So 1/4 minus one negative. Three force is the value of that integral

This question asked us to use the definition of integral to evaluate this integral. What we know is that we can write this as the limit as X approaches Infinity of six, divided by an from eyes won t end and then one plus three times negative one plus six. I divided by ad, just plugging into the definition of the integral with limit. We end up with negative 12 over end times and +108 Divide by N Squared, Time's and Time's on post one over, too, which gives us negative 12 post 54 plus 54 times one over end. So one plus three x d axe Our bounds are from negative 1 to 5. This is a prevalent to 42.


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