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

## -3

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Matt S.

July 26, 2021

How did you get the formula for the summation of i through n of i and of i^2? (timestamps 3:50 and 4:10, respectively)

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Okay, 5-22 here. Where is our introduction to definite integral? And we're using the limit definition of an integral. So this gets a little bit quite complicated with the algebra gets easier once we leave this behind, use immigration rules. But to understand the theory of what's going on and test your ability to stick with the problem, we're using this algebraic formulation. So you were given a theorem four in this section that said, hey, the definite integral is expressed by the limit as N goes to infinity of this is just the sum of the area of rectangles. So we're going to have in as a number of rectangles let in go to infinity. And so delta X, the width of each rectangle. It's just going to be the difference between the limits of integration divided by in and then exit B. These are the heights of the rectangle. So we're going to step along and say, okay, starting at the left boundary of integration and then increment one delta X two, delta X three delta X. Until we get there. Now in this problem I find it helpful just to break this up into three different integral zit breaks the algebra up into different pieces. So if we can write this as the interview from 1 to 4 of x squared dx minus the integral of 1 to 44 x dx plus the integral 124 two dx. And now we're going to come back there later. So we're going to work on each individual piece and then we're going to come right back to this spot for our final answer. So let's work on the first piece. So the integral from 1 to 4 of X squared dx. So this is going to be the limit as in approaches infinity of the some of I equals one to end. And now you look at the function is X squared. So we're going to have the limited integration starts at one. You're going to have one plus now if you look at what is delta X? So in this case delta X is 4 -1 Over in which is three over in. So this is one plus three I over in squared. And then the width of each rectangle is simply three over in. So this is the limit as in approaches infinity and let's just bring this over here of three over. In the sum I equals 1 to end. And now let's just square this binomial. So this is going to be one plus um six I over in plus nine I squared. Yeah over N squared. Yeah. Yes, Yeah. This is the limit as n approaches infinity of three over in And now let's just take a look and see what we have here. The sum from Michael one to end of one is just simply going to be in. So this is going to be in Plus six. The some from equal one to end. So the formula is the sum from my equal one to end of I is simply in N Plus 1/2. So this becomes in Yeah. And plus one Over two. N Plus and then the sum hi equal one to end of I squared as you're going to end N plus one, two, n plus one over six. So this is going to be plus nine and then you're going to have in in plus one To M-plus one over six. Yeah. Well, Mhm. Yeah. So this is yeah, the limit as in approaches infinity and I'm going to end up with when I look at the three over end times at first time I get a three and then when I look at it from the next term I get what, 18/2 which is Going to be nine. And then you have in Plus one over N. Yeah, plus and then the next term you have, was it? Um yeah, You're going to end up with 9/2, so plus 9/2. The end on the denominator here cancels. Mhm. And you are left with um Mhm. Yeah. Yeah. Yeah. Mhm. Yeah. Yeah. Mhm. Yeah. And let's see sorry about this where I had the nine I squared over in square and I'm sorry I left off in square that was taking me a little bit with the pause so the in here cancels with this one and you're gonna left with nine haves and that is N plus one, two, n plus one over N squared. Now, before we evaluate this let's do a little bit further simplification. This is the limit as in approaches infinity. Yes of three plus nine. Let's write this as one plus one over in plus nine halves. We'll write this as one plus one over in and two plus one over in. Okay? Yeah, now if you let in go to infinity when n goes to infinity, that term goes to zero, this goes to zero, this goes to zero. So this is equal to three plus nine plus 9/2 times two Which is 18, 1920 21. So that first limit is 21. So if I go back to where I started in my problem, I broke this up into three intervals. This one the value here is 21. So now let's go look at 1 to 4 of four X. So the same procedure here. How do I integrate? Yeah From 1 to 4 of four x. The X. That's going to be four Times the integral from 1 to 4 of X. DX. And then by definition this is four times the limit as N approaches infinity of the some of I equal one to end. And then if you look at your function the function here is X. And so we're gonna have one Plus and that's going to be, remember the difference 4-1 over in three over in she's gonna have three I over in that's the height of each rectangle and then the width of each rectangle Is three over in. Yeah so this gives me four times the limit as N approaches infinity. Mhm. Yeah. Okay and if I look at this I can do this some real quick. Um Let's back off. Mhm. So the sum as N goes to infinity. So the some from my equal I equal one to end of one, it's just going to be in, so you've got the width is three over in and then you're gonna be left within plus three over in. And then the sum I go into end of I By formula is just in times in plus one. Yeah, over to and so if you look at the cancellations here And over in is one, so that cancels there and there. So this is going to be four times the limit as in approaches infinity of three and then you have one plus, let's just write this three halves, one plus one over in. And then as we let in go to infinity you'll see that this term Will go to zero, so this is going to be 12. Yeah. Mhm. Yeah uh times one plus three halves, so that's going to be five halves and this is going to be 30. So now where we are in the scheme of the whole problem, I now know that this is minus 30 and then the easy one is the last one. The integral from 1 to 4 integral from 1 to 4 of um two D. X. Yeah, it's just simply going to be too comes four minus one, so two times three which is six, I mean the way you can think about this one is graphically Y core too. It's just a straight line that straight line, if it goes from 124 That means the area is three times 2 which is six. So go back up to where we had every of this plus six and so this has become 27 -30, this answer is -3. Okay, so guys, very little calculus in this, there was a lot of algebra, a lot of freak out when it comes to the submission of series and lots of places, mistakes can be made you. So I had a hiccup early on when I was doing this problem and I neglected to carry that in squared term here and it calls me a little bit of a delay in the thinking. That's why the pause was there. But anyway, that is your final answer that this definite integral is equal to negative three.

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