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Problem 23 Medium Difficulty

Use the form of the definition of the integral given in Theorem 4 to evaluate the integral.

$ \displaystyle \int^0_{-2} (x^2 + x) \, dx $

Answer

$\frac{2}{3}$

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Video Transcript

Yeah, yeah. So problem. 23 here were asked to use the limit definition of a definite integral to figure out the exact value um of this integral. So the definition is going to be the limit as in approaches infinity of the some of I equals one to end and you're summing up the area of rectangles in rectangles, but then letting in go to infinity, So the width of each rectangle is going to be zero -2 over in which is going to be too over in. So that is the width reach rectangle in this case and the height of each rectangle is going to be determined by stepping along from one to end and evaluating this function. So it's going to be um if you think about it, so X squared is the function, so you're going to have -2 is the boundary. So plus and then you step, each increment is two over in. So too I over in Yeah squared plus mm negative two plus two. I over in mm. And then the width of each rectangle as we determined was to over in. So we want to do the easiest thing to do is to break this up into two integral. So the integral from negative 2 to 0 of X squared dx plus the integral from negative 2 to 0 of X D X. And we are going to come back here for our final answer. We're going to break this problem up into pieces and do each of these integral separately. So let's do the first one. So in the first case the integral from negative 2 to 0 of X squared dx is going to be the limit as in approaches infinity of the some From I equals one to end. Now the width of each rectangle is too Excuse me? Yes, it's too over in. And then to find the height, it's going to be negative too plus two I over in and that quantity is squared. No, Now it just becomes an algebraic operation here. So this is of the limit as in approaches infinity of the some I equals one to end of two over in square this binomial and you get four and then minus eight, so for minus eight I over in and then plus four I squared over N squared. Yeah. Yeah. Now this is where we use pieces of information that the some from my goal one to end of I is in And plus 1/2. The some from I equals one to end of I squared is in in plus one, two n plus one over six. And the some I want to end of any constant is just going to be that constant times in. So this will allow us to take the some out of this equation. So this is going to give us the limit, as in approaches infinity you'll have to over in and then you will get four in -8 in N Plus one over to in plus four. And this was an in squared, sorry, and and plus one, two, n plus one Over six in squared. Yeah, so this is the limit as n approaches infinity. The first term two times four in over in is just going to give us um and eight, so this is going to be yes, yeah, eight and in the next term I see a cancellation, so I'm gonna have to over to that will go away and then an end over in that will go away. And then what you're left with in this case is just going to be eight uh times we can write this minus eight and we can write this as one plus one over in Yes. And then I'm left with in this case the end from the to ever end cancels right here and then to six is just gonna be one third, you're gonna have plus uh 4/3. Mhm And then if that ain't cancels, you're going to have this is going to leave you with Looking at the end squared here in this term, this is one plus one over in two plus one over N. Yeah. Okay, now let's let em go to infinity. If n goes to infinity, this term goes to zero, this term goes to zero, This term goes to zero and you are left with eight minus eight. Yeah, Yeah Plus 8/3 which is equal to 8/3. So we're halfway done with the problem, Go back up to our original I said we're going to break this into two integral. This first answer is 8/3. Now we need to go figure out what the value of this other integral is. This one should be a little bit easier. So the integral from mhm -2-0 of x. Dx is going to be the limit as N approaches infinity of the sun, I equal one to end with of each rectangle is still too over in plus and I'm going to increment this By two over in every time. And so now it's just a matter of figuring this out. This one's a little bit simpler. So this is the limit as in approaches infinity. Um and this is two over and I'm going to take the some out. So if you look at that -2 um that you see there so that some is just going to be minus to end And then plus two and then the sum for I is in n plus one Over two and I still have the end that is there. And now it's just a matter of simplification and taking this limit, so this is the limit as in approaches infinity of this is going to be negative for right plus and this is going to be two. And what you have here is this in cancels? Excuse me? Yeah with this in this two cancels with this too. Okay and then what you have, what you're left with is too you've got this in in this in plus one, So that's 2, 1 plus one over in. Yeah. Yeah. Yes. Now as you let in go to infinity this term goes to zero. So this is negative for plus two Which is -2. Yes, so go back up to our answer here, this is negative two, So this is 8/3 minus six thirds. Final answer is two thirds. Okay