Use the form of the definition of the integral given in Theorem 4 to evaluate the integral.
$ \displaystyle \int^2_0 (2x - x^3) \, dx $
So problem 24, we're dealing with the limit definition of an integral. So a lot of algebra involved in solving these guys, I find it easiest just to break these up. So break some of the algebra into two separate integral. So two X dx Minour the integral from 0 to 2 of x cubed dx. Let's go do the work of each of those separate and then we'll come right back here to find our answer. So let's deal with the first integral. First the integral from 0 to 2. So the integral from 0 to 2 of two x. DX by definition is going to be too times the limit as N approaches infinity of the sun of I equals one to end. Now the width of each rectangle, it's just going to be to zero over in. So the width of each rectangle is too over in the height of each rectangle is going to be this function X evaluated so X at two over in Um four over in and and so forth. You're gonna increment this. So this is gonna be too I over in right? So this is the limit or it's going to be two times the limit as N approaches infinity. And if you look at what I have here, this is going to be you can replace that i with the formula that you know the some Mhm. I go one to end of I it's just simply end N Plus 1/2. So that's going to take the some of this out, the summation sign. So I'm gonna be left with two over in and then times two in Yeah, N plus one over two in. Yeah. So all I did was replace the what you see here in N plus 1/2, that replaces the I. So in this case, if I look at what I have to simplify this guy, this is two times the limit as N approaches infinity. And so what you see here is that you've got um to end over to end and then if you look at this, so this is going to be two and if I look at this in over N plus one, I can write that as one plus one over in. Mhm. And so as N goes to infinity, this term will go to zero. So this just becomes too um times two times one plus zero, which is simply four. So if I look at where I am in the scope of this problem, I have determined that the value of this one Is four. Okay, now I could have easily gotten that one graphically, you know, what is the line, you know, y equal to X From 0 to 2. Mhm. So when you plug in a two there you get four. Oh. Mhm. So right here, so if this is too yes and this is for the area that one half based sometimes should be four. So already got that. So I know that's right now. Little bit more work to get the integral. Yes, from 0 to 2 of X cubed dx. That is going to be the limit as in approaches infinity of the some I equal one to end with of each rectangle is still too over in and then now this is going to be um two. Um Yes, I cube so to I over in cute. So what is this going to give me? Well I know how to evaluate. Um let's just go and simplify this is the limit in approaches infinity of the sum I equal 12 N two cubed is eight. So this is going to be what, 16? So to to me so 16 Over into the 4th. I cubed. Ok. And so now we need our formula that the sum I equal one to end of I cubed is 1/4 in squared. Yeah, N plus one squared. Yeah. Mhm So let me get the some on this one real quick. This is going to be the limit. Yeah. Yeah. And approaches infinity. Uh So what have we got? 16 over? End of the fourth And then the sum of I cubed. It's just going to be 1/4 and then you've got in squared N plus one squared. So this is going to give me the limit and approaches infinity. 16/4 is four. And then I can write this as so you got in squared. This term will cancel if you write that as in square times in squared And then you can write this as one plus one over in he squared. And so this answer just turns out to be four. Go back to your original, So it's 4 -4. Final answer is zero.