<p>Okay, I actually started this and I can't get it to be the right answer, 34. Does anyone know how to do this, working it out fully--</p>
<p>f(x)=27-x^3 area under curve from [1,3]
n
lim n-->infinity SIGMA f( 1+i*(2/n) ) * (2/n)
i = 1</p>
<p>I keep on getting it to equal something in the twenties, i forget--anyone know what i am doing wrong / what to do right...</p>
<p>that n and i = 1 are supposed to be with the sigma obviously</p>
<p>You have to do f(xi) * delta x. So it would be [27- (1+2i/n)^3] * 2/n.</p>
<p>well...yeah...i know that -- i just dont know where im messing up after that, unless i just dont know how to work my calculator...</p>
<p>do you HAVE to use the infinite sum, or can you use the fundamental theorem of calculus?</p>
<p>Well, the answer is 34, but you just have to figure out how to get there. That wasn't helpful. Sorry...</p>
<p>wtf you have to use infinite sum? screw it and use fund theo its soo much faster. but if you cant ask and i will help</p>
<p>you should definitely use the fund theory of calc for this one, set it up like a def integral with bounds from 3 to 1 and solve</p>