Recent content by chrischoi614

OH I am so sorry, my question is actually wrong with the definition of f(x), its supposed to be 0≤x<1 not less than or equal to 1... then will my answers be correct?

I have actually done them in the way you described it. for lower sum of P, Lp=∑mj(xj−xj−1)=m1(x1−x0)+m2(x2−x1)=... m1 = 0 , m2 = 1, then subsequently i got 0(1-0)+1(2-1) = 1 As for higher sum of P, M1=1 and M2=1 so 1(1-0)+1(2-1) = 2 is it correct?

Oh so i found out that the upper sum of P is 2 and the lower sum of P is 1, so the difference of upper sum and lower sum for P is 1. Upper sum of P: Lower sum of P: 1(1-0)+1(2-1) = 2 0(1-0)+1(2-1) = 1 Difference: 2-1=1 For upper sum of Q: Lower sum of Q...

Homework Statement comparing two partitions: Let f be defined on the interval [0,2] by f(x)=0, if 0≤x≤1 and f(x)=1 otherwise. Consider two partitions P={0,1,2} and Q={0,1-\epsilon, 1+\epsilon,2}, and calculate the difference of the upper and the lower sum with respect to each partition. By...

no... i didnt get it wrong :S... i just put the terms together...

I actually did that before, but i ended up with 2n^4 -(2n^2)(root(n^4 +1)) + 1 in the numerator, the fact that the square root is there really is making me struggle cus i don't know how to simplify it =\

Homework Statement Either the Comparison Test or Limit Comparison Test can be used to determine whether the following series converge or diverge. [i] which test you would use (CT or LCT) [ii] which series you would use in the comparison. [iii] does the series converge or not The series...