Calculus : Partitions and Riemann Integral
Let f be the function:
F(x)
1/4 x=o,
x 0<x<1
3/4 x=1
Using standard partition Pn (0,1) where n greater or equal to 4
L(f, Pn) = 2n(squared) -3n+4 all divided by 4n(squared)
U(f,Pn) = 2n(squared) +3n+4 all divided by 4n(squared)
and deduce that f is intergrable on (0,1) and evaluate
(intergral sign with 1 at top and 0 at bottom) of f
https://brainmass.com/math/derivatives/calculus-partitions-riemann-integral-7927
Solution Preview
Since,
dL/dn =0=> n = 8/3 and d^2L/dn^2 = +ve
=> L has minima at n =8/3 < 4 => increasing fn for n>= 4
therefore,
suprema(L) = lim(n->infinity) L = lim(n->inf) {2/4 -3/4n +4/4n^2}
=> ...
Solution Summary
A function is proven to be integrable over a given range.
$2.19