the variation operator and integral operator are exchangeable

Posted: January 9th, 2022

For a function defined as

show that

(1) the variation operator and integral operator are exchangeable, i.e.

(2) the variation operator and the differential operator are exchangeable, i.e.

(a) Which mesh will yield more accurate results?

(b) Which will be more computationally expensive?

(c) Suggest a way of meshing which will yield relatively accurate results and, at the same time be less computationally expensive than B?