4

Part of 2006 IberoAmerican Olympiad For University Students

Problems(1)

Alternating integral sum is zero - OIMU 2006 Problem 4

Source:

Prove that for any interval

[a,b]

of real numbers and any positive integer

n

there exists a positive integer

k

and a partition of the given interval

a = x (0) < x (1) < x (2) < \cdots < x (k-1) < x (k) = b

\int_{x(0)}^{x(1)}f(x)dx+\int_{x(2)}^{x(3)}f(x)dx+\cdots=\int_{x(1)}^{x(2)}f(x)dx+\int_{x(3)}^{x(4)}f(x)dx+\cdots

for all polynomials

f

with real coefficients and degree less than

n

calculusintegrationalgebrapolynomialfunctionreal analysisreal analysis unsolved