Question

Let a 0 and suppose that the function f is Riemann integrable on [0, a]. Prove that f(a-x) dx = 2S0[f(x) + f(a- 1 ca f(x) dx = x)j dx. Prove that f' in(1 + tan(a) tan(x)) dx = a ln(sec(a)) (0<a<T/2) Let f: [0, 1] → R be defined by f(x) = VX , 0 1 , and let x 2 n-1,2 be a partition of [0, 1]. Calculate lRll and show that lim n|Pal| = 0 Calculate L(f,%) and (f,%) , and show that f is Riemann integrable on [0, 1] with 10 Vx dx =2 3

Please give clear detailed explanation.

Answer 1

A < 0 and f is Riemann integral on [0, a] then f is bounded on [0, a] Let I = integral_0^a f(x) dx and let x = a - y then at x = 0 y = a and at x = a y = 0 and dx = -dy Then I = integral_a^0 f (a - y) (-dy) = -integral_a^0 f (a - y) dy = integral_0^a f(a - x) dx Therefore integral_0^a f(x) dx = integral_0^a f (a - x) dx - (1) integral_0^a f(x) dx = 1/2 x integral_0^a f(x) dx = 1/2 { integral_0^a f(xx) dx + integral_0^a f(a) dx = 1/2 { integral_0^a f(x) dx + integral_0^a f (a -x) dx } = 1/2 { integral_0^a (c) + f (a - x) dx } integral_0^a f(x) dx = integral_0^a f (a - x) dx = 1/2 integral_0^a (f(x) + f (a - x)] dx Let I' = integral_a^b ln (1 + tan a tan x) dx