claculas problem.Find f(1)

Question

Integral x to 0 f(t)dt=x+Intgeral 1to x tf(t)dt

Then what is value of f(1)?

Answer 1

∫₀^xf(t)dt=x+∫_x¹tf(t)dt (given);

When x=0, ∫₀^xf(t)dt=0, so ∫₀¹tf(t)dt=0 (see also later);

∫_x¹tf(t)dt=∫₀^xf(t)dt-x;

x=∫₀^xdt;

∫_x¹tf(t)dt=∫₀^xf(t)dt-x=∫₀^x(f(t)-1)dt;

∫_x¹tf(t)dt=(1-x)(∫_x¹f(t)dt)-∫_x¹f(t)dt (integration by parts);

∫₀¹tf(t)dt=∫₀¹f(t)dt-∫₀¹f(t)dt=0;

∫₀¹tf(t)dt=∫₀¹tf(t)dt+∫₀¹(f(t)-1)dt=∫₀¹(tf(t)+f(t)-1)dt=0 and, since  ∫₀¹tf(t)dt=0, ∫₀¹(f(t)-1)dt=0.

f(t)-1=0, so f(t)=1, therefore f(1)=1.