MathematicsMedium210×since 2002Q3313Let f be a non-negative function in [0, 1] and twice differentiable in (0, 1). If ∫0x1−(f′(t))2dt=∫0xf(t)dt\int_0^x {\sqrt {1 - {{(f'(t))}^2}} dt = \int_0^x {f(t)dt} }∫0x1−(f′(t))2dt=∫0xf(t)dt, 0≤x≤10 \le x \le 10≤x≤1 and f(0) = 0, then limx→01x2∫0xf(t)dt\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\int_0^x {f(t)dt}x→0limx21∫0xf(t)dt :Aequals 0Bequals 1Cdoes not existDequals 12{1 \over 2}21Check answerSkip