Let fx be a differentiable functions such that fx - x2 - -0xe-t fx-t dt- then -01fxdx

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Let fx be a...

Question

Let $f (x)$ be a differentiable functions such that $f (x) = x^{2} + \int_{0}^{x} e^{- t} f (x - t) d t$ , then $\int_{0}^{1} f (x) d x$

\frac{1}{3}

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\frac{1}{4}

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\frac{7}{12}

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\frac{5}{12}

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f : R \to R

f (x) = x^{2} + x \int 0 e^{- t} f (x - t) d t .

y = f (x)

f

f (x) = x^{2} + \int_{0}^{x} e^{- t} f (x - t) d t

f (x^{2})

f : R \to R

f (x) = x^{2} + x \int 0 e^{- t} f (x - t) d t .

1 \int 0 f (x) d x

f(x) is a differentiable function satisfying the relation $f (x) = x^{2} + \int_{0}^{x} e^{- t} f (x - t) d t$ , then

f : (0, 1) \to (0, 1)

f^{'} (x) \neq 0

x ϵ (0, 1)

f (\frac{1}{2}) = \frac{\sqrt{3}}{2} .

x, lim t \to x ⎛ ⎝ \frac{\int_{0}^{1} \sqrt{1 - (f (s))^{2}} d s - \int_{0}^{x} \sqrt{1 - (f (s))^{2}} d s}{f (t) - f (x)} ⎞ ⎠ = f (x) .

f (\frac{1}{4})

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