If f(x) =x+∫01(x+t).f(t)dt then ∫01f(x)dx=

f(x) =x(1+∫01f(t)dt)+∫01tf(t)dtf(x)=(1+a)x+ba=∫01f(t)dt=∫01((1+a)t+b)dt=1+a+2b2b=∫01t f(t)dt=2+2a+3b6 a=−7, b=−4∫01f(x)dx=1+a2+b=−7

If f(x) =x+∫01(x+t).f(t)dt then ∫01f(x)dx=

f(x) =x(1+∫01f(t)dt)+∫01tf(t)dtf(x)=(1+a)x+ba=∫01f(t)dt=∫01((1+a)t+b)dt=1+a+2b2b=∫01t f(t)dt=2+2a+3b6 a=−7, b=−4∫01f(x)dx=1+a2+b=−7

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If $f (x) = x + \int_{0}^{1} (x + t) . f (t) d t$ then $\int_{0}^{1} f (x) d x =$

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Detailed Solution

$\begin{array}{l} f (x) = x (1 + \int_{0}^{1} f (t) d t) + \int_{0}^{1} t f (t) d t \\ f (x) = (1 + a) x + b \\ a = \int_{0}^{1} f (t) d t = \int_{0}^{1} ((1 + a) t + b) d t = \frac{1 + a + 2 b}{2} \\ b = \int_{0}^{1} t f (t) d t = \frac{2 + 2 a + 3 b}{6} \\ a = - 7, b = - 4 \\ \int_{0}^{1} f (x) d x = \frac{1 + a}{2} + b = - 7 \end{array}$