Let f be a real valued continuous function on [0, 1] and f(x) = x +...

Given,

$f\left(x\right) = x\left(1 + {∫}_{0}^{1}f\left(t\right)dt\right) - {∫}_{0}^{1}tf\left(t\right)dt$

$f\left(x\right) = Ax - B . . . \left(i\right)$

A = 1 + {∫}_{0}^{1}f\left(t\right)$dt = 1 + {∫}_{0}^{1}\left(At - B\right)$dt$</p><p>$⇒ A = 2\left(1 - B\right)$. . . \left(ii\right)$

Also $B = {∫}_{0}^{1}tf\left(t\right)$dt = {∫}_{0}^{1}\left(A{t}^{2} - Bt\right)$dt$

A = \dfrac{9}{2}B . . . \left(iii\right)

From \left(ii\right)$, \left(iii\right)</p><p>A = \dfrac{18}{13}, B = \dfrac{4}{13}$

So, $f\left(x\right) = \dfrac{18}{13}x - \dfrac{4}{13}$

$⇒ f\left(6\right) = \dfrac{18 \times 6 - 4}{13} = 8$