Let f be a real valued continuous function on [0,1] and f(x) = x+_0^1(x-t)f(t),dt.

Question

Let $f$ be a real valued continuous function on $[0,1]$ and $f(x) = x+\displaystyle\int_0^1(x-t)f(t)\,dt$. Then which of the following points $(x,y)$ lies on the curve $y=f(x)$?

Accepted Answer

Given,

$$f(x) = x + \int_0^1 {(x - t)f(t)dt} $$

$$ = x + x\int_0^1 {f(t)dt - \int_0^1 {tf(t)dt} } $$

$$ = x\left( {1 + \int_0^1 {f(t)dt} } \right) - \int_0^1 {tf(t)dt} $$

Now,

let $$A = 1 + \int_0^1 {f(t)dt} $$

and $$B = \int_0^1 {tf(t)dt} $$

$$\therefore$$ $f(x) = Ax - B$

$$ \Rightarrow f(t) = At - B$$

So, $$A = 1 + \int_0^1 {f(t)dt} $$

$$ = 1 + \int_0^1 {(At - B)dt} $$

$$ = 1 + \left[ {{{A{t^2}} \over 2} - Bt} \right]_0^1$$

$$ = 1 + {A \over 2} - B$$

$$ \Rightarrow {A \over 2} = 1 - B$$ ...... (1)

$$B = \int_0^1 {tf(t)dt} $$

$$ = \int_0^1 {t(At - B)dt} $$

$$ = \int_0^1 {(A{t^2} - Bt)dt} $$

$$ = \left[ {{{A{t^3}} \over 3} - {{B{t^2}} \over 2}} \right]_0^1$$

$$ = {A \over 3} - {B \over 2}$$

$$ \Rightarrow {{3B} \over 2} = {A \over 3}$$ ....... (2)

Solving equation (1) and (2) we get,

$$A = {{18} \over {13}}$$ and $$B = {4 \over {13}}$$

$$\therefore$$ $$f(x) = {{18} \over {13}}x - {4 \over {13}}$$

By checking all the options you can see when x = 6 we get

$$y = f(x) = {{18} \over {13}} \times 6 - {4 \over {13}}$$

$$ = {{108 - 4} \over {13}} = 8$$

$$\therefore$$ Point (6, 8) lies on the curve.

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