"Let f be a real valued continuous function on [0, 1] and $f(x) = x + \int\limits_0^1 {(x - t)f(t)dt}$. Then, which of the following points (x, y) lies on the curve y = f(x) ?"

Question

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.

"

Let f be a real valued continuous function on [0, 1] and $f(x) = x + \int\limits_0^1 {(x - t)f(t)dt}$.

Then, which of the following points (x, y) lies on the curve y = f(x) ?

Solution

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Let f be a real valued continuous function on [0, 1] and $f(x) = x + \int\limits_0^1 {(x - t)f(t)dt}$. Then, which of the following points (x, y) lies on the curve y = f(x) ?

Solution

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More Definite Integration questions

Drill 25 more like these. Every day. Free.

Let f be a real valued continuous function on [0, 1] and $f(x) = x + \int\limits_0^1 {(x - t)f(t)dt}$.

Then, which of the following points (x, y) lies on the curve y = f(x) ?