Given an integral equation, integrate the function.

Question

$$f(x)=x+\int_0^1t(x+t)f(t){\rm d}t$$ Then what is $$\eta=\int_0^1f(x){\rm d}x$$ Ok you can write: $$\eta=\int_0^1\left(x+\int_0^1t(x+t)f(t)dt\right){\rm d}x=\frac12+\int_0^1\int_0^1t(x+t)f(t){\rm d}t{\rm d}x$$ How to eliminate f?

Answer 1

Reverse the order of integration.

$$\begin{align}\int_0^1 dx \, f(x) &= \frac12 + \int_0^1 dx \ \int_0^1 dt \, t(x+t) f(t)\\ &= \frac12 + \int_0^1 dx \ \int_0^1 dt \, x(t+x) f(x) \\ &= \frac12 + \int_0^1 dx \, x f(x) \left (\frac12+x \right )\\ &= \frac12 + \frac12 \int_0^1 dx \, x f(x) + \int_0^1 dx \, x^2 f(x) \end{align}$$

Note, however, that

$$f(x) = x+ x \int_0^1 dt \, t f(t) + \int_0^1 dt \, t^2 f(t) $$

so that

$$\int_0^1 dx \, f(x) = f \left ( \frac12 \right ) $$

Now, note that

$$f(x) = A x + B$$

where

$$A = 1+ \int_0^1 dt \, t f(t)$$ $$B = \int_0^1 dt \, t^2 f(t)$$

Then

$$f(x) = x + \int_0^1 dt \, x (x+t) (A t+B) = x + x \left ( \frac13 A + \frac12 B\right ) + \frac14 A + \frac13 B $$

or,

$$A x+B = \left ( 1+\frac13 A + \frac12 B \right ) x + \frac14 A + \frac13 B $$

so that

$$\frac{2}{3} A - \frac12 B = 1$$ $$\frac14 A - \frac{2}{3} B = 0$$

Then $A=\frac{48}{23}$ and $B = \frac{18}{23}$ and the integral is $f(1/2) = \frac{42}{23}$.

Answer 2

Completely expand the RHS and write $$f(x)=x+\int_0^1t(x+t)f(t){\rm d}t= x +x\int_0^1tf(t){\rm d}t + \int_0^1t^2f(t){\rm d}t =: x+ c_1x+c_2$$ with the constants $$c_1=\int_0^1tf(t){\rm d}t, \; c_2=\int_0^1t^2f(t){\rm d}t$$ Then $$\eta=\int_0^1f(x){\rm d}x = \left[\frac{1}{2}(1+c_1)x^2+c_2x\right]_0^1 = \frac{1}{2}(1+c_1)+c_2$$ Edit: To compute $c_1, c_2, \eta$ use $$c_1=\int_0^1tf(t){\rm d}t = \frac{1}{3}+\frac{1}{3}c_1+\frac{1}{2}c_2$$ $$c_2=\int_0^1t^2f(t){\rm d}t = \frac{1}{4}+\frac{1}{4}c_1+\frac{1}{3}c_2$$ From this linear system you get $$c_1= \frac{25}{23},\; c_2=\frac{18}{23}$$ and for the integral $\eta=\frac{42}{23}.$

Answer 3

This solutions makes use of the Leibniz rule of integral differentiation. Consider, \begin{align} f(x) = x + \int_{0}^{1} t(x+t) f(t) \, dt \end{align} for which differentiation leads to \begin{align} f'(x) = 1 + \int_{0}^{1} t \, f(t) \, dt. \end{align} Differentiate once again to obtain $f''(x) = 0$ which leads to $f(x) = a x + b$. Now, using the intergo-equation it is seen that \begin{align} ax + b &= x + \int_{0}^{1} t(x+t) \, (at + b) \, dt \\ &= \left( 1 + \frac{a}{3} + \frac{b}{2} \right) x + \left( \frac{a}{4} + \frac{b}{3} \right) \end{align} When the coefficients are equated it is discovered that \begin{align} f(x) = \frac{18}{23} \left( \frac{8 \, x}{3} + 1 \right). \end{align} The desired integral to obtain is \begin{align} \int_{0}^{1} f(x) \, dx = \frac{42}{23}. \end{align}