# Integral problem with equation

#### grgrsanjay

##### New member
Let $f:R \to R$ be a continuous and differential function given by

$\displaystyle f(x) = x + \int_0^1 (xy + x^2)f(y)dy$

find $\displaystyle \int_0^1 f(x)dx$ and $\displaystyle \int_0^1 xf(x)dx$

I wanted to know how i could start the problem.Please do not give full solution

It would be good if you could even help me with the LaTeX too....

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#### Jester

##### Well-known member
MHB Math Helper
Re: Calculus problem

What I might try. Expanding your integral equation gives

$\displaystyle f(x) = x + x\int_0^1 yf(y)dy + x^2\int_0^1 f(y)dy\;\;\;(1)$

Now $f(y)$ is continuous so $\displaystyle \int_0^1 f(y)dy$ and $\displaystyle \int_0^1 y f(y)$ are constant so from (1) we have a preliminary form for $f(x)$, namely

$\displaystyle f(x) = a x + bx^2$

Then use (1).

#### grgrsanjay

##### New member
Re: Calculus problem

$\displaystyle f(x) = x + \int_0^1 (xy + x^2)f(y)dy$
Let me integrate it, then $\displaystyle \int_0^1 f(x)dx$ = $\displaystyle (1+\int_0^1 yf(y)dy)$. $\displaystyle \int_0^1 xdx$ + $\displaystyle { \int_0^1 f(y)dy}. \int_0^1 x^2 dx$

So , I get the equation 4A = 3 + 3B $(Lazy$ $to$ $type$ $latex....so$ $used$ $A$ $and$ $B)$

Then what equation would i get??

#### HallsofIvy

##### Well-known member
MHB Math Helper
If you differentiate both sides of the given equation with respect to x, you get $\displaystyle f'= 1+ \int_0^1 (y+ 2x)f(y)dy$. If you differentiate again, you get $\displaystyle f'= \int_0^1 f(y)dy$ which is a constant for all x. Just call that constant A and f''= A gives f'= Ax+ B and then $\displaystyle f(x)= (A/2)x^2+ Bx+ C$, a general quadratic. Replacing f by that in the origina equation will give you three equations for A, B, and C.