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Find $f$ such that the following integral equation is satisfied: $$\int_0^x \lambda f(\lambda) ~d\lambda= \int_x^0(\lambda^2 + 1)f(\lambda) ~d\lambda + x$$

I attempted it in the following way:

$\int_0^x \lambda f(\lambda) ~d\lambda= -\int_0^x(\lambda^2 + 1)f(\lambda) ~d\lambda + x$

$\int_0^x\lambda^2f(\lambda) ~d\lambda + \int_0^x \lambda f(\lambda) ~d\lambda + \int_0^xf(\lambda) ~d\lambda - x = 0$

which kind of looks like a quadratic, but I wasn't sure how to use this property.

Can I use the fundamental theorem of calculus here somehow? I wasn't sure how to use it with the $\lambda$ terms in front fo the $f(\lambda)$ functions.

Am I on the right track or how should I approach this problem?

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up vote 2 down vote accepted

Yes. The fundamental theorem of calculus is your friend here. Note that we can rewrite what you have as $$\int_0^x \lambda f(\lambda) d \lambda + \int_0^x (\lambda^2 + 1) f(\lambda) d\lambda =x$$ $$\int_0^x (\lambda + \lambda^2 + 1) f(\lambda) d \lambda = x$$ The fundamental theorem of calculus now gives us $$(x + x^2 + 1)f(x) = 1$$ Hence, $$f(x) = \dfrac{1}{1 + x + x^2}$$

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@PeterTamaroff Could you explain your comment? "Fibonacci! Fibonacci!" – user17762 Jun 8 '12 at 5:19

I would suggest using the Fundamental Theorem of Calculus; remember that if $g(x)$ is continuous, then $$\frac{d}{dx}\int_a^x g(t)\,dt = g(x).$$

What happens if you take derivatives with respect to $x$ on both sides?

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