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?