# Stuck solving equation with integrals

I am asked to find a continuous function $f(x)$ that is defined $\forall x \in \mathbb{R}$ that satisfies the equation $$\int_{0}^{x}f(t)dt=\int_{x}^{1}t^2f(t)dt+\frac{x^{6}}{3}+\frac{x^8}{4} + C$$ I had no trouble finding $f(x)=2x^5$ But I am stuck finding C here's what I've done $$\int_{0}^{x}f(t)dt=\int_{x}^{1}t^2f(t)dt+\frac{x^{6}}{3}+\frac{x^8}{4} + C$$= $$x^5(2+2x^2-1/3x-x^{3}/4)=C$$ What do I do next ?

differentiate on both sides to find f(x)

$f(x)= -x^2f(x)+2(x^5+x^7)\implies f(x)=2x^5\dfrac{(1+x^2)}{1+x^2}=2x^5$

now, put $x=0$ in following equation along with substitution $f(t)=2t^5$ :

$\displaystyle\int_{0}^{x}f(t) dt=\displaystyle\int_{x}^{1} t^2 f(t) dt + \dfrac{x^6}{3}+\dfrac{x^8}{8}+C$

$0=2\displaystyle\int_{0}^{1} t^7 dt +C$

$C=\dfrac{-1}{4}$

You solved the difficult part. Finding $C$ is the easy part: $$\int_{0}^{x}2t^5dt=\int_{x}^{1}2t^7dt+\frac{x^{6}}{3}+\frac{x^8}{4} + C$$ $$\frac{x^{6}}{3}=\frac{1}{4}-\frac{x^8}{4}+\frac{x^{6}}{3}+\frac{x^8}{4} + C$$ Therefore $C=-\frac{1}{4}$