There exists a function $f$ wich is continuous for all $x\in R$, and satisfies the next equation:
$$\phi(t)=\int_0^x f(t)dt=\int_x^1 t^2f(t)dt+\frac{x^{16}}{8}+\frac{x^{18}}{9}+C$$
Where $C$ is a constant. Now, find the explicit formula for $f(x)$ and find the value of the constant $C$.
So first, if we need to find the explicit formula for $f(x)$, we have the antiderivative of $\phi(t)$, such that $F'(x)=f(x)$, so if we differentiate the formula, we have:
$$\frac{d}{dx}\left(\int_x^1 t^2f(t)dt+\frac{x^{16}}{8}+\frac{x^{18}}{9}+C\right)$$ $$\frac{d}{dx}\left(\int_x^1 t^2f(t)dt\right)+2x^{15}+2x^{17}$$
Now, for that integral I'm not pretty sure if I did it well: Let $g(t)=\int_x^1 t^2f(t)dt$, and for it to be possible to apply the first theorem, we change as $g(t)=-\int_1^x t^2f(t)dt$
$$\frac{d}{dx}\left(-\int_1^x t^2f(t)dt\right)$$ And because of the first fundamental theorem: $g'(x)=f(1)-x^2f(x)$
So far we got: $$f(x)=f(1)-x^2f(x)+2x^{15}+2x^{17}$$
Until know, I'm lost. Is this correct?
The first fundamental theorem says
$$\frac{d}{dx} \left(-\int_1^x t^2 f(t) dt\right) = -x^2 f(x)$$
You then get $$f(x) = -x^2f(x) +2x^{15} +2x^{17}.$$
You can continue from there.