f is continuous on x=[0,1].
Given that $\displaystyle \int_0^{\pi}\\xf(sin x)\ dx = \frac{\pi}{2}\int_0^{\pi}\\f(sinx) dx$
Show that $\displaystyle F(x)=\int_0^1\\f(x+t) dt$ is differentiable and that $\displaystyle F\prime(x)=f(x+1)-f(x)$.
f is continuous on x=[0,1].
Given that $\displaystyle \int_0^{\pi}\\xf(sin x)\ dx = \frac{\pi}{2}\int_0^{\pi}\\f(sinx) dx$
Show that $\displaystyle F(x)=\int_0^1\\f(x+t) dt$ is differentiable and that $\displaystyle F\prime(x)=f(x+1)-f(x)$.
i think the first relation has nothing to do here.
put $\displaystyle u=x+t$ in your integral and then $\displaystyle F(x)=\int_x^{x+1}f(u)\,du,$ and $\displaystyle f$ was given as continuous, so $\displaystyle F$ is differentiable and then $\displaystyle F'(x)=f(x+1)-f(x).$