Help with showing integral is differentiable

**owq** · October 18th 2009, 06:34 AM

f is continuous on x=[0,1].

Given that $\int_0^{\pi}\\xf(sin x)\ dx = \frac{\pi}{2}\int_0^{\pi}\\f(sinx) dx$

Show that $F(x)=\int_0^1\\f(x+t) dt$ is differentiable and that $F\prime(x)=f(x+1)-f(x)$ .

**Krizalid** · October 18th 2009, 10:04 AM

i think the first relation has nothing to do here.

put $u=x+t$ in your integral and then $F(x)=\int_x^{x+1}f(u)\,du,$ and $f$ was given as continuous, so $F$ is differentiable and then $F'(x)=f(x+1)-f(x).$

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