Let f:[0,π/4]→Rf:[0,\pi/4]\to\mathbb{R}f:[0,π/4]→R be continuous. Then there exists c∈(0,π/4)c\in(0,\pi/4)c∈(0,π/4) such that
f(c)=∫0π/4f(t) dtf(c) = \int_0^{\pi/4} f(t)\,dtf(c)=∫0π/4f(t)dt
f(c)=2cos(2c) ∫0π/4f(t) dtf(c) = 2\cos(2c)\,\int_0^{\pi/4} f(t)\,dtf(c)=2cos(2c)∫0π/4f(t)dt
f(c)=sin(2c)f(c) = \sin(2c)f(c)=sin(2c)
f(c)=sin(2c) ∫0π/4f(t) dtf(c) = \sin(2c)\,\int_0^{\pi/4} f(t)\,dtf(c)=sin(2c)∫0π/4f(t)dt
