Average value of $f(x)=\int_x^1 \cos(t^2) \,dt$ on the interval $[0,1]$. Find the average value of the function 

$$f(x)=\int_x^1 \cos(t^2) \,dt
$$ 

on the interval $[0,1]$.

One may recall that if $f(x)$ is continuous on $[a,b]$, then the average value of $f(x)$ on $[a,b]$ is defined to be
$$I=\frac1{b-a}\int_a^b f(x)\,dx.$$
Applying it here gives
$$
\begin{align}
I=\int_0^1 \left(\int_x^1 \cos(t^2) \,dt\right)\,dx&=\left.x\int_x^1 \cos(t^2) \,dt \right|_0^1+\int_0^1 x \cos(x^2)\,dx
\\\\&=0+\int_0^1 x \cos(x^2)\,dx
\end{align}$$ where we have performed an integration by parts, obtaining 

$$
I=\frac12\int_0^1 \cos(u)\,du=\color{red}{\frac{ \sin 1}2}
$$ 

to be the average value of $f$ over $[0,1]$.
