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

Question

Find the average value of the function

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

on the interval $[0,1]$.

Answer 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 $[0,1]$.

Answer 2

Hint 1: The average value of an integrable function $f(x)$ on the interval $[a,b]$ is $$\frac{\int_a^b f(x)\,dx}{b-a}_.$$

Hint 2: $f(x)$ is closely related to the "Fresnel Integral".

Answer 3

\begin{align} \overline{f} & =\frac 1 {1-0} \int_0^1 \int_x^1\cos (t^2) \; dt \, dx = \int_0^1 \int_0^t\cos(t^2)\;dx\,dt \\[10pt] & =\int_0^1t\cos(t^2)\,dt = \left[\frac 1 2 \sin(t^2)\right]_0^1 = \frac 1 2 \sin1 \end{align}