If we have a signal
$$x(t)=\begin{cases} t &0\leq t < 1 \\ 0.5+0.5\cos(2 \pi t) &1 \leq t < 2 \\ 3-t &2\leq t<3 \\ 0 &\text{elsewhere} \end{cases}$$
It's energy is $E(x)=\frac{25}{24}\approx 1.0417$. Calculate energy for $y(t)=\frac 13 x(2t)$
I did my work like this, does it seem correct?
$$E=\int_{\mathbb{R}}|y(t)|^2 dt$$ now $$x(2t)=\begin{cases} 2t &0\leq 2t < 1 \\ 0.5+0.5\cos(4 \pi t) &1 \leq 2t < 2 \\ 3-2t &2\leq 2t <3 \\ 0 &\text{elsewhere} \end{cases}$$ then
$$|y(t)|^2=\frac 19 \times\begin{cases} 4t^2 &0\leq 2t < 1 \\ (0.5+0.5\cos(4 \pi t))^2 &1 \leq 2t < 2 \\ (3-2t)^2 &2\leq 2t <3 \\ 0 &\text{elsewhere} \end{cases}$$
So the integral would be $$E=\frac 19 \big( \overbrace{\int_0^{1/2} 4t^2 dt}^{A_1} + \overbrace{\int_{1/2}^1 \frac 14+ \frac 12 \cos (4\pi t)+\frac 14 \cos^2 (4 \pi t)dt}^{A_2}+\overbrace{\int_1^{3/2} 9-12t+4t^2 dt}^{A_3}\big)$$
\begin{align} A_1&=\int_0^{1/2} 4t^2 dt = \frac 16 \\ A_2&=\int_{1/2}^1 \frac 14+ \frac 12 \cos (4\pi t)+\frac 14 \cos^2 (4 \pi t)dt \\ &=\int_{1/2}^1 \frac 14+ \frac 12 \cos (4\pi t) +\frac 18 \cos(8 \pi t) + \frac 18 dt =\frac {3}{16} \\ A_3&=\int_1^{3/2} 9-12t+4t^2 dt = \frac{1}{6} \end{align} So the energy is $E=\frac 19 (A_1+A_2+A_3)=\frac{25}{432} \neq \frac {1}{18} E(x)$. Did I mess up the integral?
I think the function y is
$2t$ for $0\le 2t <1$
$\frac{1+cos(4\pi t)}2$ for $1\le2t<2$
$3-2t$ for $2\le2t<3$
and $0$ elsewhere