Let $$x(t) =\begin{cases}2&-2\lt t \lt 0 \\ -2 &0\lt t \lt 1\\ 0& \text{otherwise}\end{cases}$$ and $$h(t) =\begin{cases}4(1-|t|)&|t|\le1 \\0& \text{otherwise}\end{cases}$$ Compute $x*h=\int_{-\infty}^\infty x(u)h(t - u)du$ .
Using graphical interpretation of convolution we get:
$$y(t) =\begin{cases}0&t \lt -3 \\ 4(t+3)^2 &-3\lt t \lt -2\\ -4(t^2+2t-1) &-2\lt t \lt -1& \\ -8t(t+2) &-1\lt t \lt 0 \\ 12t^2-16t & 0 \lt t \lt 1\\-4(t-2)^2 &1 \lt t \lt 2 \\ 0 & t \gt 2 \end{cases}$$ This answer was verified with Mathematica.
My attempt:
$$x*h=\int_{-\infty}^\infty x(t-u)h(u)du = \int_{-1}^1 x(t-u)h(u)du = \int_{-1}^1 4x(t-u)(1-|u|)du =\int_{-1}^0 4x(t-u)(1+u)du + \int_{0}^1 4x(t-u)(1-u)du = I_1 + I_2$$
Then we have $$x(t-u) =\begin{cases}2&-2\lt t-u \lt 0 \\ -2 &0\lt t-u \lt 1\\ 0& \text{otherwise}\end{cases}$$
If we consider $I_1$ then $-1\lt u \lt 0$ and $I_2$ then $0\lt u \lt 1$. So there are two separate cases: $$-1 \lt u \lt 0 \implies x(t-u) =\begin{cases}2&-3\lt t \lt 0 \\ -2 &-1 \lt t \lt 1\\ 0& \text{otherwise}\end{cases}$$
and similarly $$0 \lt u \lt 1 \implies x(t-u) =\begin{cases}2&-2\lt t \lt 1 \\ -2 &0 \lt t \lt 2\\ 0& \text{otherwise}\end{cases}$$ then evaluating integrals leads to
$$I_1 =\begin{cases}4&-3\lt t \lt 0 \\ -4 &-1 \lt t \lt 1\\ 0& \text{otherwise}\end{cases}$$ and
$$I_2 =\begin{cases}4&-2\lt t \lt 1 \\ -4 &0 \lt t \lt 2\\ 0& \text{otherwise}\end{cases}$$
Is my calculations so far correct? If so, what's the next step? I don't understand why the values of the integrals are constant. They should be functions of $t$ to form such that solution.