Absolute Value in Integrals

Question

I'm trying to solve this integral problem but I'm a bit confused as to how to evaluate the absolute value portion of it. For context, $a$ is a positive constant.

$$ \int_{-a}^{a}\left(\sqrt{a^{2}-t^{2}}+|a+3 t|+a\right) d t $$

Answer 1

$| a+ 3t|= \begin{cases} 3t+a & 3t+a\geq 0, (t \geq -a/3)\\ -(3t+a)& 3t+a<0, (t < -a/3) \end{cases}$

Now you divide the integral

\begin{align} &\int_{-a}^{a}\left(\sqrt{a^{2}-t^{2}}+|a+3t|+a\right) d t =\\ & \int_{-a}^{-a/3}\left(\sqrt{a^{2}-t^{2}}-a-3 t+a\right) d t +\int_{-a/3}^{a}\left(\sqrt{a^{2}-t^{2}}+a+3 t+a\right) d t\\ &= \int_{-a}^{-a/3}\left(\sqrt{a^{2}-t^{2}}-3 t\right) d t +\int_{-a/3}^{a}\left(\sqrt{a^{2}-t^{2}}+3 t+2a\right) d t \end{align}

you complete the solution now.