Compare an arcsine integral with an absolute-value area

$\int_1^2x^2\ln x\,dx=[x^3\ln x/3-x^3/9]_1^2=8\ln2/3-7/9$.

$3+2x-x^2=4-(x-1)^2$, so the antiderivative is $9\sin^{-1}((x-1)/2)+C$ on $-1<x<3$. The left side is $9[\sin^{-1}(1/2)-\sin^{-1}(-1/2)]=3\pi$.

From $-1/2$ to $1/2$, the graph of $|1-2x|$ is a triangle of base $1$ and height $2$, so its integral is $1$. The cumulative integral from $-1/2$ is increasing; to reach $3$ it must continue past $1/2$. Thus $a>1/2$, before removing the absolute value on the remaining interval.

$1+\int_{1/2}^a(2x-1)\,dx=3$ gives $1+[x^2-x]_{1/2}^a=3$, hence $a^2-a-7/4=0$. Thus $a=(1\pm2\sqrt2)/2$; only $a=(1+2\sqrt2)/2$ satisfies $a>1/2$.