Compare modulus integrals using algebra and geometry

The Teacher advises: draw a diagram. The V-shaped graph has vertex $(c,0)$, with $|x-c|=c-x$ for $x<c$ and $|x-c|=x-c$ for $x\ge c$. Since $c>0$, the interval $[-c,0]$ lies wholly to the left of the vertex, while $[0,2c]$ crosses it.

$\int_{-c}^0|x-c|\,dx=\int_{-c}^0(c-x)\,dx=[cx-x^2/2]_{-c}^0=c^2+\frac12c^2=\frac32c^2$.

$\int_0^{2c}|x-c|\,dx=\int_0^c(c-x)\,dx+\int_c^{2c}(x-c)\,dx=[cx-x^2/2]_0^c+[x^2/2-cx]_c^{2c}=\frac12c^2+\frac12c^2=c^2$. Hence $\frac32c^2=kc^2$, and $k=3/2$ because $c>0$.

The source labels the left trapezium $A_1$ and the two right triangles $A_2,A_3$. Their areas give $\frac12c(2c+c)=k[\frac12c(c)+\frac12c(c)]$, again $k=3/2$. All three areas are nonnegative; the left trapezium is above the x-axis.