Showing that $\mathrm{Var}(X)=\mathrm{Var}(Y)$ Let $X$ and $Y$ be continuous random variables with the following PDF $(b>a)$:

Show that $\mathrm{Var}(X)=\mathrm{Var}(Y)$.
Here is my attempt:
\begin{align*}
\mathrm{Var}(X)&=\mathbb{E}(X^2)-(\mathbb{E}(X))^2 = \int_1^2 bx^2 \, dx + \int_2^4 ax^2 \, dx - \left(\int_1^2 bx \, dx + \int_2^4 ax \, dx\right)^2\\
&=\frac{7b+56a}{3}-\frac{9b^2}{4}-18ba-36a^2\\
\mathrm{Var}(Y)&=\mathbb{E}(Y^2)-(\mathbb{E}(Y))^2 = \int_2^4 ay^2\, dy - \int_4^5 by^2\, dy - \left(\int_2^4 ay\, dy - \int_4^5 by\, dy\right)^2\\
&=\frac{56a-61b}{3}-36a^2+54ab-\frac{81b^2}{4}
\end{align*}

$Var(X)$ is correct but there are mistakes in signs while calculation $Var(Y)$.
\begin{align*}
\mathrm{Var}(X)&=\mathbb{E}(X^2)-(\mathbb{E}(X))^2 = \int_1^2 bx^2 \, dx + \int_2^4 ax^2 \, dx - \left(\int_1^2 bx \, dx + \int_2^4 ax \, dx\right)^2\\
&=\frac{7b+56a}{3}- \left(6a+\frac{3b}{2}\right)^2\\
\mathrm{Var}(Y)&=\mathbb{E}(Y^2)-(\mathbb{E}(Y))^2 = \int_2^4 ay^2\, dy + \int_4^5 by^2\, dy - \left(\int_2^4 ay\, dy + \int_4^5 by\, dy\right)^2\\
&=\frac{56a+61b}{3}- \left(6a+\frac{9b}{2}\right)^2
\end{align*}
$ \displaystyle Var(Y) - Var(X) = 18b + \left(6a+\frac{3b}{2}\right)^2 - \left(6a+\frac{9b}{2}\right)^2$
$ \displaystyle = 18b - 36 ab - 18b^2$
Now given the pdf, we also note that $2a + b = 1 \implies 2a = 1 - b$
So, $ \displaystyle Var(Y) - Var(X) = 18b - 18 b (1-b) - 18b^2 = 0$
