Always keep track of yourself.

\(\displaystyle \int_{1}^{2}6\cdot\left[\frac{1}{4}-\left(x-\frac{3}{2}\right)^{2}\right]\;dx\;=\;1\)

This is a proper pdf

When you apply the exponent, 3, you get...

\(\displaystyle \int_{1}^{2}\left{6\cdot\left[\frac{1}{4}-\left(x-\frac{3}{2}\right)^{2}\right]\right}^{3}\;dx\;=\;\frac{54}{35}\;>\;1\)

Clearly, this is no good.

Try this:

\(\displaystyle \int_{1.5}^{2}6\cdot\left[\frac{1}{4}-\left(x-\frac{3}{2}\right)^{2}\right]\;dx\;=\;\frac{1}{2}\)

After that, it's a Binomial problem with p = 1/2 and n = 3. Let's see what you get.