Problem: For a fixed $c \in \mathbb{R} $, let
\[
\alpha = \int _0^2 \left( 9x^2 - 5cx^4 \right) \mathrm{d} x.
\]
If the value of $\int _0^2 \left( 9x^2 - 5cx^4 \right) \mathrm{d} x$ obtained by using the Trapezoidal rule is equal to $\alpha $, then the value of $c$ is ______________ (rounded off to 2 decimal places).
Solution: The given integral will be
\begin{align*}
\alpha & = \int_0^2 \left( 9x^2 - 5c x^4 \right) \mathrm{d} x \\
& = \left[ 3x^3 - c x^5 \right] _0^2 \\
& = 24 - 32 c.
\end{align*}
It is also given that the value of the integral by the Trapezoidal rule is equal to $\alpha $. We recall that the Trapezoidal rule is given by
\[
\int_a^b f(x)\mathrm{d} x = \frac{b-a}{2}\left[ f(b) - f(a) \right] .
\]
So, we have
\begin{align*}
& 24 - 32c = \int_0^2 \left( 9x^2 - 5cx^4 \right) \mathrm{d} x \\
\implies & 24 - 32c = \frac{2+0}{2}\left[ 36 - 80c \right] \\
\implies & 24 - 32c = 36 - 80c \\
\implies & 48c = 12 \\
\implies & c = \frac{12}{48} = \frac{1}{4} = 0.25.
\end{align*}
Therefore, the value of $c$ will be $0.25$.