The absolute symbol [imath]\displaystyle | \cdot \cdot \cdot \cdot \cdot|[/imath] looks cute, however it is the chaos of [imath]\displaystyle 99[/imath]% of students' mistakes. I myself still solve many problems falsely because of it.
We have:
[imath]\displaystyle f(x) = \left|6x^2-33x + 45\right|[/imath]
When I was in the Duck class in the kindergarten, I have learnt that [imath]\displaystyle g(x) = |x| = \pm x [/imath] has two signs. Therefore, to avoid making mistakes, find when the function [imath]\displaystyle f(x) [/imath] has a positive sign and when it has a negative sign.
Let [imath]\displaystyle h(x) = 6x^2-33x + 45[/imath].
I will leave it for you as an exercise to show that:
[imath]\displaystyle 6x^2-33x + 45 = (2x - 5)(3x - 9)[/imath]
We want to find the values of [imath]\displaystyle x[/imath] when [imath]\displaystyle (2x - 5)(3x - 9) = 0[/imath].
[imath]\displaystyle x = 2.5, \ \ x = 3[/imath]
I will also leave it for you as an exercise to show that the function [imath]\displaystyle h(x)[/imath] is negative between [imath]\displaystyle x = 2.5 \ \text{and} \ x = 3[/imath], and positive everywhere else.
This information will give us the piecewise form of [imath]\displaystyle f(x)[/imath].
[imath]\displaystyle f(x) =\begin{cases} \ \ \ (6x^2-33x + 45), & \ \text{everywhere else} \ \\-(6x^2-33x + 45), & \ \ 2.5 < x < 3\end{cases}[/imath]
Now we can safely set correct integrals:
[imath]\displaystyle\int_{0}^{5} f(x) \ dx = \int_{0}^{5} \left|6x^2-33x + 45\right| \ dx[/imath]
[imath]= \displaystyle \int_{0}^{2.5} \left(6x^2-33x + 45\right) \ dx + \int_{2.5}^{3} -\left(6x^2-33x + 45\right) \ dx + \int_{3}^{5} \left(6x^2-33x + 45\right) \ dx = 62.75[/imath]
The wrong approach was:
[imath]\displaystyle\int_{0}^{5} \left(6x^2-33x + 45\right) \ dx = 62.5[/imath]
