Answer
$$
\int_{1}^{4}\left(2 x^{2}-3 x+1\right) d x =\frac{45}{2}=22.5
$$
Work Step by Step
$$
\begin{aligned}
\int_{1}^{4}\left(2 x^{2}-3 x+1\right) d x &=2 \int_{1}^{4} x^{2} d x-3 \int_{1}^{4} x d x+\int_{1}^{4} 1 d x \\
& \,\,\,\,\,\,\,\,\,\ \text{ Use the results of Exercises 27 and 28} \\
&=2 \cdot \frac{1}{3}\left(4^{3}-1^{3}\right)-3 \cdot \frac{1}{2}\left(4^{2}-1^{2}\right)+1(4-1)\\
&=\frac{45}{2}=22.5
\end{aligned}
$$
