Step 1:

Using our knowledge of the integral power rule, we can find the integral as follows:

[latex]\large{\int_{0}^{2}(3x^2-2x+1) dx}[/latex]

[latex]\large{ = (x^3-x^2+x)|_{0}^{2}}[/latex]

Step 2:

Now, we can apply the fundamental theorem of calculus. This states that:

[latex]\large{\int_{a}^{b}f(x) dx = F(b) – F(a)}[/latex]

Thus, we have:

[latex]\large{\int_{0}^{2}(3x^2-2x+1) dx }[/latex]

[latex]\small{= [(2)^3 – (2)^2 + 2] – [(0)^3-(0)^2+0 ] }[/latex]

[latex]\large{ = 6 \:-\: 0 = 6}[/latex]

Step 3:

So, our final answer is:

[latex]\large{\int_{0}^{2}(3x^2-2x+1)dx = 6}[/latex]

Solution Complete!