Step 1:

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

[latex]\large{\int_{3}^{1}(12t^4 – 3t^3 + 4t – 3) dt }[/latex]

[latex]\large{= (\frac{12}{5}t^5-\frac{3}{4}t^4+2t^2 – 3t)|_{3}^{1}}[/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_{3}^{1}(12t^4-3t^3+4t-3) dt}[/latex]

[latex]= [\frac{12}{5}(1)^5-\frac{3}{4}(1)^4+2(1)^2-3(1)] [/latex]

[latex] – \:[\frac{12}{5}(3)^5-\frac{3}{4}(3)^4+2(3)^2-3(3)][/latex]

[latex]\large{ =  [0.65]\:-\: [531.45]}[/latex]

[latex]\large{= -530.8}[/latex]

Step 3:

So, our final answer is:

[latex]\int_{3}^{1}(12t^4 – 3t^3 + 4t – 3) dt = -530.8[/latex]

Solution Complete!