Since the function we want to integrate is a polynomial, we use the following general rule:
[tex]\int ax^ndx=a\cdot\frac{x^{n+1}}{n+1}[/tex]
Then, in this case we have:
[tex]\begin{gathered} \int ^4_{-1}(4x^2-3x+7)dx=4\cdot\frac{x^3}{3}-3\cdot\frac{x^2}{2}+7x \\ =\frac{4}{3}x^3-\frac{3}{2}x^2+7x \end{gathered}[/tex]
Now we must evaluate F(b)-F(a). We have that a=-1 and b=4, then:
[tex]\begin{gathered} F(x)=\frac{4}{3}x^2-\frac{3}{2}x^2+7x \\ a=-1 \\ b=4 \\ F(-1)=\frac{4}{3}\cdot(-1)^3-\frac{3}{2}\cdot(-1)^2+7(-1)=-\frac{4}{3}-\frac{3}{2}-7=-\frac{59}{6} \\ F(4)=\frac{4}{3}\cdot(4)^3-\frac{3}{2}\cdot(4)^2+7\cdot(4)=\frac{256}{3}-24+28=\frac{256}{3}+4 \\ =\frac{268}{3} \\ \Rightarrow F(b)-F(a)=\frac{268}{3}-(-\frac{59}{6})=\frac{268}{3}+\frac{59}{6}=\frac{536+59}{6}=\frac{595}{6} \end{gathered}[/tex]
Therefore, the integral of 4x^2-3x+7 evaluated from -1 to 4 is 595/6
