Answer:
[tex]\int\limits^5_1 {(x^2-x+1)} \, dx= \frac{100}{3}\approx33.33units^2[/tex]
Step-by-step explanation:
To determine the area of the definite integral, we take each term and find its corresponding integral. We know that [tex]\int\limits {x^n} \, dx =\frac{x^{n+1}}{n+1} +C[/tex], so therefore we rewrite the expression as [tex]\frac{x^3}{3}-\frac{x^2}{2}+x[/tex].
Now, we plug in each limit into the expression and find the difference between them:
[tex](\frac{5^3}{3}-\frac{5^2}{2}+5)-(\frac{1^3}{3}-\frac{1^2}{2}+1)[/tex]
[tex](\frac{125}{3}-\frac{25}{2}+5)-(\frac{1}{3}-\frac{1}{2}+1)[/tex]
[tex](\frac{250}{6}-\frac{75}{6}+\frac{30}{6} )-(\frac{2}{6}-\frac{3}{6}+\frac{6}{6})[/tex]
[tex](\frac{205}{6})-(\frac{5}{6})[/tex]
[tex]\frac{200}{6}[/tex]
[tex]\frac{100}{3}[/tex]
Therefore, [tex]\int\limits^5_1 {(x^2-x+1)} \, dx= \frac{100}{3}\approx33.33units^2[/tex]
