Question

18. Use limits to find the area between the graph of the function and the x axis given by the definite integral. ∫_1^5·(x^2-x+1)dx

Answer 1

Answer:

$\int\limits^5_1 {(x^2-x+1)} \, dx= \frac{100}{3}\approx33.33units^2$

Step-by-step explanation:

To determine the area of the definite integral, we take each term and find its corresponding integral. We know that $\int\limits {x^n} \, dx =\frac{x^{n+1}}{n+1} +C$, so therefore we rewrite the expression as $\frac{x^3}{3}-\frac{x^2}{2}+x$.

Now, we plug in each limit into the expression and find the difference between them:

$(\frac{5^3}{3}-\frac{5^2}{2}+5)-(\frac{1^3}{3}-\frac{1^2}{2}+1)$

$(\frac{125}{3}-\frac{25}{2}+5)-(\frac{1}{3}-\frac{1}{2}+1)$

$(\frac{250}{6}-\frac{75}{6}+\frac{30}{6} )-(\frac{2}{6}-\frac{3}{6}+\frac{6}{6})$

$(\frac{205}{6})-(\frac{5}{6})$

$\frac{200}{6}$

$\frac{100}{3}$

Therefore, $\int\limits^5_1 {(x^2-x+1)} \, dx= \frac{100}{3}\approx33.33units^2$