# How do you use limits to find the area between the curve y=x^2-x+1 and the x axis from [0,3]?

Dec 19, 2016

The area is $\frac{15}{2} = 7.5$.

#### Explanation:

In general, for a "nice" (continuous) function $f$ whose graph is above the $x$-axis for $a \le x \le b$, the area under the graph can be computed using limits as $\setminus {\int}_{a}^{b} f \left(x\right) \mathrm{dx} = {\lim}_{n \to \infty} \setminus {\sum}_{k = 1}^{n} f \left({x}_{k}\right) \setminus \Delta x$, where $\Delta x = \frac{b - a}{n}$ and ${x}_{k} = a + k \cdot \Delta x$.

For the given situation, $a = 0$ and $b = 3$ so that $\Delta x = \frac{3}{n}$ and ${x}_{k} = 0 + k \cdot \frac{3}{n} = \frac{3 k}{n}$. We then get $f \left({x}_{k}\right) = {\left(\frac{3 k}{n}\right)}^{2} - \frac{3 k}{n} + 1 = \frac{9 {k}^{2}}{n} ^ 2 - \frac{3 k}{n} + 1$.

This leads to ${\sum}_{k = 1}^{n} f \left({x}_{k}\right) \setminus \Delta x = {\sum}_{k = 1}^{n} \left(\frac{27 {k}^{2}}{n} ^ 3 - \frac{9 k}{n} ^ 2 + \frac{3}{n}\right)$

$= \frac{27}{n} ^ 3 \cdot {\sum}_{k = 1}^{n} {k}^{2} - \frac{9}{n} ^ 2 \cdot {\sum}_{k = 1}^{n} k + \frac{3}{n} \cdot {\sum}_{k = 1}^{n} 1$.

Now ${\sum}_{k = 1}^{n} 1 = n$, ${\sum}_{k = 1}^{n} k = \frac{n \left(n + 1\right)}{2}$, and
${\sum}_{k = 1}^{n} {k}^{2} = \frac{n \left(2 n + 1\right) \left(n + 1\right)}{6}$ (you can look these facts up...for example, see http://mathforum.org/library/drmath/view/56920.html ).

Thus, ${\sum}_{k = 1}^{n} f \left({x}_{k}\right) \setminus \Delta x$

$= \frac{27 n \left(2 n + 1\right) \left(n + 1\right)}{6 {n}^{3}} - \frac{9 n \left(n + 1\right)}{2 {n}^{2}} + 3$

Therefore, the area is

$\setminus {\int}_{a}^{b} f \left(x\right) \mathrm{dx} = {\lim}_{n \to \infty} \setminus {\sum}_{k = 1}^{n} f \left({x}_{k}\right) \setminus \Delta x$

$= \frac{54}{6} - \frac{9}{2} + 3 = \frac{15}{2}$