# How do you use the limit process to find the area of the region between the graph y=-2x+3 and the x-axis over the interval [0,1]?

##### 1 Answer
May 19, 2017

${\int}_{0}^{1} \setminus - 2 x + 3 \setminus \mathrm{dx} = 2$

#### Explanation:

By definition of an integral, then

${\int}_{a}^{b} \setminus f \left(x\right) \setminus \mathrm{dx}$

represents the area under the curve $y = f \left(x\right)$ between $x = a$ and $x = b$. We can estimate this area under the curve using thin rectangles. The more rectangles we use, the better the approximation gets, and calculus deals with the infinite limit of a finite series of infinitesimally thin rectangles.

That is

${\int}_{a}^{b} \setminus f \left(x\right) \setminus \mathrm{dx} = {\lim}_{n \rightarrow \infty} \frac{b - a}{n} {\sum}_{i = 1}^{n} \setminus f \left(a + i \frac{b - a}{n}\right)$

Here we have $f \left(x\right) = - 2 x + 3$ and we partition the interval $\left[0 , 1\right]$ using:

$\Delta = \left\{0 , 0 + 1 \cdot \frac{1}{n} , 0 + 1 \cdot \frac{2}{n} , \ldots , 0 + 1 \cdot \frac{n}{n}\right\}$
$\setminus \setminus \setminus = \left\{0 , \frac{1}{n} , \frac{2}{n} , \ldots , \frac{n}{n}\right\}$

And so:

$I = {\int}_{0}^{1} \setminus \left(- 2 x + 3\right) \setminus \mathrm{dx}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{1}{n} {\sum}_{i = 1}^{n} \setminus f \left(0 + 1 \cdot \frac{i}{n}\right)$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{1}{n} {\sum}_{i = 1}^{n} \setminus f \left(\frac{i}{n}\right)$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{1}{n} {\sum}_{i = 1}^{n} \setminus - 2 \left(\frac{i}{n}\right) + 3$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{1}{n} {\sum}_{i = 1}^{n} \setminus \frac{- 2 i}{n} + 3$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{1}{n} \left\{- \frac{2}{n} {\sum}_{i = 1}^{n} i + {\sum}_{i = 1}^{n} 3\right\}$

Using the standard summation formula:

${\sum}_{r = 1}^{n} r = \frac{1}{2} n \left(n + 1\right)$

we have:

$I = {\lim}_{n \rightarrow \infty} \frac{1}{n} \left\{- \frac{2}{n} \frac{1}{2} n \left(n + 1\right) + 3 n\right\}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{1}{n} \left\{- \left(n + 1\right) + 3 n\right\}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{1}{n} \left\{- n - 1 + 3 n\right\}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{1}{n} \left\{2 n - 1\right\}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \left\{2 - \frac{1}{n}\right\}$
$\setminus \setminus = 2$

Using Calculus

If we use Calculus and out knowledge of integration to establish the answer, for comparison, we get:

${\int}_{0}^{1} \setminus - 2 x + 3 \setminus \mathrm{dx} = {\left[- {x}^{2} + 3 x\right]}_{0}^{1}$
$\text{ } = \left(- 1 + 3\right) - \left(0 + 0\right)$
$\text{ } = 2$