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

Apr 18, 2017

$I = {\int}_{-} {1}^{1} \setminus \left(1 - {x}^{3}\right) \setminus \mathrm{dx} = 2$

#### 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 $f \left(x\right) = 1 - {x}^{3}$ and we partition the interval $\left[- 1 , 1\right]$ using $\Delta = \left\{- 1 , - 1 + 2 \cdot \frac{1}{n} , - 1 + 2 \cdot \frac{2}{n} , \ldots , - 1 + 2 \cdot \frac{n}{n}\right\}$

And so:

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

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{2}{n} {\sum}_{i = 1}^{n} \setminus \left(2 - \frac{6 i}{n} + \frac{12 {i}^{2}}{n} ^ 2 - \frac{8 {i}^{3}}{n} ^ 3\right)$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{2}{n} \left\{{\sum}_{i = 1}^{n} 2 - \frac{6}{n} {\sum}_{i = 1}^{n} i + \frac{12}{n} ^ 2 {\sum}_{i = 1}^{n} {i}^{2} - \frac{8}{n} ^ 3 {\sum}_{i = 1}^{n} {i}^{3}\right\}$

Using the standard summation formula:

${\sum}_{r = 1}^{n} r \setminus = \frac{1}{2} n \left(n + 1\right)$
${\sum}_{r = 1}^{n} {r}^{2} = \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right)$
 sum_(r=1)^n r^3 = 1/4n&2(n+1)^2

we have:

$I = {\lim}_{n \rightarrow \infty} \frac{2}{n} \left\{2 n - \frac{6}{n} \frac{1}{2} n \left(n + 1\right) + \frac{12}{n} ^ 2 \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right) - \frac{8}{n} ^ 3 \frac{1}{4} {n}^{2} {\left(n + 1\right)}^{2}\right\}$

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

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

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

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

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{2}{n} ^ 2 \left\{- {n}^{2} - 3 n + 4 {n}^{2} + 6 n + 2 - 2 {n}^{2} - 4 n - 2\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{2}{n} ^ 2 \left\{{n}^{2} - n\right\}$

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