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

Aug 7, 2017

${\int}_{- 1}^{0} \setminus {x}^{2} - {x}^{3} \setminus \mathrm{dx} = \frac{7}{12}$

#### 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 $f \left(x\right) = {x}^{2} - {x}^{3}$ and we partition the interval $\left[- 1 , 0\right]$ using:

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

And so:

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

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

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

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

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{1}{n} {\sum}_{i = 1}^{n} \setminus \left\{2 - \frac{5 i}{n} + \frac{4 {i}^{2}}{n} ^ 2 - {i}^{3} / {n}^{3}\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{1}{n} \left\{2 {\sum}_{i = 1}^{n} 1 - \frac{5}{n} {\sum}_{i = 1}^{n} i + \frac{4}{n} ^ 2 {\sum}_{i = 1}^{n} {i}^{2} - \frac{1}{n} ^ 3 {\sum}_{i = 1}^{n} {i}^{3}\right\}$

Using the standard summation formula:

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

we have:

$I = {\lim}_{n \rightarrow \infty} \frac{1}{n} \left\{2 n - \frac{5}{n} \frac{1}{2} n \left(n + 1\right) + \frac{4}{n} ^ 2 \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right) - \frac{1}{n} ^ 3 \frac{1}{4} {n}^{2} {\left(n + 1\right)}^{2}\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{1}{n} \left\{2 n - \frac{5}{2} \left(n + 1\right) + \frac{2}{3 n} \left(n + 1\right) \left(2 n + 1\right) - \frac{1}{4 n} {\left(n + 1\right)}^{2}\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{1}{12 {n}^{2}} \left\{24 {n}^{2} - 30 n \left(n + 1\right) + 8 \left(n + 1\right) \left(2 n + 1\right) - 3 {\left(n + 1\right)}^{2}\right\}$

$\setminus \setminus = \frac{1}{12} {\lim}_{n \rightarrow \infty} \frac{1}{{n}^{2}} \left\{24 {n}^{2} - 30 {n}^{2} - 30 n + 16 {n}^{2} + 24 n + 8 - 3 {n}^{2} - 6 n - 3\right\}$

$\setminus \setminus = \frac{1}{12} {\lim}_{n \rightarrow \infty} \frac{1}{{n}^{2}} \left\{7 {n}^{2} - 6 n + 5\right\}$
$\setminus \setminus = \frac{1}{12} {\lim}_{n \rightarrow \infty} \left\{7 - \frac{6}{n} + \frac{5}{n} ^ 2\right\}$

And this is now a trivial limit to evaluate, so:

$L = \frac{1}{12} \left\{7 - 0 + 0\right\}$
$\setminus \setminus = \frac{7}{12}$

Using Calculus

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

${\int}_{- 1}^{0} \setminus {x}^{2} - {x}^{3} \setminus \mathrm{dx} = {\left[{x}^{3} / 3 - {x}^{4} / 4\right]}_{- 1}^{0}$
$\text{ } = \left(0 - 0\right) - \left(- \frac{1}{3} - \frac{1}{4}\right)$
$\text{ } = \frac{7}{12}$