# How do you use the limit process to find the area of the region between the graph y=64-x^3 and the x-axis over the interval [1,4]?

May 25, 2017

${\int}_{1}^{4} \setminus 64 - {x}^{3} \setminus \mathrm{dx} = \frac{513}{4}$

#### 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 $f \left(x\right) = 64 - {x}^{3}$ and we partition the interval $\left[1 , 4\right]$ using:

$\Delta = \left\{1 , 1 + 1 \cdot \frac{3}{n} , 1 + 2 \cdot \frac{3}{n} , \ldots , 1 + n \cdot \frac{3}{n}\right\}$

And so:

$I = {\int}_{1}^{4} \setminus \left(64 - {x}^{3}\right) \setminus \mathrm{dx}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{3}{n} {\sum}_{i = 1}^{n} \setminus f \left(1 + i \cdot \frac{3}{n}\right)$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{3}{n} {\sum}_{i = 1}^{n} \setminus f \left(1 + \frac{3 i}{n}\right)$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{3}{n} {\sum}_{i = 1}^{n} \setminus \left\{64 - {\left(1 + \frac{3 i}{n}\right)}^{3}\right\}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{3}{n} {\sum}_{i = 1}^{n} \setminus \left\{64 - \left(1 + 3 \left(\frac{3 i}{n}\right) + 3 {\left(\frac{3 i}{n}\right)}^{2} + {\left(\frac{3 i}{n}\right)}^{3}\right)\right\}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{3}{n} {\sum}_{i = 1}^{n} \setminus \left\{63 - \frac{9 i}{n} - \frac{27 {i}^{2}}{n} ^ 2 - \frac{27 {i}^{3}}{n} ^ 3\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{3}{n} \left\{{\sum}_{i = 1}^{n} 63 - \frac{9}{n} {\sum}_{i = 1}^{n} i - \frac{27}{n} ^ 2 {\sum}_{i = 1}^{n} {i}^{2} - \frac{27}{n} ^ 3 {\sum}_{i = 1}^{n} {i}^{3}\right\}$

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:

$I = {\lim}_{n \rightarrow \infty} \frac{3}{n} \left\{63 n - \frac{9}{n} \frac{1}{2} n \left(n + 1\right) - \frac{27}{n} ^ 2 \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right) - \frac{27}{n} ^ 3 \frac{1}{4} {n}^{2} {\left(n + 1\right)}^{2}\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{3}{n} \left\{63 n - \frac{9}{2} \left(n + 1\right) - \frac{9}{2 n} \left(2 {n}^{2} + 3 n + 1\right) - \frac{27}{4 n} \left({n}^{2} + 2 n + 1\right)\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{3}{n} \frac{1}{4 n} \left\{252 {n}^{2} - 18 n \left(n + 1\right) - 18 \left(2 {n}^{2} + 3 n + 1\right) - 27 \left({n}^{2} + 2 n + 1\right)\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{3}{4 {n}^{2}} \left\{252 {n}^{2} - 18 {n}^{2} - 18 n - 36 {n}^{2} - 54 n - 18 - 27 {n}^{2} - 54 n - 27\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{3}{4 {n}^{2}} \left\{171 {n}^{2} - 123 n - 27\right\}$

$\setminus \setminus = \frac{3}{4} \setminus {\lim}_{n \rightarrow \infty} \left\{171 - \frac{123}{n} - \frac{27}{n} ^ 2\right\}$

$\setminus \setminus = \frac{3}{4} \setminus \left(171 - 0 - 0\right)$

$\setminus \setminus = \frac{513}{4}$

Using Calculus

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

${\int}_{1}^{4} \setminus 64 - {x}^{3} \setminus \mathrm{dx} = {\left[64 x - \frac{1}{4} {x}^{4}\right]}_{1}^{4}$
$\text{ } = \left(256 - 64\right) - \left(64 - \frac{1}{4}\right)$
$\text{ } = 192 - \frac{255}{4}$
$\text{ } = \frac{513}{4}$