How do you use limits to evaluate int8x^3dx from [3,5]?

Sep 13, 2017

$1088$

Explanation:

we need to use the power rule

$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + c , \text{ } n \ne - 1$

then apply the limits given

${\int}_{3}^{5} 8 {x}^{3} \mathrm{dx}$

$= 8 {\left[{x}^{4} / 4\right]}_{3}^{5}$

$= 2 {\left[{x}^{4}\right]}_{3}^{5}$

$= 2 \left\{{\left[{x}^{4}\right]}^{5} - {\left[{x}^{4}\right]}_{3}\right\}$

$2 \left({5}^{4} - {3}^{4}\right)$

$= 2 \left(625 - 81\right)$

$= 1088$

Sep 13, 2017

${\int}_{3}^{5} \setminus 8 {x}^{3} \setminus \mathrm{dx} = 1088$

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

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

And we partition the interval $\left[a , b\right]$ equally spaced using:

$\Delta = \left\{a + 0 \left(\frac{b - a}{n}\right) , a + 1 \left(\frac{b - a}{n}\right) , \ldots , a + n \left(\frac{b - a}{n}\right)\right\}$
$\setminus \setminus \setminus = \left\{a , a + 1 \left(\frac{b - a}{n}\right) , a + 2 \left(\frac{b - a}{n}\right) , \ldots , b\right\}$

Here we have $f \left(x\right) = 8 {x}^{3}$ and so we partition the interval $\left[3 , 5\right]$ using:

$\Delta = \left\{3 , 3 + \frac{2}{n} , 3 + 2 \frac{2}{n} , 3 + 3 \frac{2}{n} , \ldots , 5\right\}$

And so:

$I = {\int}_{3}^{5} \setminus 8 {x}^{3} \setminus \mathrm{dx}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} 8 \cdot \frac{2}{n} {\sum}_{i = 1}^{n} \setminus f \left(3 + i \cdot \frac{2}{n}\right)$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{16}{n} {\sum}_{i = 1}^{n} \setminus {\left(3 + \frac{2 i}{n}\right)}^{3}$
 \ \ = lim_(n rarr oo) 16/n sum_(i=1)^n \ {(3)^3+3(3)^2((2i)/n)^1 + 3(3)((2i)/n)^2 + ((2i)/n)^3

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{16}{n} {\sum}_{i = 1}^{n} \setminus \left\{27 + 54 \frac{i}{n} \setminus + 36 {i}^{2} / {n}^{2} + 8 {i}^{3} / {n}^{3}\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{16}{n} \left\{{\sum}_{i = 1}^{n} 27 + {\sum}_{i = 1}^{n} 54 \frac{i}{n} + {\sum}_{i = 1}^{n} 36 {i}^{2} / {n}^{2} + {\sum}_{i = 1}^{n} 8 {i}^{3} / {n}^{3}\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{16}{n} \left\{27 {\sum}_{i = 1}^{n} 1 + \frac{54}{n} {\sum}_{i = 1}^{n} i + \frac{36}{n} ^ 2 {\sum}_{i = 1}^{n} {i}^{2} + \frac{8}{n} ^ 3 {\sum}_{i = 1}^{n} {i}^{3}\right\}$

Using the standard summation formula:

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

we have:

 I = 16 lim_(n rarr oo) 1/n { 27n + 54/n 1/2n(n+1) + 36/n^2 1/6n(n+1)(2n+1) + 8/n^3 1/4n^2(n+1)^2

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{16}{n} \left\{27 n + 27 \left(n + 1\right) + \frac{6}{n} \left(n + 1\right) \left(2 n + 1\right) + \frac{2}{n} {\left(n + 1\right)}^{2}\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{16}{n} ^ 2 \left\{27 {n}^{2} + 27 n \left(n + 1\right) + 6 \left(n + 1\right) \left(2 n + 1\right) + 2 {\left(n + 1\right)}^{2}\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{16}{n} ^ 2 \left\{27 {n}^{2} + 27 {n}^{2} + 27 n + 6 \left(2 {n}^{2} + 3 n + 1\right) + 2 \left({n}^{2} + 2 n + 1\right)\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{16}{n} ^ 2 \left\{27 {n}^{2} + 27 {n}^{2} + 27 n + 12 {n}^{2} + 18 n + 6 + 2 {n}^{2} + 4 n + 2\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{16}{n} ^ 2 \left\{68 {n}^{2} + 49 n + 8\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \left\{1088 + \frac{784}{n} + \frac{8}{n} ^ 2\right\}$

$\setminus \setminus = 1088 + 0 + 0$

$\setminus \setminus = 1088$

Using Calculus

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

${\int}_{3}^{5} \setminus 8 {x}^{3} \setminus \mathrm{dx} = {\left[8 {x}^{4} / 4\right]}_{3}^{5}$
$\text{ } = 2 \left({5}^{4} - {3}^{4}\right)$
 " " = 2(625-81
$\text{ } = 2 \cdot 544$
$\text{ } = 1088$