Write the Riemann sum to find the area under the graph of the function f(x) = x^2 from x = 1 to x = 5?

Feb 8, 2018

${\int}_{1}^{5} \setminus {x}^{2} \setminus \mathrm{dx} = \frac{124}{3}$

Explanation:

$I = {\int}_{1}^{5} \setminus {x}^{2} \setminus \mathrm{dx}$

Using Riemann sums. 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:

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

Here we have $f \left(x\right) = {x}^{2}$ and so we partition the interval $\left[1 , 5\right]$ using:

$\Delta = \left\{1 , 1 + 1 \frac{4}{n} , 1 + 2 \frac{4}{n} , 1 + 3 \frac{4}{n} , \ldots , 4\right\}$

And so:

$I = {\int}_{1}^{5} \setminus \left({x}^{2}\right) \setminus \mathrm{dx}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{4}{n} {\sum}_{i = 1}^{n} \setminus f \left(1 + \frac{4 i}{n}\right)$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{4}{n} {\sum}_{i = 1}^{n} \setminus {\left(1 + \frac{4 i}{n}\right)}^{2}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{4}{n} {\sum}_{i = 1}^{n} \setminus {\left(\frac{n + 4 i}{n}\right)}^{2}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{4}{n} {\sum}_{i = 1}^{n} \setminus \frac{1}{n} ^ 2 {\left(n + 4 i\right)}^{2}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{4}{n} ^ 3 {\sum}_{i = 1}^{n} \setminus \left({n}^{2} + 8 n i + 16 {i}^{2}\right)$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{4}{n} ^ 3 \left\{{n}^{2} {\sum}_{i = 1}^{n} 1 + 8 n {\sum}_{i = 1}^{n} i + 16 {\sum}_{i = 1}^{n} {i}^{2}\right)$

Using the standard summation formula:

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

we have:

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

 \ \ = lim_(n rarr oo) 4/n^3 \ n/3 {3n^2 + 12n(n+1)+ 8(n+1)(2n+1)

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{4}{3 {n}^{2}} \left(3 {n}^{2} + 12 {n}^{2} + 12 n + 16 {n}^{2} + 24 n + 8\right)$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{4}{3 {n}^{2}} \left(31 {n}^{2} + 36 n + 8\right)$
$\setminus \setminus = \frac{4}{3} {\lim}_{n \rightarrow \infty} \left(31 + \frac{36}{n} + \frac{8}{n} ^ 2\right)$
$\setminus \setminus = \frac{4}{3} \left(31 + 0 + 0\right)$
$\setminus \setminus = \frac{124}{3}$

Using Calculus

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

$I = {\int}_{1}^{5} \setminus {x}^{2} \setminus \mathrm{dx}$
$\setminus \setminus = {\left[{x}^{3} / 3\right]}_{1}^{5}$
$\setminus \setminus = \frac{125}{3} - \frac{1}{3}$
$\setminus \setminus = \frac{124}{3}$