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

Sep 7, 2017

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

#### 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

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

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

$\Delta = \left\{0 , \frac{3}{n} , 2. \frac{3}{n} , 3. \frac{3}{n} , \ldots , 3\right\}$

And so:

$I = {\int}_{0}^{3} \setminus {x}^{2} + 1 \setminus \mathrm{dx}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{3}{n} {\sum}_{i = 1}^{n} \setminus f \left(0 + i \cdot \frac{3}{n}\right)$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{3}{n} {\sum}_{i = 1}^{n} \setminus f \left(\frac{3 i}{n}\right)$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{3}{n} {\sum}_{i = 1}^{n} \setminus \left\{{\left(\frac{3 i}{n}\right)}^{2} + 1\right\}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{3}{n} {\sum}_{i = 1}^{n} \setminus \left\{\frac{9 {i}^{2}}{n} ^ 2 + 1\right\}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{3}{n} \left\{{\sum}_{i = 1}^{n} \setminus \frac{9 {i}^{2}}{n} ^ 2 + {\sum}_{i = 1}^{n} \setminus 1\right\}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{3}{n} \left\{\frac{9}{n} ^ 2 {\sum}_{i = 1}^{n} \setminus {i}^{2} + {\sum}_{i = 1}^{n} \setminus 1\right\}$

Using the standard summation formula:

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

we have:

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

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

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

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{3}{2 {n}^{2}} \left\{6 {n}^{2} + 9 n + 3 + 2 {n}^{2}\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{3}{2 {n}^{2}} \left\{8 {n}^{2} + 9 n + 3\right\}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{24 {n}^{2} + 27 n + 9}{2 {n}^{2}}$

$\setminus \setminus = {\lim}_{n \rightarrow \infty} \left(12 + \frac{27}{2 n} + \frac{9}{2 {n}^{2}}\right)$

$\setminus \setminus = 12 + 0 + 0$

$\setminus \setminus = 12$

Using Calculus

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

${\int}_{0}^{3} \setminus {x}^{2} + 1 \setminus \mathrm{dx} = {\left[{x}^{3} / 3 + x\right]}_{0}^{3}$
$\text{ } = \left(9 + 3\right) - 0$
$\text{ } = 12$