# How do you use limits to evaluate int (x+2)dx from [1,4]?

Apr 17, 2017

Perform the integration as if the integral were indefinite but without a constant of integration.
Subtract the expression evaluated at the lower limit from the expression evaluated at the upper limit.

#### Explanation:

Given: ${\int}_{1}^{4} \left(x + 2\right) \mathrm{dx}$

Perform the integration as if the integral were indefinite but without a constant of integration:

${\int}_{1}^{4} \left(x + 2\right) \mathrm{dx} = {\left.{x}^{2} / 2 + 2 x\right]}_{1}^{4}$

Subtract the resulting expression evaluated at the lower limit from the expression evaluated at the upper limit:

${\int}_{1}^{4} \left(x + 2\right) \mathrm{dx} = \left({4}^{2} / 2 + 2 \left(4\right)\right) - \left({1}^{2} / 2 + 2 \left(1\right)\right)$

${\int}_{1}^{4} \left(x + 2\right) \mathrm{dx} = 8 + 8 - \frac{1}{2} - 2$

${\int}_{1}^{4} \left(x + 2\right) \mathrm{dx} = 13.5$

Apr 18, 2017

${\int}_{1}^{4} \setminus \left(x + 2\right) \setminus \mathrm{dx} = 13.5$

#### 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 $f \left(x\right) = x + 2$ and we partition the interval $\left[1 , 4\right]$ using $\Delta = \left\{1 , 1 + 3 \cdot \frac{1}{n} , 1 + 3 \cdot \frac{2}{n} , \ldots , 1 + 3 \cdot \frac{n}{n}\right\}$

And so:

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

Using the standard summation formula:

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

we have:

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

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

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

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

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

$\setminus \setminus = 13.5$