# How do you evaluate the definite integral by the limit definition given int 4dx from [0,3]?

May 19, 2017

$12$

#### Explanation:

${\int}_{0}^{3} 4 \mathrm{dx} = {\left[4 x\right]}_{0}^{3}$

$= \left[4 \cdot 3\right] - \left[4 \cdot 0\right] = 12$

May 22, 2017

see below

#### Explanation:

${\int}_{0}^{3} 4 \mathrm{dx}$,

Integral by limit definition is given by

${\lim}_{n \to \infty} {\sum}_{i}^{n} f \left({c}_{i}\right) \delta x$

therefore,

$\delta x = \frac{3 - 0}{n} = \frac{3}{n}$
${c}_{i} = a + \delta x = 0 + \frac{3}{n} i = \frac{3}{n} i$

${\int}_{0}^{3} 4 \mathrm{dx} = {\lim}_{n \to \infty} {\sum}_{i}^{n} f \left({c}_{i}\right) \delta x$,

$= {\lim}_{n \to \infty} {\sum}_{i}^{n} f \left(\frac{3}{n} i\right) \frac{3}{n}$

for $i = 1 , 2 , 3 , \ldots , n \to f \left(x\right) = 4$

$= {\lim}_{n \to \infty} {\sum}_{i}^{n} 4 \cdot \frac{3}{n} = {\lim}_{n \to \infty} {\sum}_{i}^{n} \frac{12}{n}$

$= {\lim}_{n \to \infty} n \frac{12}{n} = {\lim}_{n \to \infty} 12 = 12$