  ### Home > APCALC > Chapter 7 > Lesson 7.2.3 > Problem7-83

7-83.

The limit below represents a definite integral. Fill in the blanks:

$\lim\limits _ { n \rightarrow \infty } \frac { 6 } { n } [ ( \frac { 1 } { n } + 3 ) +$ $( \frac { 2 } { n } + 3 ) + \ldots +$$( \frac { n } { n } + 3 ) ] = \int _ { \boxed{\text{_}} } ^ { \boxed{\text{_}} } \boxed{\text{___}} d x$

To make this easier, factor the $6$ out of the limit.
The $6$ will be factored out of the integral as well.
We will now write all $n$-terms as $x$-terms.

Find $dx$:

Observe that: $\lim\limits_{\Delta x\rightarrow 0}=\lim\limits_{n\rightarrow 0}\frac{1}{n}.$

...so $\frac{1}{n}\text{ represents }\Delta x, \text{ and }\lim\limits_{n\rightarrow \infty }\frac{1}{n}=dx$

Find the integrand:
$x$ is a variable. So find the part of the sum that changes:

$\frac{1}{n},\frac{2}{n},\frac{3}{n},...\frac{n}{n}.$

So the integrand must be $f(x) = x + 3$.

$\color{red}6\lim\limits_{n\rightarrow \infty }\left [ \color{red}{\frac{1}{n}}\left ( \frac{1}{n}+3 \right )+ \left( \frac{2}{n}+3 \right )+...+\left( \frac{n}{n}+3 \right )\right ]=\color{red}6\int_{a}^{b}f(x)dx$

$6\lim\limits_{n\rightarrow \infty }\color{red}{\Delta x}\left [ \left ( \frac{1}{n}+3 \right )+ \left( \frac{2}{n}+3 \right )+...+\left( \frac{n}{n}+3 \right )\right ]=6\int_{a}^{b}f(x)dx$

$6\lim\limits_{n\rightarrow \infty }\Delta x\color{red}{(x+3)}=6\int_{a}^{b}(x+3)dx { \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$

$6\lim\limits_{n\rightarrow \infty }\Delta x(x+3)=6\color{red}{\int_{0}^{1}}(x+3)dx { \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$

Find the bounds:
Look at the original sum.

$\text{Recall that the variable }x \text{ was defined by the changing terms: }\frac{1}{n},\frac{2}{n},\frac{3}{n},...$

$\text{The smallest value of }x \text{ was }\frac{1}{n},\text{ so }\lim\limits_{n\rightarrow \infty }\frac{1}{n}=0, \text{ that's the lower bound.}$

$\text{The largest value of }x \text{ was }\frac{n}{n},\text{ so }\lim\limits_{n\rightarrow \infty }\frac{n}{n}=1, \text{ that's the upper bound.}$