# Thread: writing a limit as a integral on an interval

1. ## writing a limit as a integral on an interval

The question asks to write a definite integral on the integral [1,5] for the following limit.

$\displaystyle \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\left( {\frac{4}{n} \times \frac{1}{{\left( {2 + \frac{{4k}}{n}} \right)}}} \right)}$

I understand how to evaluate a Riemann sum on an interval for n strips etc.

But I am at a complete loss on how to work limits into an integral.
In desperate helps for some understanding on these, I've gone through my notes and text several time and cannot understand it.

2. Originally Posted by Craka
The question asks to write a definite integral on the integral [1,5] for the following limit.

$\displaystyle \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\left( {\frac{4}{n} \times \frac{1}{{\left( {2 + \frac{{4k}}{n}} \right)}}} \right)}$

I understand how to evaluate a Riemann sum on an interval for n strips etc.

But I am at a complete loss on how to work limits into an integral.
In desperate helps for some understanding on these, I've gone through my notes and text several time and cannot understand it.
If you think about it you'll see that you're approximating the integral of $\displaystyle f(x) = \frac{1}{x+1}$ over [1, 5] using right rectangles of width 4/n.

3. That is an approximating sum for the integral $\displaystyle \int_1^5 {\frac{1}{{1 + x}}} dx$.

4. Originally Posted by Plato
That is an approximating sum for the integral $\displaystyle \int_1^5 {\frac{1}{{2 + x}}} dx$.
I think it might be $\displaystyle f(x) = \frac{1}{x + {\color{red}1}}$.

The first right-rectangle has a width of $\displaystyle \frac{4}{n}$ and a height of $\displaystyle f\left(1 + \frac{4}{n}\right) = \frac{1}{1 + \frac{4}{n} + 1} = \frac{1}{2 + \frac{4}{n}}$.

The second right-rectangle has a width of $\displaystyle \frac{4}{n}$ and a height of $\displaystyle f\left(1 + 2 \cdot \frac{4}{n}\right) = \frac{1}{1 + 2 \cdot \frac{4}{n} + 1} = \frac{1}{2 + 2 \cdot \frac{4}{n}}$.

The kth right-rectangle has a width of $\displaystyle \frac{4}{n}$ and a height of $\displaystyle f\left(1 + k \cdot \frac{4}{n}\right) = \frac{1}{1 + k \cdot \frac{4}{n} + 1} = \frac{1}{2 + \frac{4k}{n}}$.

5. Thanks Mr. Fantastic

I can see how thing worked in the last post, which had a routine. Is there away to apply a routine to the initial problem?