# Thread: FUNdamental Theorem of calculus

1. ## FUNdamental Theorem of calculus

Express the limit, $\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \frac{e^{\frac {i}{n}}}{n}$ as a definite integral and evaluate it using the Fundamental Theorem of the Calculus

I have no idea what it's asking me to do haha
And if your wondering I do know the the Fundamental Theorem of Calculus\
Thanks : )

2. Originally Posted by qzno
Express the limit, $\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \frac{e^{\frac {i}{n}}}{n}$ as a definite integral and evaluate it using the Fundamental Theorem of the Calculus

I have no idea what it's asking me to do haha
And if your wondering I do know the the Fundamental Theorem of Calculus\
Thanks : )
If you were to approximate the area under $y = e^x$ on $[0,1]$ using n rectangles of equal thickness you would use the following

$\sum_{i=1}^n f(c_i) \Delta x,$

where (if using the right endpoint of the subinterval)

$c_i = \frac{i}{n},\;\;\; \Delta x = \frac{1}{n}$

or

$\sum_{i=1}^n e^{i/n} \, \frac{1}{n} \Delta x,$.

In the limit as $n \to \infty$ the answer would becomes exact. So the answer would be

$\int_0^1 e^x\, dx = \lim_{n \rightarrow \infty} \sum_{i=1}^{n} \frac{e^{\frac {i}{n}}}{n}$.

I'm guessing you can take it from here.

3. Originally Posted by qzno
Express the limit, $\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \frac{e^{\frac {i}{n}}}{n}$ as a definite integral and evaluate it using the Fundamental Theorem of the Calculus

I have no idea what it's asking me to do haha
And if your wondering I do know the the Fundamental Theorem of Calculus\
Thanks : )
$\sum_{i=1}^n e^{\frac{i}{n}} \cdot \frac{1}{n} = (e^{\frac{1}{n}}+e^{\frac{2}{n}}+e^{\frac{3}{n}}+ ... + e^{1}) \cdot \frac{1}{n}$

$\int_0^1 e^x dx$

4. why is it from 0 to 1?

5. look at the exponents for each term of the sum ...

least term is $e^{\frac{1}{n}} \approx e^0$

greatest term is $e^1$

these are the y-values of each infinitesimal rectangle in the Riemann sum

note also the factor $\frac{1}{n}$ ... a horizontal distance of 1 divided into n partitions.

6. Originally Posted by qzno
why is it from 0 to 1?
In general, the Riemann sum would be

$\sum_{i=1}^n f(c_i)\, \Delta x$ where $\Delta x = \frac{b-a}{n}\;\;\; \text{and}\;\;\; c_i = a + \frac{b-a}{n}\;i$. Now compare with what you have

$
\sum_{i=1}^{n} \frac{e^{\frac {i}{n}}}{n}
$

so $a + \frac{b-a}{n}\;i = \frac{i}{n}$ giving $a = 0,\;\;\; b = 1.$

7. thanks guys : )