How To Find the Definite Integral Of This:

**AlphaRock** · January 12th 2010, 02:15 PM

Thanks. This is useful.

**galactus** · January 12th 2010, 04:02 PM

It is given as Riemann sum.

The subintervals are $\frac{6-0}{n}=\frac{6}{n}$

The integral is:

$\int_{0}^{6}e^{4+x}dx$

**abender** · January 12th 2010, 06:53 PM

On top of what the last poster offered, this is what I wrote up (and forgot to send a long time ago because I got enormously side-tracked):

$ \lim_{n\to\infty} \sum_{i=1}^n e^{4+\frac{6i}{n}} \cdot \frac{6}{n} =$ $ \lim_{n\to\infty} \sum_{i=1}^n e^4\cdot e^{\frac{6i}{n}}\cdot\frac{6}{n} =$ $ \lim_{n\to\infty} e^4\cdot \sum_{i=1}^n e^{\frac{6i}{n}}\cdot\frac{6}{n} =$ $ e^4 \cdot \lim_{n\to\infty}\sum_{i=1}^n e^{\frac{6i}{n}}\cdot\frac{6}{n} =$ $ e^4 \cdot \lim_{n\to\infty}\sum_{i=1}^n e^{(\frac{6}{n})i}\cdot\frac{6}{n} =$ $ e^4 \cdot \lim_{n\to\infty}\sum_{i=1}^n \bigg(e^{(\frac{6}{n})i}\bigg)\bigg(\frac{6}{n}\bi gg)$

Let $x_i^*$ be the right-hand endpoint of the $i^{\text{th}}$ subinterval; i.e.,

$x_i^* = \frac{6i}{n}$ , for $i= 1,2,...,n$

Thus,

$\Delta x_i = \frac{6}{n}$ , $a =x_0 = \frac{6 \cdot 0}{n}=0$ , and $b =x_1= \frac{6 \cdot n}{n}=6$

And we have:

$ \lim_{n\to\infty} \sum_{i=1}^n e^{4+\frac{6i}{n}} \cdot \frac{6}{n} = e^4 \cdot \int_{0}^6 = e^x dx = \int_{0}^6 = e^4 \cdot e^x dx = \int_{0}^6 = e^{x+4} dx $

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