Calculate the definite integral of $\lim\limits_{n\to\infty}\sum\limits_{i=1}^{n} (1+\frac{3i}{n})e^{2(1+\frac{3i}{n})}(\frac{3}{n})$

Question

I have the following Riemann sum and I have to convert it to definite integral: $$\lim_{n\to\infty}\sum_{i=1}^{n} (1+\frac{3i}{n})e^{2(1+\frac{3i}{n})}(\frac{3}{n})$$

I know $\Delta x = \frac{3}{n}, so:$

$$\lim_{n\to\infty}\sum_{i=1}^{n} (1+i\Delta x)e^{2(1+i\Delta x)}\Delta x$$

I know that $\Delta x = \frac{b-a}{n}$, and so $\frac{3}{n} = \frac{b-a}{n} \to 3 = b-a$

From $1 + i\Delta x$, we know $a=1$, and so $b=4$. We also know that $x_i = a + i\Delta x$, so:

$$\lim_{n\to\infty}\sum_{i=1}^{n} (x_i)e^{2(x_1)}\Delta x$$

So far in the definite integral we have:

$$\int_{1}^{4} f(x) dx$$

How do I calculate $f(x)$?

Answer 1

As you go to infinity that $\Delta x$ becomes $dx$ and the $x_i$'s become just x's. So this is the definite integral $\int_{1}^{4}xe^{2x}\text{d}x$

Answer 2

Rewrite : $$\lim_{n\to\infty}\sum_{i=1}^{n} \left(1+\frac{3i}{n}\right)e^{2\left(1+\frac{3i}{n}\right)}\left(\frac{3}{n}\right) = \lim_{n\to\infty} \left(\frac{b-a}{n}\right)\sum_{i=1}^{n} f\left(a + i\frac{b-a}{n}\right) $$

where $f : x \longmapsto xe^{2x}$, $a = 1$ and $b = 4$ .

Thus $$\lim_{n\to\infty}\sum_{i=1}^{n} \left(1+\frac{3i}{n}\right)e^{2\left(1+\frac{3i}{n}\right)}\left(\frac{3}{n}\right) = \int_1^4 xe^{2x}dx$$

Answer 3

hint

Your function is $f(x)=xe^{2x}$ cause the Riemann sum here is

$$\sum_{i=1}^{n}f(x_i)\Delta x=\sum_{i=1}^n x_ie^{2x_i}\Delta x$$

to compute the integral, use by parts method.