If `lim_(nrarroo)Sigma_(r=1)^(2n)(3r^(2))/(n^(3))e^((r^(3))/(n^(3)))=e^(a)-e^(b)`, then `a+b` is equal to

Answer

To solve the limit

\[ \lim_{n \to \infty} \sum_{r=1}^{2n} \frac{3r^2}{n^3} e^{\frac{r^3}{n^3}}, \]

we will convert the summation into an integral using the properties of limits and Riemann sums.

### Step 1: Rewrite the limit as a Riemann sum

We can express the limit as:

\[ \lim_{n \to \infty} \sum_{r=1}^{2n} \frac{3r^2}{n^3} e^{\frac{r^3}{n^3}} = \lim_{n \to \infty} \sum_{r=1}^{2n} \frac{3}{n} \left(\frac{r}{n}\right)^2 e^{\left(\frac{r}{n}\right)^3}. \]

Here, we notice that as \( n \to \infty \), \( \frac{r}{n} \) approaches a continuous variable \( x \) ranging from \( 0 \) to \( 2 \).

### Step 2: Identify the function and limits

Let \( x = \frac{r}{n} \). Then, as \( r \) goes from \( 1 \) to \( 2n \), \( x \) goes from \( \frac{1}{n} \) to \( 2 \). The sum can be approximated by the integral:

\[ \int_{0}^{2} 3x^2 e^{x^3} \, dx. \]

### Step 3: Evaluate the integral

We will perform a substitution to evaluate the integral. Let \( t = x^3 \), then \( dt = 3x^2 \, dx \) or \( dx = \frac{dt}{3x^2} \). The limits change as follows:

- When \( x = 0 \), \( t = 0^3 = 0 \).
- When \( x = 2 \), \( t = 2^3 = 8 \).

Thus, the integral becomes:

\[ \int_{0}^{8} e^t \, dt. \]

### Step 4: Calculate the integral

The integral of \( e^t \) is simply \( e^t \). Therefore,

\[ \int_{0}^{8} e^t \, dt = e^8 - e^0 = e^8 - 1. \]

### Step 5: Relate to the original expression

From the problem statement, we have:

\[ e^a - e^b = e^8 - 1. \]

This implies:

- \( e^a = e^8 \) (thus \( a = 8 \)),
- \( e^b = 1 \) (thus \( b = 0 \)).

### Step 6: Find \( a + b \)

Now, we can find \( a + b \):

\[ a + b = 8 + 0 = 8. \]

Thus, the final answer is:

\[ \boxed{8}. \]