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 problem, we need to evaluate the limit:

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

and express it in the form \( e^a - e^b \) to find \( a + b \).

### Step 1: Rewrite the Sum
We can rewrite the sum as follows:

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

Here, we have factored out \( \frac{1}{n} \) and expressed \( r^2 \) in terms of \( \frac{r}{n} \).

### Step 2: Identify the Integral
As \( n \to \infty \), the sum approaches the integral:

\[ \int_0^2 3x^2 e^{x^3} \, dx \]

where \( x = \frac{r}{n} \) and the limits change from \( \frac{1}{n} \) to \( \frac{2n}{n} = 2 \).

### Step 3: Evaluate the Integral
To evaluate the integral, we can use substitution. Let:

\[ t = e^{x^3} \implies dt = 3x^2 e^{x^3} \, dx \]

Thus, we have:

\[ dx = \frac{dt}{3x^2 e^{x^3}} = \frac{dt}{3t} \]

Now we need to change the limits of integration. When \( x = 0 \), \( t = e^{0^3} = 1 \), and when \( x = 2 \), \( t = e^{2^3} = e^8 \).

The integral becomes:

\[ \int_1^{e^8} dt = e^8 - 1 \]

### Step 4: Relate to the Given Expression
The limit we evaluated is:

\[ \lim_{n \to \infty} \sum_{r=1}^{2n} \frac{3r^2}{n^3} e^{\frac{r^3}{n^3}} = e^8 - 1 \]

We can express this as:

\[ e^8 - e^0 \]

### Step 5: Identify \( a \) and \( b \)
From the expression \( e^a - e^b \), we have:

- \( a = 8 \)
- \( b = 0 \)

### Step 6: Calculate \( a + b \)
Thus, we find:

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

### Final Answer
The value of \( a + b \) is:

\[ \boxed{8} \]