Evaluate : `int_(0)^(1)a^(2-3x)dx` , as limit of sums .

Answer

To evaluate the integral \( \int_{0}^{1} a^{2-3x} \, dx \) as a limit of sums, we will follow these steps:

### Step 1: Define the integral as a limit of sums
The integral can be expressed as:
\[ \int_{0}^{1} a^{2-3x} \, dx = \lim_{n \to \infty} \sum_{i=0}^{n-1} f(x_i) \Delta x \]
where \( \Delta x = \frac{b-a}{n} = \frac{1-0}{n} = \frac{1}{n} \) and \( x_i = 0 + i \Delta x = \frac{i}{n} \).

### Step 2: Substitute into the function
The function \( f(x) = a^{2-3x} \). Therefore, we can write:
\[ f(x_i) = a^{2 - 3\left(\frac{i}{n}\right)} = a^{2 - \frac{3i}{n}} \]

### Step 3: Write the Riemann sum
The Riemann sum becomes:
\[ \sum_{i=0}^{n-1} f(x_i) \Delta x = \sum_{i=0}^{n-1} a^{2 - \frac{3i}{n}} \cdot \frac{1}{n} \]

### Step 4: Simplify the sum
This can be rewritten as:
\[ \frac{1}{n} \sum_{i=0}^{n-1} a^{2 - \frac{3i}{n}} = \frac{1}{n} \sum_{i=0}^{n-1} a^2 \cdot a^{-\frac{3i}{n}} = a^2 \cdot \frac{1}{n} \sum_{i=0}^{n-1} \left(a^{-\frac{3}{n}}\right)^i \]

### Step 5: Recognize the geometric series
The sum \( \sum_{i=0}^{n-1} \left(a^{-\frac{3}{n}}\right)^i \) is a geometric series with first term 1 and common ratio \( a^{-\frac{3}{n}} \):
\[ \sum_{i=0}^{n-1} \left(a^{-\frac{3}{n}}\right)^i = \frac{1 - \left(a^{-\frac{3}{n}}\right)^n}{1 - a^{-\frac{3}{n}}} \]

### Step 6: Evaluate the limit
As \( n \to \infty \), \( a^{-\frac{3}{n}} \to 1 \) and \( \left(a^{-\frac{3}{n}}\right)^n \to a^{-3} \):
\[ \frac{1 - a^{-3}}{1 - a^{-\frac{3}{n}}} \to \frac{1 - a^{-3}}{1 - 1} \text{ (which is indeterminate)} \]
Using L'Hôpital's Rule or recognizing the limit gives:
\[ \lim_{n \to \infty} \frac{1 - a^{-3}}{-\frac{3}{n} a^{-\frac{3}{n}}} = \frac{1 - a^{-3}}{-\frac{3}{n}} \to \frac{n(1 - a^{-3})}{-3} \to -\frac{1 - a^{-3}}{3} \]

### Final Step: Combine results
Thus, the integral evaluates to:
\[ \int_{0}^{1} a^{2-3x} \, dx = a^2 \cdot \left(-\frac{1 - a^{-3}}{3}\right) = \frac{a^2}{3} (1 - a^{-3}) = \frac{a^2 - a^{-1}}{3} \]

### Final Answer
\[ \int_{0}^{1} a^{2-3x} \, dx = \frac{a^2 - a^{-1}}{3} \]