Find the value of : `int_0^(10)e^(2x-[2x])d(x-[x])w h e r e[dot]` denotes the greatest integer function).

Explanation

To solve the integral

\[ I = \int_0^{10} e^{2x - [2x]} d(x - [x]) \]

where \([x]\) denotes the greatest integer function, we can follow these steps:

### Step 1: Understanding the Integral
The expression \(d(x - [x])\) represents the differential of the fractional part of \(x\), which is \(x - [x]\). The fractional part of \(x\) is periodic with a period of 1.

### Step 2: Splitting the Integral
Since \(x\) ranges from 0 to 10, we can split the integral into intervals of length 1:

\[ I = \sum_{n=0}^{9} \int_n^{n+1} e^{2x - [2x]} d(x - [x]) \]

In each interval \([n, n+1)\), \([2x] = 2n\) for \(x \in [n, n+1)\) when \(n\) is an integer. Thus, we can rewrite \(I\) as:

\[ I = \sum_{n=0}^{9} \int_n^{n+1} e^{2x - 2n} d(x - [x]) \]

### Step 3: Evaluating the Integral
The differential \(d(x - [x])\) is equal to \(dx\) since \(x - [x]\) is differentiable in each interval. Therefore, we can write:

\[ I = \sum_{n=0}^{9} \int_n^{n+1} e^{2x - 2n} dx \]

### Step 4: Simplifying the Integral
Now, we can simplify the integral:

\[ I = \sum_{n=0}^{9} e^{-2n} \int_n^{n+1} e^{2x} dx \]

The integral \(\int e^{2x} dx\) is:

\[ \int e^{2x} dx = \frac{1}{2} e^{2x} + C \]

Thus, we evaluate:

\[ \int_n^{n+1} e^{2x} dx = \left[ \frac{1}{2} e^{2x} \right]_n^{n+1} = \frac{1}{2} (e^{2(n+1)} - e^{2n}) = \frac{1}{2} e^{2n} (e^2 - 1) \]

### Step 5: Putting It All Together
Substituting this back into our expression for \(I\):

\[ I = \sum_{n=0}^{9} e^{-2n} \cdot \frac{1}{2} e^{2n} (e^2 - 1) \]

This simplifies to:

\[ I = \frac{(e^2 - 1)}{2} \sum_{n=0}^{9} 1 = \frac{(e^2 - 1)}{2} \cdot 10 \]

Thus, we have:

\[ I = 5(e^2 - 1) \]

### Final Answer
So the final value of the integral is:

\[ I = 5(e^2 - 1) \]

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