The value of the integral `I=int_(1)^(2)t^([{t}]+t)(1+ln t)dt` is equal to (`[.] and {.}` denotes the greatest integer and fractional part function respectively)

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the integral

\[ I = \int_{1}^{2} t^{\{t\} + t} (1 + \ln t) \, dt \]

where \(\{t\}\) denotes the fractional part of \(t\) and \([t]\) denotes the greatest integer function, we can proceed as follows:

### Step 1: Understand the functions involved
The fractional part \(\{t\}\) is defined as \(t - [t]\). For \(t\) in the interval \([1, 2)\), we have:
- \([t] = 1\) (the greatest integer less than or equal to \(t\))
- \(\{t\} = t - 1\)

Thus, for \(t\) in \([1, 2)\):
\[ \{t\} + t = (t - 1) + t = 2t - 1 \]

### Step 2: Rewrite the integral
Now we can rewrite the integral \(I\) as:
\[ I = \int_{1}^{2} t^{2t - 1} (1 + \ln t) \, dt \]

### Step 3: Simplify the expression
We can separate the integral into two parts:
\[ I = \int_{1}^{2} t^{2t - 1} \, dt + \int_{1}^{2} t^{2t - 1} \ln t \, dt \]

### Step 4: Change of variable
Let's denote \(x = t^t\). Then, taking the natural logarithm of both sides gives:
\[ \ln x = t \ln t \]
Differentiating both sides:
\[ \frac{1}{x} \frac{dx}{dt} = \ln t + 1 \]
Thus,
\[ dx = x (\ln t + 1) dt \]

### Step 5: Change limits of integration
When \(t = 1\):
\[ x = 1^1 = 1 \]
When \(t = 2\):
\[ x = 2^2 = 4 \]

### Step 6: Substitute into the integral
Now, substituting \(x\) into the integral:
\[ I = \int_{1}^{4} 1 \, dx \]

### Step 7: Evaluate the integral
The integral of \(1\) from \(1\) to \(4\) is simply:
\[ I = x \bigg|_{1}^{4} = 4 - 1 = 3 \]

Thus, the value of the integral \(I\) is:
\[ \boxed{3} \]

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