Choose the correct Answer of the Following Questions : `int_0^e logx dx=`

Options

A. 1
B. e-1
C. e+1
D. 0

Answer

To solve the definite integral \( \int_0^e \log x \, dx \), we will follow these steps:

### Step 1: Set up the integral
We need to evaluate the integral from 0 to \( e \):
\[ \int_0^e \log x \, dx \]

### Step 2: Find the indefinite integral of \( \log x \)
To find the indefinite integral of \( \log x \), we will use integration by parts. We set:
- \( u = \log x \) (thus \( du = \frac{1}{x} \, dx \))
- \( dv = dx \) (thus \( v = x \))

Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
we get:
\[ \int \log x \, dx = x \log x - \int x \cdot \frac{1}{x} \, dx \]
This simplifies to:
\[ \int \log x \, dx = x \log x - \int 1 \, dx = x \log x - x + C \]

### Step 3: Evaluate the definite integral
Now we will evaluate the definite integral using the limits from 0 to \( e \):
\[ \int_0^e \log x \, dx = \left[ x \log x - x \right]_0^e \]

### Step 4: Calculate the upper limit
Substituting the upper limit \( e \):
\[ = \left[ e \log e - e \right] = e \cdot 1 - e = e - e = 0 \]

### Step 5: Calculate the lower limit
Now substituting the lower limit \( 0 \):
\[ = \lim_{x \to 0^+} \left[ x \log x - x \right] \]
As \( x \to 0^+ \), \( \log x \to -\infty \) and \( x \log x \to 0 \). Thus:
\[ \lim_{x \to 0^+} (x \log x) = 0 \]
So, the lower limit evaluates to:
\[ = 0 - 0 = 0 \]

### Step 6: Combine the results
Now, we combine the results from the upper and lower limits:
\[ \int_0^e \log x \, dx = 0 - 0 = 0 \]

### Final Answer
Thus, the value of the definite integral \( \int_0^e \log x \, dx \) is:
\[ \boxed{0} \]

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