The mean value of the function `f(x)= 2/(e^x+1)` in the interval [0,2] is

Options

A. `log.(2)/(e^(2)+1)`
B. `1+log (2)/(e^(2)+1)`
C. `2+log (2)/(e^(2)+1)`
D. `2+log (e^(2)+1)`

Answer

To find the mean value of the function \( f(x) = \frac{2}{e^x + 1} \) in the interval \([0, 2]\), we will follow these steps:

### Step 1: Write the formula for the mean value of a function
The mean value \( M \) of a function \( f(x) \) over the interval \([a, b]\) is given by the formula:
\[ M = \frac{1}{b - a} \int_a^b f(x) \, dx \]
In our case, \( a = 0 \) and \( b = 2 \).

### Step 2: Substitute the function and limits into the formula
Substituting \( f(x) \) and the limits into the formula, we get:
\[ M = \frac{1}{2 - 0} \int_0^2 \frac{2}{e^x + 1} \, dx = \int_0^2 \frac{2}{e^x + 1} \, dx \]

### Step 3: Simplify the integral
We can simplify the integral:
\[ M = 2 \int_0^2 \frac{1}{e^x + 1} \, dx \]

### Step 4: Use substitution to evaluate the integral
Let \( t = e^x \). Then, \( dt = e^x \, dx \) or \( dx = \frac{dt}{t} \).
Now, we need to change the limits:
- When \( x = 0 \), \( t = e^0 = 1 \)
- When \( x = 2 \), \( t = e^2 \)

Thus, the integral becomes:
\[ M = 2 \int_1^{e^2} \frac{1}{t + 1} \cdot \frac{1}{t} \, dt = 2 \int_1^{e^2} \frac{1}{t(t + 1)} \, dt \]

### Step 5: Split the integrand
We can use partial fractions to split the integrand:
\[ \frac{1}{t(t + 1)} = \frac{A}{t} + \frac{B}{t + 1} \]
Multiplying through by \( t(t + 1) \) and solving for \( A \) and \( B \):
\[ 1 = A(t + 1) + Bt \]
Setting \( t = 0 \) gives \( A = 1 \). Setting \( t = -1 \) gives \( B = -1 \). Thus:
\[ \frac{1}{t(t + 1)} = \frac{1}{t} - \frac{1}{t + 1} \]

### Step 6: Integrate
Now, we can integrate:
\[ M = 2 \left( \int_1^{e^2} \left( \frac{1}{t} - \frac{1}{t + 1} \right) dt \right) \]
This gives:
\[ M = 2 \left( \left[ \ln t - \ln(t + 1) \right]_1^{e^2} \right) \]
Calculating the limits:
\[ = 2 \left( \left( \ln(e^2) - \ln(e^2 + 1) \right) - \left( \ln(1) - \ln(2) \right) \right) \]
Since \( \ln(1) = 0 \):
\[ = 2 \left( 2 - \ln(e^2 + 1) + \ln(2) \right) \]

### Step 7: Final simplification
Thus, we have:
\[ M = 2 \left( 2 + \ln(2) - \ln(e^2 + 1) \right) = 4 + 2\ln(2) - 2\ln(e^2 + 1) \]
Using the property of logarithms:
\[ M = 4 + 2\ln\left(\frac{2}{e^2 + 1}\right) \]

### Final Answer
The mean value of the function \( f(x) = \frac{2}{e^x + 1} \) in the interval \([0, 2]\) is:
\[ M = 4 + 2\ln\left(\frac{2}{e^2 + 1}\right) \]