Let `p(x)` be a function defined on `R` such that `p'(x)=p'(1-x)` for all `x epsilon[0,1],p(0)=1,` and `p(1)=41`.
Then `int_(0)^(1)p(x)dx` is equal to

Options

A. `42`
B. `sqrt(41)`
C. `21`
D. `41`

Explanation

To solve the problem, we need to find the value of the integral \( \int_0^1 p(x) \, dx \) given the conditions on the function \( p(x) \).

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We have the condition \( p'(x) = p'(1-x) \) for all \( x \in [0, 1] \). This implies that the derivative of \( p(x) \) is symmetric about \( x = \frac{1}{2} \).

2. **Integrating the Condition**:
Integrate both sides of the equation \( p'(x) = p'(1-x) \):
\[ \int p'(x) \, dx = \int p'(1-x) \, dx \]
By changing the variable in the right integral, let \( u = 1 - x \) which gives \( du = -dx \):
\[ \int p'(x) \, dx = -\int p'(u) \, du \]
This leads to:
\[ p(x) = -p(1-x) + C \]
where \( C \) is a constant of integration.

3. **Using Boundary Conditions**:
We know \( p(0) = 1 \) and \( p(1) = 41 \). Plugging \( x = 0 \) into the equation:
\[ p(0) = -p(1) + C \implies 1 = -41 + C \implies C = 42 \]
Thus, we have:
\[ p(x) = -p(1-x) + 42 \]

4. **Finding \( p(1-x) \)**:
From the earlier derived equation, we can express \( p(1-x) \):
\[ p(1-x) = -p(x) + 42 \]

5. **Setting Up the Integral**:
Now, we can set up the integral:
\[ I = \int_0^1 p(x) \, dx \]
Using the property of the integral:
\[ I = \int_0^1 p(1-x) \, dx \]
Substitute \( p(1-x) \):
\[ I = \int_0^1 (-p(x) + 42) \, dx \]
This can be separated into two integrals:
\[ I = -\int_0^1 p(x) \, dx + \int_0^1 42 \, dx \]
The second integral evaluates to:
\[ \int_0^1 42 \, dx = 42 \]
Thus, we have:
\[ I = -I + 42 \]

6. **Solving for \( I \)**:
Rearranging gives:
\[ 2I = 42 \implies I = 21 \]

### Final Answer:
\[ \int_0^1 p(x) \, dx = 21 \]