if `f(x)` is a differential function such that `f(x)=int_(0)^(x)(1+2xf(t))dt&f(1)=e`, then
Q. `int_(0)^(1)f(x)dx=`

Options

A. `(e-1)/(2)`
B. is less that 1
C. `(e+1)/(2)`
D. is greater than 1

Explanation

To solve the problem, we need to find the value of the integral \( \int_0^1 f(x) \, dx \) given that \( f(x) = \int_0^x (1 + 2x f(t)) \, dt \) and \( f(1) = e \).

### Step-by-Step Solution:

1. **Start with the given function:**
\[ f(x) = \int_0^x (1 + 2x f(t)) \, dt \]

2. **Differentiate both sides with respect to \( x \):**
Using the Leibniz rule for differentiation under the integral sign, we have:
\[ f'(x) = \frac{d}{dx} \left( \int_0^x (1 + 2x f(t)) \, dt \right) = 1 + 2x f(x) \]

3. **Rearranging the equation:**
We can rearrange the equation to isolate \( f'(x) \):
\[ f'(x) - 2x f(x) = 1 \]

4. **This is a first-order linear differential equation:**
We can solve it using an integrating factor. The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{-\int 2x \, dx} = e^{-x^2} \]

5. **Multiply through by the integrating factor:**
\[ e^{-x^2} f'(x) - 2x e^{-x^2} f(x) = e^{-x^2} \]

6. **Recognize the left-hand side as a derivative:**
\[ \frac{d}{dx}(e^{-x^2} f(x)) = e^{-x^2} \]

7. **Integrate both sides:**
\[ e^{-x^2} f(x) = \int e^{-x^2} \, dx + C \]

8. **Solve for \( f(x) \):**
\[ f(x) = e^{x^2} \left( \int e^{-x^2} \, dx + C \right) \]

9. **Use the condition \( f(1) = e \) to find \( C \):**
We know \( f(1) = e \), so we substitute \( x = 1 \):
\[ e^{1^2} \left( \int_0^1 e^{-t^2} \, dt + C \right) = e \]
This simplifies to:
\[ \int_0^1 e^{-t^2} \, dt + C = 1 \]
Thus, \( C = 1 - \int_0^1 e^{-t^2} \, dt \).

10. **Substituting back to find \( f(x) \):**
\[ f(x) = e^{x^2} \left( \int e^{-t^2} \, dt + 1 - \int_0^1 e^{-t^2} \, dt \right) \]

11. **Now calculate \( \int_0^1 f(x) \, dx \):**
\[ \int_0^1 f(x) \, dx = \int_0^1 x e^{x^2} \, dx \]
Let \( u = x^2 \), then \( du = 2x \, dx \) or \( dx = \frac{du}{2x} \):
\[ \int_0^1 x e^{x^2} \, dx = \frac{1}{2} \int_0^1 e^u \, du = \frac{1}{2} (e - 1) \]

12. **Final answer:**
\[ \int_0^1 f(x) \, dx = \frac{1}{2} (e - 1) \]