Let the straight line x= b divide the area enclosed by `y=(1-x)^(2),y=0, and x=0` into two parts `R_(1)(0lexleb) and R_(2)(blexle1)` such that `R_(1)-R_(2)=(1)/(4).` Then b equals

Options

A. `3//4`
B. `1//2`
C. `1//3`
D. `1//4`

Explanation

To solve the problem, we need to find the value of \( b \) such that the area \( R_1 \) (from \( x = 0 \) to \( x = b \)) minus the area \( R_2 \) (from \( x = b \) to \( x = 1 \)) equals \( \frac{1}{4} \).

### Step-by-step Solution:

1. **Identify the Area Under the Curve**:
The area under the curve \( y = (1 - x)^2 \) from \( x = 0 \) to \( x = 1 \) can be calculated using integration:
\[ A = \int_0^1 (1 - x)^2 \, dx \]

2. **Calculate the Total Area**:
First, we compute the total area:
\[ A = \int_0^1 (1 - x)^2 \, dx = \int_0^1 (1 - 2x + x^2) \, dx \]
\[ = \left[ x - x^2 + \frac{x^3}{3} \right]_0^1 = \left( 1 - 1 + \frac{1}{3} \right) - 0 = \frac{1}{3} \]

3. **Set Up the Areas \( R_1 \) and \( R_2 \)**:
The area \( R_1 \) from \( x = 0 \) to \( x = b \) is:
\[ R_1 = \int_0^b (1 - x)^2 \, dx \]
The area \( R_2 \) from \( x = b \) to \( x = 1 \) is:
\[ R_2 = \int_b^1 (1 - x)^2 \, dx \]

4. **Calculate \( R_1 \)**:
\[ R_1 = \int_0^b (1 - x)^2 \, dx = \left[ x - x^2 + \frac{x^3}{3} \right]_0^b = \left( b - b^2 + \frac{b^3}{3} \right) - 0 = b - b^2 + \frac{b^3}{3} \]

5. **Calculate \( R_2 \)**:
\[ R_2 = \int_b^1 (1 - x)^2 \, dx = \int_0^1 (1 - x)^2 \, dx - \int_0^b (1 - x)^2 \, dx \]
\[ = \frac{1}{3} - R_1 = \frac{1}{3} - \left( b - b^2 + \frac{b^3}{3} \right) \]

6. **Set Up the Equation**:
We are given that \( R_1 - R_2 = \frac{1}{4} \):
\[ R_1 - R_2 = \left( b - b^2 + \frac{b^3}{3} \right) - \left( \frac{1}{3} - (b - b^2 + \frac{b^3}{3}) \right) = \frac{1}{4} \]

7. **Simplify the Equation**:
This simplifies to:
\[ 2(b - b^2 + \frac{b^3}{3}) - \frac{1}{3} = \frac{1}{4} \]
\[ 2b - 2b^2 + \frac{2b^3}{3} - \frac{1}{3} = \frac{1}{4} \]

8. **Combine Terms**:
Rearranging gives:
\[ 2b - 2b^2 + \frac{2b^3}{3} = \frac{1}{4} + \frac{1}{3} \]
\[ = \frac{3 + 4}{12} = \frac{7}{12} \]

9. **Final Equation**:
Multiply through by 12 to eliminate the fraction:
\[ 24b - 24b^2 + 8b^3 = 7 \]
Rearranging gives:
\[ 8b^3 - 24b^2 + 24b - 7 = 0 \]

10. **Solve for \( b \)**:
Using numerical methods or graphing, we find that \( b = \frac{1}{2} \) is a solution.

### Final Answer:
Thus, the value of \( b \) is:
\[ b = \frac{1}{2} \]