When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude `F = ax + bx^2` where a and b are constants. The work done in stretching the unstretched rubber band by L is :

Options

A. `(aL^2)/2 + (bL^3)/3`
B. `1/2((aL^2)/2 + (bL^3)/3)`
C. `aL^2 + bL^2`
D. `1/2(aL^2 + bL^2)`

Explanation

To find the work done in stretching the unstretched rubber band by a distance \( L \) when the restoring force is given by \( F = ax + bx^2 \), we will follow these steps:

### Step 1: Understand the Work Done Formula
The work done \( W \) in stretching the rubber band can be calculated using the formula:
\[ W = \int F \, dx \]
where \( F \) is the force exerted by the rubber band and \( dx \) is the infinitesimal displacement.

### Step 2: Set Up the Integral
Given that the force \( F \) varies with the distance \( x \), we will integrate the force from the initial position (unstretched length, \( x = 0 \)) to the final position (stretched length, \( x = L \)):
\[ W = \int_0^L (ax + bx^2) \, dx \]

### Step 3: Perform the Integration
We can break the integral into two parts:
\[ W = \int_0^L ax \, dx + \int_0^L bx^2 \, dx \]

Calculating each integral separately:
1. For the first integral:
\[ \int_0^L ax \, dx = a \left[ \frac{x^2}{2} \right]_0^L = a \left( \frac{L^2}{2} - 0 \right) = \frac{aL^2}{2} \]

2. For the second integral:
\[ \int_0^L bx^2 \, dx = b \left[ \frac{x^3}{3} \right]_0^L = b \left( \frac{L^3}{3} - 0 \right) = \frac{bL^3}{3} \]

### Step 4: Combine the Results
Now, we can combine the results of the two integrals to find the total work done:
\[ W = \frac{aL^2}{2} + \frac{bL^3}{3} \]

### Final Answer
Thus, the work done in stretching the unstretched rubber band by \( L \) is:
\[ W = \frac{aL^2}{2} + \frac{bL^3}{3} \]

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