Let I be the purchase value of an equipment and V(t) be the value after it has been used for t years. The value V(t) depreciates at a rate given by differential equation `(d V(t)/(dt)=-k(T-t)` , where `k"">""0` is a constant and T is the total life in years of the equipment. Then the scrap value V(T) of the equipment is :
(1) `T^2-1/k`
(2) `I-(k T^2)/2`
(3) `I-(k(T-t)^2)/2`
(4) `e^(-k T)`

Options

A. `e^(-kT)`
B. `T^(2)-I/k`
C. `I-(kT^(2))/2`
D. `I-(k(T-t)^(2))/(2)`

Explanation

To solve the problem, we start with the given differential equation that describes the depreciation of the value of the equipment:

\[ \frac{dV(t)}{dt} = -k(T - t) \]

where \( k > 0 \) is a constant, \( T \) is the total life of the equipment in years, and \( I \) is the initial purchase value of the equipment.

### Step 1: Rearranging the Differential Equation

We can rearrange the equation to isolate \( dV(t) \):

\[ dV(t) = -k(T - t) dt \]

### Step 2: Integrating Both Sides

Next, we integrate both sides. We will integrate from \( t = 0 \) to \( t = T \) for the left side and from \( V(0) = I \) to \( V(T) = V(T) \) for the right side:

\[ \int_{I}^{V(T)} dV(t) = \int_{0}^{T} -k(T - t) dt \]

### Step 3: Solving the Left Side

The left side integrates to:

\[ V(T) - I \]

### Step 4: Solving the Right Side

Now, we need to solve the right side:

\[ \int_{0}^{T} -k(T - t) dt \]

This can be simplified as follows:

\[ -k \int_{0}^{T} (T - t) dt \]

Calculating the integral:

\[ \int (T - t) dt = Tt - \frac{t^2}{2} \]

Evaluating from \( 0 \) to \( T \):

\[ \left[ Tt - \frac{t^2}{2} \right]_{0}^{T} = \left( T \cdot T - \frac{T^2}{2} \right) - \left( 0 - 0 \right) = T^2 - \frac{T^2}{2} = \frac{T^2}{2} \]

Thus, the right side becomes:

\[ -k \cdot \frac{T^2}{2} \]

### Step 5: Setting the Two Integrals Equal

Now we set the two results equal to each other:

\[ V(T) - I = -k \cdot \frac{T^2}{2} \]

### Step 6: Solving for \( V(T) \)

Rearranging gives us:

\[ V(T) = I - \frac{k T^2}{2} \]

### Conclusion

The scrap value \( V(T) \) of the equipment after \( T \) years is:

\[ V(T) = I - \frac{k T^2}{2} \]

Thus, the correct answer is option (2):

\[ \text{(2) } I - \frac{k T^2}{2} \]