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Step-by-Step Solution
Step 1: Rewrite the given condition as a differential equation
We have the relation:
f'(x) + f(x) = \int_{0}^{2} f(t)\,dt.
Note that the right-hand side is a constant (independent of x ). Let
k = \int_{0}^{2} f(t)\,dt.
Then the equation becomes
f'(x) + f(x) = k.
Step 2: Solve the homogeneous equation
First, consider the homogeneous equation:
f'(x) + f(x) = 0.
Its general solution is
f_h(x) = C \, e^{-x},
where C is an arbitrary constant.
Step 3: Find a particular solution
Because k is a constant, assume a constant particular solution f_p(x) = K_0. Substituting into
f'(x) + f(x) = k yields
0 + K_0 = k \quad \Rightarrow \quad K_0 = k.
Hence, a particular solution is f_p(x) = k.
Step 4: Form the general solution
By combining the homogeneous and particular solutions, the general solution is:
f(x) = C \, e^{-x} + k.
Step 5: Determine the value of k using the integral condition
We know:
k = \int_{0}^{2} f(t)\,dt = \int_{0}^{2} \bigl(C \, e^{-t} + k \bigr)\,dt.
Break it into two integrals:
k = \int_{0}^{2} C \, e^{-t}\,dt \;+\; \int_{0}^{2} k \,dt.
\displaystyle \int_{0}^{2} C \, e^{-t}\,dt = C \Bigl[-e^{-t}\Bigr]_{0}^{2} = C\bigl(-e^{-2}+1\bigr) = C\bigl(1 - e^{-2}\bigr).
\displaystyle \int_{0}^{2} k \,dt = k \,\times (2 - 0) = 2k.
Thus,
k = C \bigl(1 - e^{-2}\bigr) + 2k.
Rearrange to solve for C :
k - 2k = C \bigl(1 - e^{-2}\bigr) \quad \Rightarrow \quad -k = C \bigl(1 - e^{-2}\bigr).
Therefore,
C = -\,\frac{k}{1 - e^{-2}}.
Step 6: Apply the initial condition f(0) = e^{-2}
From f(x) = C\,e^{-x} + k , when x=0 :
f(0) = C \, e^{0} + k = C + k = e^{-2}.
Hence,
C + k = e^{-2}.
Step 7: Solve for k and C
We already have
C = -\dfrac{k}{1 - e^{-2}}.
Substitute this expression into C + k = e^{-2} :
-\frac{k}{1 - e^{-2}} + k = e^{-2}.
Let \alpha = 1 - e^{-2} . Then the equation becomes:
-\frac{k}{\alpha} + k = e^{-2}, \quad \text{so} \quad k\Bigl(1 - \frac{1}{\alpha}\Bigr) = e^{-2}.
Notice
1 - \tfrac{1}{\alpha} = \tfrac{\alpha - 1}{\alpha} = \tfrac{(1 - e^{-2}) - 1}{\alpha} = \tfrac{-\,e^{-2}}{\alpha}.
Hence,
k \,\times \bigl(-\tfrac{e^{-2}}{\alpha}\bigr) = e^{-2} \quad \Rightarrow \quad k = -\,e^{-2} \,\frac{\alpha}{e^{-2}} = -\,\bigl(1 - e^{-2}\bigr).
Thus,
k = e^{-2} - 1.
From C + k = e^{-2} , we get
C = e^{-2} - k = e^{-2} - (e^{-2} - 1) = 1.
Step 8: Final form of the function f(x)
Hence,
f(x) = C\,e^{-x} + k = 1 \cdot e^{-x} + \bigl(e^{-2} - 1 \bigr) = e^{-x} + e^{-2} - 1.
Step 9: Calculate 2f(0) - f(2)
Compute f(0) :
f(0) = e^{-0} + e^{-2} - 1 = 1 + e^{-2} - 1 = e^{-2}.
Compute f(2) :
f(2) = e^{-2} + e^{-2} - 1 = 2e^{-2} - 1.
Therefore,
2f(0) - f(2) = 2\bigl(e^{-2}\bigr) - \bigl(2e^{-2} - 1\bigr) = 2e^{-2} - 2e^{-2} + 1 = 1.
So, the required value is
1.
