Question

Let $f(x)$ be a differentiable function satisfying \[ f(x)=e^x+\int_0^1 (y+xe^x)f(y)\,dy \] Find $f(0)+e$, where $e$ is Napier's constant.

Hint:

When integrals involve unknown constants, assume them as constants and solve using consistency conditions.

Answer 1

Correct Answer: 2

Step 1: Express the function in a simplified form

Given,

f(x) = e^x + ∫₀¹ y f(y) dy + x e^x ∫₀^{1 f(y) dy}

Let,

A = ∫₀¹ y f(y) dy,   B = ∫₀¹ f(y) dy

Then,

f(x) = e^x + A + Bx e^x

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Step 2: Evaluate A

A = ∫₀¹ y [ e^y + A + B y e^y ] dy

= ∫₀¹ y e^y dy + A ∫₀¹ y dy + B ∫₀¹ y² e^y dy

∫₀¹ y e^y dy = 1

∫₀¹ y dy = 1/2

∫₀¹ y² e^y dy = e − 2

Hence,

A = 1 + A/2 + B(e − 2)

⇒ A/2 + B(2 − e) = 1  ……(1)

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Step 3: Evaluate B

B = ∫₀¹ [ e^y + A + B y e^y ] dy

= ∫₀¹ e^y dy + A ∫₀¹ dy + B ∫₀¹ y e^y dy

= (e − 1) + A + B

Thus,

A = 1 − e

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Step 4: Find B using equation (1)

Substitute A = 1 − e in (1):

(1 − e)/2 + B(2 − e) = 1

Solving gives,

B = 1

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Step 5: Write the explicit form of f(x)

f(x) = e^x + (1 − e) + x e^x

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Step 6: Evaluate f(0) + e

f(0) = 1 + 1 − e = 2 − e

f(0) + e = 2

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Final Answer:

The value of f(0) + e is  
2