Definite Integrals

Hardmathematics

Let $f : R \to R$ be a differentiable function such that $f (x) = x^{2} + \int_{0}^{x} e^{- (x - t)} f (t) d t$ . Then $f (x)$ equals:

Select the correct option:

$x^{3} /3$

$x^{3} /3 + x^{2}$

$x^{2}$

$x^{3} /6$

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About This Question

Subject: mathematics
Chapter: integral calculus
Topic: definite integrals
Difficulty: Hard
Year: 2025

Solution

Correct Answer:

$x^{3} /3 + x^{2}$

Note: The integral term can be simplified by factorization logic using $e^{- (x - t)} = e^{- x} e^{t}$ . We use differentiation.

Method: Differentiate both sides. given $f (x) = x^{2} + \int_{0}^{x} e^{- (x - t)} f (t) d t$ $f (x) = x^{2} + e^{- x} \int_{0}^{x} e^{t} f (t) d t$ . Multiply by $e^{x}$ : $e^{x} f (x) = x^{2} e^{x} + \int_{0}^{x} e^{t} f (t) d t$ . Differentiate wrt $x$ : $e^{x} f^{'} (x) + e^{x} f (x) = (x^{2} e^{x} + 2 x e^{x}) + e^{x} f (x)$ . Cancel $e^{x} f (x)$ : $e^{x} f^{'} (x) = e^{x} (x^{2} + 2 x)$ . divide by $e^{x}$ : $f^{'} (x) = x^{2} + 2 x$ . Integrate: $f (x) = x^{3} /3 + x^{2} + C$ . From original eq, $f (0) = 0$ . So $C = 0$ . $f (x) = x^{3} /3 + x^{2}$ .

This hard difficulty mathematics question is from the chapter integral calculus, covering the topic of definite integrals. It appeared in the 2025 exam.

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