Let f : [0, \infty) → R be a continuous function such that \text{f} \left(\right. \text{x} \left.\right) \textrm{ } = \textrm{ }\text{1}\textrm{ } – \textrm{ }\text{2x}\textrm{ } + \int_{0}^{x} e^{x - t} f \left(\right. t \left.\right) \textrm{ } dt for all x \in [0, \infty). Then, which of the following statement(s) is(are) TRUE?

Engineering

Mathematics

Introduction to Area

Question

Let f : [0, $\infty$ ) → R be a continuous function such that $f (x) = 1 - 2x + \int_{0}^{x} e^{x - t} f (t) dt$ for all x $\in$ [0, $\infty$ ). Then, which of the following statement(s) is(are) TRUE?

The curve y = f(x) passes through the point (1, 2).

The area of the region ${(x, y) \in [0, 1] \times R : f (x) \leq y \leq \sqrt{1 - x^{2}}}^{} is . \frac{π - 2}{4}$

The area of the region ${(x, y) \in [0, 1] \times R : f (x) \leq y \leq \sqrt{1 - x^{2}}}^{} is \frac{π - 1}{4} .$

The curve y = f(x) passes through the point (2, – 1).

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$f (x) = 1 - 2x + e^{x} \int_{0}^{x} e^{- t} \cdot f (t) dt$

$f' (x) = - 2 + e^{x} \int_{0}^{x} e^{- t} \cdot f (t) \cdot dt + e^{x} \cdot e^{- x} \cdot f (x)$

$\frac{dy}{dx} = - 2 + (y + 2x - 1) + y$

$\frac{dy}{dx} - 2y = 2x - 3$

$y \times e^{- 2x} = \int e^{- 2 x} (2 x - 3) dx$

$y \times e^{- 2x} = \frac{e^{- 2 x}}{- 2} (2 x - 3) + \int e^{- 2 x} \cdot dx$

$= \frac{e^{- 2 x}}{- 2} (2 x - 3) - \frac{e^{- 2 x}}{2} + C$

2y = – (2x – 3) – 1 + (2c) × e^2x ( $∵$ f (0) = 1, then c = 0)

y = – x + 1

$Ar = \frac{π}{4} - \frac{1}{2}$