First slide

Evaluation of definite integrals

Question

$let f : R \to R be non-constant differentiable function and satisfies.$ $f (x) = x^{2} - \int_{0}^{1} (x + f (t))^{2} d t . Then f (4) is equal to$

Difficult

A

1
B

2
C

9
D

4

Solution

$\begin{array}{l} Given that f (x) = x^{2} - \int_{0}^{1} (x + f (t))^{2} d t \\ \Rightarrow f (x) = x^{2} - \int_{0}^{1} [x^{2} + f^{2} (t) + 2 x f (t)] d t \\ = x^{2} - x^{2} (t)_{0}^{1} - \int_{0}^{1} f^{2} (t) d t + 2 x \int_{0}^{1} f (t) d t \end{array}$
$\begin{array}{l} \Rightarrow f (x) = - \int_{0}^{1} f^{2} (t) d t - 2 x \int_{0}^{1} f (t) d t \\ ∴ f' (x) = - 2 \int_{0}^{1} f (t) d t \dots \dots . . (i) \end{array}$
= constant
$\Rightarrow$ f must be linear.
$\begin{array}{l} So, let f (x) = a x + b \\ ∴ a = - 2 \int_{0}^{1} (a t + b) d t \\ & b = - \int_{0}^{1} (a t + b)^{2} d t (\sin c e f (0) = a (0) + b = b) \\ \Rightarrow a = 3, b = - 3 \\ \Rightarrow f (x) = 3 (x - 1) \\ So, f (4) = 3 (4 - 1) = 9 \end{array}$

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