First slide

Evaluation of definite integrals

Question

$Let f (x) be continuous positive function for x \geq 0 and satisfying$ $\int_{0}^{x} f (t) d t = x \sqrt{f (x)} with f (1) = \frac{1}{2} . Then which of the following is/are FALSE$

Difficult

A

$f (x) = \frac{1}{{(1 + (\sqrt{2} + 1) x)}^{2}}$
B

$f (\tan 22 {\frac{1}{2}}^{0}) = \frac{1}{4}$
C

$f (2) = \frac{1}{4}$
D

$Exactly two values of x satisfying f (x) = 1$

Solution

$\begin{array}{l} \int_{0}^{x} f (t) d t = x \sqrt{f (x)} \\ \Rightarrow f (x) = \frac{x \cdot f^{1} (x)}{2 \sqrt{f (x)}} + \sqrt{f (x)} \\ Let \sqrt{f (x)} = y \Rightarrow y^{2} - y = x \cdot \frac{d y}{d x} \\ \Rightarrow \int \frac{d t}{y^{2} - y} = \int \frac{d x}{x} \end{array}$
$\Rightarrow \int_{}^{} (\frac{1}{y - 1} - \frac{1}{y}) d y = \log x c \Rightarrow \log \frac{y - 1}{y} = \log x c \Rightarrow \frac{y - 1}{y} = x c$
$\Rightarrow f (1) = \frac{1}{2} \Rightarrow y = \sqrt{\frac{1}{2}} w h e n x = 1$
$\Rightarrow c = 1 - \sqrt{2}$
$\Rightarrow \frac{y - 1}{y} = (1 - \sqrt{2}) x$
$\Rightarrow f (x) = \frac{1}{(1 + (\sqrt{2} - 1) x)^{2}}$

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