First slide

Evaluation of definite integrals

Question

Let $f (x) = \int_{0}^{x} \sqrt{6 - u^{2}} d u .$ Then the real roots of the equation
$x^{2} - f^{'} (x) = 0$ are

Moderate

A

$x = \pm \sqrt{6}$
B

$x = \pm \sqrt{3}$
C

$x = \pm \sqrt{2}$
D

$x = \pm 1$

Solution

Since $f^{'} (x) = \sqrt{6 - x^{2}},$ the equation $x^{2} - f^{'} (x) = 0$ becomes
$\begin{array}{l} x^{2} - \sqrt{6 - x^{2}} = 0 \Rightarrow x^{4} + x^{2} - 6 = 0 \\ \Rightarrow (x^{2} - 2) (x^{2} + 3) = 0 \Rightarrow x^{2} - 2 = 0 (x^{2} + 3 \neq 0) \\ \Rightarrow x = \pm \sqrt{2} . \end{array}$

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