First slide

Evaluation of definite integrals

Question

Let $f (x) = \int_{1}^{x} \sqrt{2 - t^{2}} d t .$ Then the real roots of the equation $x^{2} - f^{'} (x) = 0$ are

Easy

A

± 1
B

$\pm 1 / \sqrt{2}$
C

± 1/2
D

0 and 1

Solution

Using Property 18, we have $f^{'} (x) = \sqrt{2 - x^{2}}$
and thus the given equation reduces to $x^{2} - \sqrt{2 - x^{2}} = 0$
$\Rightarrow (x^{2} + 2) (x^{2} - 1) = 0 .$ Thus the real roots are given by $x = \pm 1 .$

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