The area bounded by the curves $y = s i n x$ , $y = c o s x$ and the $y$ -axis and the first point of intersection in sq units is

\sqrt{2}

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\sqrt{2} - 1

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2 + \sqrt{2}

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Solution

The correct option is D $\sqrt{2} - 1$
$\begin{matrix} A r e a o f s h a d e d p a r t i o n = \int_{0}^{π / 4} (cos x - sin x) d x = {(sin x + cos x)}_{0}^{π / 4} = (- \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - 1 = \frac{2}{\sqrt{2}} - 1 = (\sqrt{2} - 1) O p t i o n B i s c o r r e c t a n s w e r . \end{matrix}$

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