JEE MAIN - Mathematics (2023 - 13th April Morning Shift - No. 1)
The area of the region enclosed by the curve
$$f(x)=\max \{\sin x, \cos x\},-\pi \leq x \leq \pi$$ and the $$x$$-axis is
$$2 \sqrt{2}(\sqrt{2}+1)$$
4
$$2(\sqrt{2}+1)$$
$$4(\sqrt{2})$$
Explanation
_13th_April_Morning_Shift_en_1_1.png)
Area =
= $$ \begin{aligned} &\left|\int_{-\pi}^{-3 \pi / 4} \sin x d x\right|+\left|\int_{\pi / 4}^\pi \sin x d x\right|+\left|\int_{-3 \pi / 4}^{-\pi / 2} \cos x d x\right| +\left|\int_{-\pi / 2}^{\pi / 4} \cos x d x\right| \end{aligned} $$
$$ \begin{aligned} & =\left|\frac{1}{\sqrt{2}}-1\right|+\left|1+\frac{1}{\sqrt{2}}\right|+\left|-1+\frac{1}{\sqrt{2}}\right|+\left|\frac{1}{\sqrt{2}}+1\right| \\\\ & = 2+\frac{2}{\sqrt{2}}+2-\frac{2}{\sqrt{2}}=4 \end{aligned} $$
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