JEE Mains · Maths · STD 12 - 8. Application and integration
Let the area of the region enclosed by the curve \(\mathrm{y}=\min \{\sin \mathrm{x}, \cos \mathrm{x}\}\) and the \(\mathrm{x}\)-axis between \(\mathrm{x}=-\pi\) to \(\mathrm{x}=\pi\) be \(\mathrm{A}\). Then \(\mathrm{A}^2\) is equal to ...........
- A \(16\)
- B \(17\)
- C \(18\)
- D \(19\)
Answer & Solution
Correct Answer
(A) \(16\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & y=\min \{\sin x, \cos x\} \\ & x-\text { axis } \quad x-\pi \quad x=\pi\end{aligned}\) \( \int_0^{\pi / 4} \sin x=(\cos x)_{\pi / 4}^0=1-\frac{1}{\sqrt{2}} \) \( \int_{-\pi}^{-3 \pi / 4}(\sin x-\cos x)=(-\cos x-\sin x)_{-\pi}^{-3 \pi / 4} \)…
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