ExamBro
ExamBro
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 ...........

  1. A \(16\)
  2. B \(17\)
  3. C \(18\)
  4. D \(19\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(16\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & y=\min \{\sin x, \cos x\} \\ & x-\text { axis } \quad x-\pi \quad x=\pi\end{aligned}\) \( \int_0^{\pi / 4} \sin x=(\cos x)_{\pi / 4}^0=1-\frac{1}{\sqrt{2}} \) \( \int_{-\pi}^{-3 \pi / 4}(\sin x-\cos x)=(-\cos x-\sin x)_{-\pi}^{-3 \pi / 4} \)…