JEE Mains · Maths · STD 12 - 8. Application and integration
The area, enclosed by the curves \(y=\sin x+\cos x\) and \(\mathrm{y}=|\cos \mathrm{x}-\sin \mathrm{x}|\) and the lines \(\mathrm{x}=0, \mathrm{x}=\frac{\pi}{2}\) is:
- A \(2 \sqrt{2}(\sqrt{2}-1)\)
- B \(2(\sqrt{2}+1)\)
- C \(4(\sqrt{2}-1)\)
- D \(2 \sqrt{2}(\sqrt{2}+1)\)
Answer & Solution
Correct Answer
(A) \(2 \sqrt{2}(\sqrt{2}-1)\)
Step-by-step Solution
Detailed explanation
\(\mathrm{A}=\int_{0}^{\pi / 2}((\sin \mathrm{x}+\cos \mathrm{x})-|\cos \mathrm{x}-\sin \mathrm{x}|) \mathrm{dx}\) \(\mathrm{A}=\int_{0}^{\pi / 2}((\sin \mathrm{x}+\cos \mathrm{x})-(\cos \mathrm{x}-\sin \mathrm{x})) \mathrm{dx}\)…
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