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  • Let A1 be the area of the region bounded by the curves  and y-axis in the first quadrant. AAlso, le

Let A1 be the area of the region bounded by the curves y= \sin x, y= \cos x and y-axis in the first quadrant. AAlso, let A2 br the area of the region bounded by the curves y= \sin x, y= \cos x x-axis and x =\frac{\pi}{2} in the first quadrant. Then,
Option: 1 2A_1=A_2\text{ and } A_1 +A_2=1 +\sqrt{2}
Option: 2 A_{1}:A_2=1:2\text{ and }A_{1}+A_2{}=1
Option: 3 A_1:A_2=1:\sqrt{2}\text{ and }A_1+A_2=1
Option: 4 A_1=A_2\text{ and } A_1 +A_{2}=\sqrt{2}

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\\\mathrm{A}_{1}=\int_{0}^{\pi / 4}(\cos \mathrm{x}-\sin \mathrm{x}) \mathrm{d} \mathrm{x} \\ \mathrm{A}_{1}=(\sin \mathrm{x}+\cos \mathrm{x})_{0}^{\pi / 4}=\sqrt{2}-1 \\ \mathrm{~A}_{2}=\int_{0}^{\pi / 4} \sin \mathrm{x} \mathrm{d} \mathrm{x}+\int_{\pi / 4}^{\pi / 2} \cos \mathrm{x} \mathrm{d} \mathrm{x}

\\=(-\cos \mathrm{x})_{0}^{\pi / 4}+(\sin \mathrm{x})_{\pi / 4}^{\pi / 2} \\ \mathrm{~A}_{2}=\sqrt{2}(\sqrt{2}-1) \\ \mathrm{A}_{1}: \mathrm{A}_{2}=1: \sqrt{2}, \mathrm{~A}_{1}+\mathrm{A}_{2}=1

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himanshu.meshram

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