CUET Preparation Today
The area bounded by the \(y-\) axis, \(y=\cos x\) and \(y=\sin x\) when \(0\leq x\leq \frac{\pi}{2}\) is |
\(2(\sqrt{2}-1)\) \(\sqrt{2}-1\) \(\sqrt{2}+1\) \(\sqrt{2}\) |
\(\sqrt{2}-1\) |
they intersect at $x=\frac{\pi}{4}$ So required area = $\int\limits_{0}^{\frac{\pi}{4}}\cos x-\sin xdx$ $=\left[\sin x+\cos x\right]_{0}^{\frac{\pi}{4}}=(\sqrt{2}-1)$ |
