Question

The area bounded by the curves$y=\cos x\text{,}y=\sin x$ between the ordinates $x=0$ and $x=\frac{3\pi }{2}$ is

• A. $(4\sqrt{2}–2)$sq units
• B. $(4\sqrt{2}+2)$sq units
• C. $(4\sqrt{2}-1)$ sq units
• D. $(4\sqrt{2}+1)$sq units

Answer 1

The correct option is A

$(4\sqrt{2}–2)$sq units

Explanation for the correct option:

Area enclosed between two curves:

Finding the area bounded by the curves$y=\cos x\text{,}y=\sin x$ between the ordinates $x=0$ and $x=\frac{3\pi }{2}$ is

Given: $y=\cos x\text{,}y=\sin x$

Required area A is given by
$$\begin{gathered} A={\int }_0^{\frac{3\pi }{2}}\left|\cos x-\sin x\right|dx \\ \Rightarrow A={\int }_0^{\frac{\pi }{4}}\left|\cos x-\sin x\right|dx+{\int }_{\frac{\pi }{4}}^{\frac{5\pi }{4}}\left|\cos x-\sin x\right|dx+{\int }_{\frac{5\pi }{4}}^{\frac{3\pi }{2}}\left|\cos x-\sin x\right|dx \\ \Rightarrow A={\int }_0^{\frac{\pi }{4}}(\cos x-\sin x)dx+{\int }_{\frac{\pi }{4}}^{\frac{5\pi }{4}}(\sin x-\cos x)dx+{\int }_{\frac{5\pi }{4}}^{\frac{3\pi }{2}}(\cos x-\sin x)dx \\ \Rightarrow A={[\sin x+\cos x]}_0^{\frac{\pi }{4}}+{[-\cos x-\sin x]}_{\frac{\pi }{4}}^{\frac{5\pi }{4}}+{[\sin x+\cos x]}_{\frac{5\pi }{4}}^{\frac{3\pi }{2}} \\ \Rightarrow A=(\sqrt{2}-1)+(\sqrt{2}+\sqrt{2})+(-1+\sqrt{2}) \\ \Rightarrow A=4\sqrt{2}-2. \end{gathered}$$

Hence, the correct answer is option (A)