Question

Find the area bounded by the curve y = cos x, x-axis and the ordinates x = 0 and x = 2π.

Answer 1

\[\text{ The shaded region is the required area bound by the curve }y = \cos x , x\text{ axis and }x = 0 , x = 2\pi\]
\[\text{ Consider a vertical strip of length }= \left| y \right| \text{ and width }= dx \text{ in the first quadrant }\]
\[\text{ Area of the approximating rectangle }= \left| y \right| dx\]
\[\text{ The approximating rectangle moves from }x = 0\text{ to }x = 2\pi\]
\[\text{ Now }, 0 \leq x \leq \frac{\pi}{2}\text{ and }\frac{3\pi}{2} \leq x \leq 2\pi , y > 0 \Rightarrow \left| y \right| = y\]
\[\frac{\pi}{2} \leq x \leq \frac{3\pi}{2}, y < 0 \Rightarrow \left| y \right| = - y\]
\[ \Rightarrow\text{ Area of the shaded region }= \int_0^{2\pi} \left| y \right| dx\]
\[ \Rightarrow A = \int_0^\frac{\pi}{2} \left| y \right| dx + \int_\frac{\pi}{2}^\frac{3\pi}{2} \left| y \right| dx + \int_\frac{3\pi}{2}^{2\pi} \left| y \right| dx\]
\[ \Rightarrow A = \int_0^\frac{\pi}{2} y dx + \int_\frac{\pi}{2}^\frac{3\pi}{2} - y dx + \int_\frac{3\pi}{2}^{2\pi} y dx\]
\[ \Rightarrow A = \int_0^\frac{\pi}{2} \cos x dx + \int_\frac{\pi}{2}^\frac{3\pi}{2} - \cos x dx + \int_\frac{3\pi}{2}^{2\pi} \cos x dx\]
\[ \Rightarrow A = \left[ \sin x \right]_0^\frac{\pi}{2} + \left[ - \sin x \right]_\frac{\pi}{2}^\frac{3\pi}{2} + \left[ - \sin x \right]_\frac{3\pi}{2}^{2\pi} \]
\[ \Rightarrow A = 1 + \left( 1 + 1 \right) + \left( 0 - \left( - 1 \right) \right)\]
\[ \Rightarrow A = 4\text{ sq . units }\]
\[\text{ Area bound by the curve }y = \cos x, x -\text{ axis and }x = 0, x = 2\pi = 4\text{ sq . units }\]