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

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Solution

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

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