The area bounded by the curves y = 1 - cos(pi | vedclass

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(D) The area $A$ is bounded by $y_1 = 1 - \cos(\pi x)$ and $y_2 = -x^2$ between $x = -\frac{1}{2}$ and $x = \frac{1}{2}$.Since both curves are symmetr

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$\frac{13}{12} - \frac{2}{\pi}$

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(D) The area $A$ is bounded by $y_1 = 1 - \cos(\pi x)$ and $y_2 = -x^2$ between $x = -\frac{1}{2}$ and $x = \frac{1}{2}$.Since both curves are symmetric about the $y$-axis,the total area is $2 	imes \int_{0}^{1/2} (y_1 - y_2) dx$.$A = 2 \int_{0}^{1/2} (1 - \cos(\pi x) - (-x^2)) dx$$A = 2 \int_{0}^{1/2} (1 - \cos(\pi x) + x^2) dx$$A = 2 \left[ x - \frac{\sin(\pi x)}{\pi} + \frac{x^3}{3} ight]_{0}^{1/2}$$A = 2 \left[ (\frac{1}{2} - \frac{\sin(\pi/2)}{\pi} + \frac{(1/2)^3}{3}) - (0 - 0 + 0) ight]$$A = 2 \left[ \frac{1}{2} - \frac{1}{\pi} + \frac{1}{24} ight]$$A = 2 \left[ \frac{12 + 1}{24} - \frac{1}{\pi} ight] = 2 \left[ \frac{13}{24} - \frac{1}{\pi} ight]$$A = \frac{13}{12} - \frac{2}{\pi}$.

The area bounded by the curves $y = 1 - \cos(\pi x)$,$y = -x^2$ and the lines $x = \frac{1}{2}$ and $x = -\frac{1}{2}$ is:

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