A) \[\frac{1}{4}\]
B) \[\frac{1}{2}\]
C) 2
D) 4
Correct Answer: D
Solution :
We know that ac = (semi-minor axis)2 = 4 Now \[\int\limits_{-4}^{4}{\{2x\}\,dx=}16\,\int\limits_{0}^{1/2}{\{2x\}\,dx\,=\int\limits_{-8\times \frac{1}{2}}^{8\times \frac{1}{2}}{\{2x\}\,dx}}\] \[=16\int\limits_{0}^{1/2}{\{2x\}dx\,=16}\,\int\limits_{0}^{1/2}{2x\,dx=32}\,\int\limits_{0}^{1/2}{x\,dx}\] \[=32\left( \frac{{{x}^{2}}}{2} \right)_{0}^{\frac{1}{2}}\,=32\times \frac{1}{8}=4\]
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