A) 0
B) \[2\int_{0}^{1}{\frac{\sin x}{3-|x|}\,dx}\]
C) \[2\int_{0}^{1}{\frac{-{{x}^{2}}}{3-|x|}}\,dx\]
D) \[2\int_{0}^{1}{\frac{\sin x-{{x}^{2}}}{3-|x|}\,dx}\]
Correct Answer: C
Solution :
\[I=\int_{-1}^{1}{\,\frac{\sin x-{{x}^{2}}}{3\,\,-|x|}\,}dx=\int_{-1}^{1}{\,\frac{\sin x}{3-|x|}}\,dx-\int_{-1}^{1}{\,\frac{{{x}^{2}}}{3-|x|}}\,dx\] Here, \[f(x)=\frac{\sin x}{3-|x|}\] is an odd function but \[f(x)=\frac{{{x}^{2}}}{3-|x|}\] is an even function \[\therefore \,\,I=-\int_{-1}^{1}{\frac{{{x}^{2}}}{3-|x|}\,}dx=-2\int_{0}^{1}{\frac{{{x}^{2}}}{3-|x|}}\,dx\]\[=2\int_{0}^{1}{\frac{-{{x}^{2}}}{3-|x|}\,}dx\].
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