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  4. If f(x) is defined [-2,2] by f(x)=4x2-3x+1 and g(x)=(f(-x)-f(x)/x2+3), then ∫-22g(x)dx is equal to

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Q. If $ f(x) $ is defined $ [-2,2] $ by $ f(x)=4{{x}^{2}}-3x+1 $ and $ g(x)=\frac{f(-x)-f(x)}{{{x}^{2}}+3}, $ then $ \int_{-2}^{2}{g(x)}dx $ is equal to

JamiaJamia 2008

Solution:

Given that, $ f(x)=4{{x}^{2}}-3x+1,g(x)=\frac{f(-x)-f(x)}{{{x}^{2}}+3} $ $ \therefore $ $ g(x)=\frac{(4{{x}^{2}}+3x+1)-(4{{x}^{2}}-3x+1)}{{{x}^{2}}+3} $ $ =\frac{6x}{{{x}^{2}}+3} $ Now, $ g(-x)=-\frac{6x}{{{x}^{2}}+3} $ $ =-g(x) $ Which is an odd function $ \therefore $ $ \int_{-2}^{2}{g(x)}\,dx=0 $