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f(x)=ex1+ex,I1=f(a)∫f(−a)xg(x(1−x))dx,I2=f(a)∫f(−a)g(x(1−x))dx,thenI2I1= |
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Answer» f(x)=ex1+ex,I1=f(a)∫f(−a)xg(x(1−x))dx,I2=f(a)∫f(−a)g(x(1−x))dx,thenI2I1= |
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