Question #4194

Prove that integral(bottom=a, top=b) (x[sup]2[/sup]+1)e[sup]-x^2[/sup]& dx & >= & (e[sup]-a^2& [/sup]- e[sup]-b^2[/sup])

Expert's answer

Denote x_1 = 1/2

x_2 = -1/4

x_3 = 1/8

x_4 = -1/16

Then

x_{1} = 1/2 = 1/(2^{1}) = (-1)^{1+1}/(2^{1})x_{2} = -1/4 = -1/(2^{2}) = (-1)^{2+1}/(2^{2})x_{3} = 1/8 = 1/(2^{3}) = (-1)^{3+1}/(2^{3})x_{4} = 1/16 = 1/(2^{4}) = (-1)^{4+1}/(2^{4})Thus the general formula for x_n is the following

x_{n} = -1^{n+1}/ 2^{n}

x_2 = -1/4

x_3 = 1/8

x_4 = -1/16

Then

x

x

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