It is known that a > b > 0, and the proof is ea + e-A > EB + E-B

It is known that a > b > 0, and the proof is ea + e-A > EB + E-B


Let f (x) = ex + E-X, method 1: ∧ f (a) = EA + e-A, f (b) = EB + E-B, ∧ f (a) - f (b) = EA + 1ea-eb-1eb = (EA − EB) (EAEB − 1) EAEB, ∧ a > b > 0, ∧ for molecule: EA > EB, EAEB > 1, so the molecule is greater than 0, for denominator: EAEB > 0, ∧ f (a) - f (b) >



Find the area of the figure enclosed by the curve y = e ^ x, y = e ^ - X and the straight line x = 1


Area = ∫ e ^ xdx + ∫ e ^ (- x) DX
=[e^x]│+[-e^(-x)]│
=(1-0)+(1-1/e)
=2-1/e

Elden Ring Premiere

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