prove2

1
Prove that âˆ« √(1+x)(1+x³)dt cant exceed
0
√15/8

2 Answers

341

Hari Shankar ·2009-01-08 08:26:50

That is a consequence of the Schwarz Bunyakovsky Inequality

(₀âˆ«¹ f(x) g(x) dx)² â‰¤ ₀âˆ«¹ f²(x) dx ₀âˆ«¹ g²(x) dx

Hence (₀âˆ«¹√(1+x)(1+x³) dx)² â‰¤ ₀âˆ«¹ (1+x) dx ₀âˆ«¹ 1+x³ dx = 15/8

good one dude.. i was thinking this one for a few minutes.. could not get a solution.. :)

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