Cho Hàm Số Fx Có F0=0 Và F'x=cosx. Cos22x,∀x∈ℝ . Khi đó ∫0πfxdx ...
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Cho hàm số fx có f0=0 và f'x=cosx. cos22x,∀x∈ℝ . Khi đó ∫0πfxdx bằng A.1042225 . B.208225 . C.242225 . D.149225 . Đáp án và lời giải Đáp án:C Lời giải:Lời giải Chọn C Ta có f'x=cosx. cos22x,∀x∈ℝ nên fx là một nguyên hàm của f'x . Có ∫f'xdx=∫cosx. cos22xdx=∫cosx. 1+cos4x2dx=∫cosx2dx+∫cosx. cos4x2dx =12∫cosxdx+14∫cos5x+cos3xdx=12sinx+120sin5x+112sin3x+C . Suy ra fx=12sinx+120sin5x+112sin3x+C,∀x∈ℝ . Mà f0=0⇒C=0 . Do đó fx=12sinx+120sin5x+112sin3x,∀x∈ℝ . Khi đó: ∫0πfxdx=∫0π12sinx+120sin5x+112sin3xdx=−12cosx−1100cos5x−136cos3x0π=242225 .Bạn có muốn?
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