∫(-π/2→π/2) (cos²2x + 8) dx
= ∫(0→π/2) (1 + cos4x) dx + 8∫(-π/2→π/2) dx
= (x + 1/4*sin4x) |(0→π/2) + 8π
= π/2 + 8π
= 17π/2
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x + 3 = 0,x∈[-4,0]
=> x = -3
x∈[-4,-3],x + 3 < 0
x∈[-3,0],x + 3 > 0
∫(-4→0) |x + 3| dx
= ∫(-4→-3) -(x + 3) dx + ∫(-3→0) (x + 3) dx
= 1/2 + 9/2
= 5您好,第二題為什么我前面與您相同,答案卻是四,您可以寫一下∫(-4→-3) -(x + 3) dx + ∫(-3→0) (x + 3) dx后面那步嗎?∫(-4→-3) (- x - 3) dx= -(x²/2 + 3x) |(-4→-3)= - { [(- 3)²/2 + 3(- 3)] - [(- 4)²/2 + 3(- 4)] }= - [ (9/2 - 9) - (8 - 12) ]= - [ - 9/2 - (- 4)] = - (- 1/2) = 0.5∫(-3→0) (x + 3) dx= (x²/2 + 3x) |(-3→0)= 0 - [ (-3)²/2 + 3(-3)]= - (9/2 - 9) = - (- 4.5) = 4.5