设f(x)=x^2 (x≤0) f(x)=cosx-1 (x>0) 试求∫ (上π/2下-1)f(X)dx
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设f(x)=x^2 (x≤0) f(x)=cosx-1 (x>0) 试求∫ (上π/2下-1)f(X)dx
f(x) = { x²,x ≤ 0
{ cosx - 1,x > 0
∫(-1→π/2) f(x) dx
= ∫(-1→0) f(x) dx + ∫(0→π/2) f(x) dx
= ∫(-1→0) x² dx + ∫(0→π/2) (cosx - 1) dx
= x³/3:(-1→0) + (sinx - x):(0→π/2)
= [0 - (- 1/3)] + [(1 - π/2) - 0]
= 1/3 + 1 - π/2
= 4/3 - π/2
{ cosx - 1,x > 0
∫(-1→π/2) f(x) dx
= ∫(-1→0) f(x) dx + ∫(0→π/2) f(x) dx
= ∫(-1→0) x² dx + ∫(0→π/2) (cosx - 1) dx
= x³/3:(-1→0) + (sinx - x):(0→π/2)
= [0 - (- 1/3)] + [(1 - π/2) - 0]
= 1/3 + 1 - π/2
= 4/3 - π/2
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