大一微积分求解答
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F(3)=∫(3,0) f(t) dt
=∫(2,0) f(t) dt+∫(3,2) f(t)dt
=π(2/2)²-π(1/2)²
=3π/4
F(-2)=∫(-2,0) f(t) dt
=-∫(0,-2) f(t)dt
=π(2/2)²
=π
F(-3)=∫(-3,0) f(t)dt
=-∫(0,-3) f(t)dt
=-[∫(0,-2) f(t) dt+∫(-2,-3) f(t)dt
=π(2/2)²-π(1/2)²
=3π/4
F(2)=∫(2,0) f(t)dt
=π
=∫(2,0) f(t) dt+∫(3,2) f(t)dt
=π(2/2)²-π(1/2)²
=3π/4
F(-2)=∫(-2,0) f(t) dt
=-∫(0,-2) f(t)dt
=π(2/2)²
=π
F(-3)=∫(-3,0) f(t)dt
=-∫(0,-3) f(t)dt
=-[∫(0,-2) f(t) dt+∫(-2,-3) f(t)dt
=π(2/2)²-π(1/2)²
=3π/4
F(2)=∫(2,0) f(t)dt
=π
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