高等数学不定积分计算题,谢谢
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发布时间：2022-09-25 19:03
共1个回答
热心网友
时间：2023-09-20 12:25
x^2-2x-3 <0
(x-3)(x+1)<0
-1<x<3
∫(-2->4) |x^2-2x-3| dx
=∫(-2->-1) (x^2-2x-3) dx - ∫(-1->3) (x^2-2x-3) dx + ∫(3->4) (x^2-2x-3) dx
=[(1/3)x^3-x^2-3x]|(-2->-1) -[(1/3)x^3-x^2-3x]|(-1->3) +[(1/3)x^3-x^2-3x]|(3->4)
=[ (-1/3-1+3)-(-8/3-4+6)] -[(9-9-9)-(-8/3-4+6)] +[(64/3-16-12)-(9-9-9)]
=(5/3 +2/3) -(-9-5/3) +(-20/3+9)
=2/3 +2(5/3) +2(9) - 20/3
=2/3 +10/3 +18 -20/3
=4+18 -20/3
=22-20/3
=46/3
(8)
lim(x->0) {∫(0->x) [e^(t^2) -2 ] dt + x }/x^3
洛必达
=lim(x->0) [e^(x^2) -2 + 1 ]/(3x^2)
=lim(x->0) [e^(x^2) - 1 ]/(3x^2)
=lim(x->0) x^2/(3x^2)
=1/3
热心网友
时间：2023-09-20 12:25
x^2-2x-3 <0
(x-3)(x+1)<0
-1<x<3
∫(-2->4) |x^2-2x-3| dx
=∫(-2->-1) (x^2-2x-3) dx - ∫(-1->3) (x^2-2x-3) dx + ∫(3->4) (x^2-2x-3) dx
=[(1/3)x^3-x^2-3x]|(-2->-1) -[(1/3)x^3-x^2-3x]|(-1->3) +[(1/3)x^3-x^2-3x]|(3->4)
=[ (-1/3-1+3)-(-8/3-4+6)] -[(9-9-9)-(-8/3-4+6)] +[(64/3-16-12)-(9-9-9)]
=(5/3 +2/3) -(-9-5/3) +(-20/3+9)
=2/3 +2(5/3) +2(9) - 20/3
=2/3 +10/3 +18 -20/3
=4+18 -20/3
=22-20/3
=46/3
(8)
lim(x->0) {∫(0->x) [e^(t^2) -2 ] dt + x }/x^3
洛必达
=lim(x->0) [e^(x^2) -2 + 1 ]/(3x^2)
=lim(x->0) [e^(x^2) - 1 ]/(3x^2)
=lim(x->0) x^2/(3x^2)
=1/3
