What is the integration of (2x^3)sqrt(4-x^4) dx, u = 4-x^4?

👨🏻‍🏫 學術台數學

MandM

2011-10-20 3:31 am

What is the integration of (2x^3)sqrt(4-x^4) dx, u = 4-x^4?

i don't get why the answer starts with (-1/3)?

How?

更新1:

i dont get the second step

更新2:

du/-2?

更新3:

du/-2?

更新4:

please also explain -1/2 * [2/3 * u^(3/2)] + C -1/3*(4 - x^4)^(3/2) + C

回答 (2)

[匿名]

2011-10-20 3:37 am

✔ 最佳答案

-1/3 (4-x^4)^(3/2) + C

Edit: actually its du/-4 and then you divide by two and it becomes du/-2.

it come from U = 4-x^4, then du - -4x^3, dx = -1/4du

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[匿名]

2011-10-20 3:37 am

∫(2x^3)*√(4 - x^4) dx

u = 4 - x^4

du/-2 = 2x^3 dx

-1/2*∫√u du

-1/2 * [2/3 * u^(3/2)] + C

-1/3*(4 - x^4)^(3/2) + C

Does that perhaps answer your question?

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