Question

求∫⁰₁(3-2x)⁵dx

Answer 1

我们可以使用换元法来求解该定积分。丛森具体来说，我们可以令 u = 3 - 2x，则有 du/dx = -2，从而可以得到 dx = -du/2。将其代入原式得：蚂搭
int_{0}^{1} (3-2x)^5 dx = -(1/2) * int_{3}^{1} u^5 du
= (1/4) * int_{1}^{3} u^5 du
= (1/4) * [(u^6)/6]_{1}^{3}
= (1/4) * [(3^6 - 1^6)/6]
= 121/渗物亩160
因此，所求定积分的值为 int_{0}^{1} (3-2x)^5 dx = 121/160。

Answer 2

∫芦祥(0->1) (3-2x)^5 dx
=(-1/2)∫陪好搏(0->袜闹1) (3-2x)^5 d(3-2x)
=(-1/12) [(3-2x)^6]|(0->1)
=(1/12)( 3^6 - 1)
=728/12
=182/3