Question

定积分大学数学？

第一第二题 解析 谢谢

Answer 1

(1)

∫(0->x) f(t) dt = 2x^3

f(x) = 6x^2

∫(0->π/2) cosx. f(-sinx) dx

=∫(0->π/2) cosx. 6(-sinx)^2 dx

=6∫(0->π/2) (sinx)^2 dsinx

=2[(sinx)^3]|(0->π/2)

=2

ans : C

(2)

let

u=-x

du=-dx

x=-t, u=t

x=t, u=-t

∫(-t->t) f(-x) dx

=∫(t->-t) f(u) (-du)

=∫(-t->t) f(u) du

=∫(-t->t) f(x) dx

ans : C