定积分问题.1、∫(cos2x)^8dx积分上限为π/4,下限为0.2、设f（x）在a,b上连续,且∫f(x)dx=1(积分上限为b,下限为a.)求∫f(a+b-x)dx(积分上限为b,下限为a)

Question

定积分问题.
1、∫(cos2x)^8dx 积分上限为π/4,下限为0.
2、设f（x）在【a,b】上连续,且∫f(x)dx=1 (积分上限为b,下限为a.) 求∫f(a+b-x)dx (积分上限为b,下限为a)

Answer 1

1.
令y = 2x、dy = 2 dx
∫(0→π/4) cos⁸2x dx
= (1/2)∫(0→π/2) cos⁸y dy、根据Wallis公式、这个你自己去查查吧
= 1/2 * 7!/8! * π/2
= 1/2 * (7 * 5 * 3 * 1)/(8 * 6 * 4 * 2) * π/2
= 35π/512
2.
令y = a + b - x、dy = - dx
∫(a→b) f(a + b - x) dx
= ∫(b→a) f(y) * (- dy)
= ∫(a→b) f(y) dy
= ∫(a→b) f(x) dx = 1