Question

一道积分题目

y=(1-x)^0.5+x在[-1,1]的定积分怎样求

Answer 1

y= (1-x)^0.5+x在[-1,1]的定积分怎样求
原式的不定积分 = ∫[√(1-x) + x] dx
= ∫√(1-x) dx + ∫x dx
= ∫√(1-x) d(1 - x) + ∫x dx
= - 2/3* (1 - x)^(3/2) + x²/2 + C
代入[-1,1] 即可
原式的定积分 = 0 + 2/3* 2^(3/2) = 4√2 / 3

方法2：原定积分 = = ∫[√(1-x) + x] dx {x = - 1→ 1}
= ∫√(1-x) dx + ∫x dx
= 2∫√(1-x) dx + 0 {x = 0 → 1}
= 下略

Answer 2

∫[√(1-x) + x] dx {x = - 1→ 1}
= ∫√(1-x) dx + ∫x dx
= 2∫√(1-x) dx + 0 {x = 0 → 1}
= 下略

Answer 3

我是文盲我不会！等待续集……