还是高数下,还是感谢大神们给出过程(≧▽≦)
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/07/15 20:07:14
还是高数下,还是感谢大神们给出过程(≧▽≦)
![](http://img.wesiedu.com/upload/d/bd/dbd8011236fa00ddb5e9ae6bb3385a3f.jpg)
![](http://img.wesiedu.com/upload/d/bd/dbd8011236fa00ddb5e9ae6bb3385a3f.jpg)
![还是高数下,还是感谢大神们给出过程(≧▽≦)](/uploads/image/z/18622264-40-4.jpg?t=%E8%BF%98%E6%98%AF%E9%AB%98%E6%95%B0%E4%B8%8B%2C%E8%BF%98%E6%98%AF%E6%84%9F%E8%B0%A2%E5%A4%A7%E7%A5%9E%E4%BB%AC%E7%BB%99%E5%87%BA%E8%BF%87%E7%A8%8B%28%E2%89%A7%E2%96%BD%E2%89%A6%29)
记f(x)在 下限a上限b 的积分为∫[a.b] f(x)dx
∫[0,x] f(t)dt = x+∫[0,x] t f(x-t)dt
换元令m =x -t
∫[0,x] f(t)dt = x - ∫[x,0] (x-m )f(m)dm
∫[0,x] f(t)dt = x +∫[0,x] (x-m )f(m)dm
∫[0,x] f(t)dt = x +x∫[0,x] f(m)dm +∫[0,x] (-m) f(m)dm
求导
f(x)= 1 +∫[0,x] f(m)dm + xf(x) -xf(x)
f(x)= 1 +∫[0,x] f(m)dm
f(x)可微
f '(x) = f(x)
f(0) =1
所以f(x) 应该是e^x
f '(x) = e^x
再问: 谢谢你了!
∫[0,x] f(t)dt = x+∫[0,x] t f(x-t)dt
换元令m =x -t
∫[0,x] f(t)dt = x - ∫[x,0] (x-m )f(m)dm
∫[0,x] f(t)dt = x +∫[0,x] (x-m )f(m)dm
∫[0,x] f(t)dt = x +x∫[0,x] f(m)dm +∫[0,x] (-m) f(m)dm
求导
f(x)= 1 +∫[0,x] f(m)dm + xf(x) -xf(x)
f(x)= 1 +∫[0,x] f(m)dm
f(x)可微
f '(x) = f(x)
f(0) =1
所以f(x) 应该是e^x
f '(x) = e^x
再问: 谢谢你了!