Question

高数，设f(x)=∫0→x2 xsint dt，求f(x)″

设f(x)=∫0→x2 xsint dt，求f(x)″
积分后面那是0到x的平方 ，答案是6xsinx^2+4x^3cosx^2

Answer 1

解：因为f(x)=< 0→x²>∫xsintdt，所以
f(x)=-xcost+c|< 0→x²>
=-xcos(x²)+c-(-xcos0+c)
=x-xcos(x²)

所以：f'(x)=1-cos(x²)+2x²sin(x²)
f"(x)=[-cos(x²)]’+[2x²sin(x²)]’
=sin(x²)*2x+[2x²]’sin(x²)+2x²[sin(x²)]’
=2xsin(x²)+4xsin(x²)+2x²cos(x²)[(x²)]’
=2xsin(x²)+4xsin(x²)+4x³cos(x²)
=6xsin(x²)+4x³cos(x²)

Answer 2

解：因为f(x)=∫xsintdt
所以：
f(x)=-xcost+c（0→x2）
=-xcos(x2)+c-(-xcos0+c)
=-xcos(x2)+x
=x[1-cos(x2)]

所以：f'(x)=1-cos(x2)
f"(x)=0。

补充答案：
早说呀！还以为是常数x2呢！还我费了半天劲！
解：因为f(x)=∫xsintdt
所以：
f(x)=-xcost+c（0→x^2）
=-xcos(x^2)+c-(-xcos0+c)
=-xcos(x^2)+x
=x-xcos(x^2)

所以：
f'(x)=1-[cos(x^2)-2(x^2)sin(x^2)]
=1-cos(x^2)+2(x^2)sin(x^2)

f"(x)=[f'(x)]'
=2xsin(x^2)+4xsin(x^2)+4(x^3)cos(x^2)
=6xsin(x^2)+4(x^3)cos(x^2)

Answer 3

此积分是上限x^2的函数,而x^2是x的函数,由复合函数求导链式法则:f'(x)=xsin(x^2)*(2x)=2x^2sinx^2,f"(x)=2(2xsinx^2+x^2(cosx^2)*2x)=4(xsinx^2+x^3cosx^2)