高数,设f(x)=∫0→x2 xsint dt,求f(x)″
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因为f(x)=∫xsintdt,所以
f(x)=-xcost+c|
=-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)
f(x)=-xcost+c|
=-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)
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