Question

高数,定积分1.设f(x)是连续函数,且∫f(t)dt=x,求f(x)2.设f(x)=∫arctan(1+t²？

Answer 1

1.∵∫f(t)dt=x
==>(x³-1)'*f(x³-1)=1 (根据参数积分求导公式,对等式两端求导)
==>3x²f(x³-1)=1
==>f(x³-1)=1/(3x²)
令t=x³-1,则x=(t+1)^(1/3)
∴f(t)=f(x³-1)=1/(3x²)=(1/3)(t+1)^(-2/3)
故f(x)=(1/3)(x+1)^(-2/3)；
2.∵f(x)=∫arctan(1+t²)dt
==>f'(x)=(sinx)'arctan(1+sin²x)-(-x)'arctan(1+(-x)²)
==>f'(x)=cosx*arctan(1+sin²x)+arctan(1+x²)
∴f'(0)=1*arctan(1+0)+arctan(1+0)
=arctan1+arctan1
=π/4+π/4
=π/2.,2,如图： ,1,高数,定积分
1.设f(x)是连续函数,且∫f(t)dt=x,求f(x)
2.设f(x)=∫arctan(1+t²)dt,求f '(0)