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.