设f(x)在[0,π]上二阶连续可导,且f(π)=2满足∫π0(f(x)+f″(x))sinxdx=5,试计算f(0)的值.
由于∫π0f(x)sinxdx=∫π0f(x)(-cosx)′dx=f(π)+f(0)+∫π0f′(x)cosθdx,∫π0f″(x)sinxdx=f′(x)sinx|π0-∫π0f′(x)cosxdx=-∫π0f′(x)cosxdx,∴5=f(π)+f(0)=2+f(0),∴f(0)=3.
举一反三
- 设f(x)在[a,+∞)内二阶可导,f A: =A>0,f'(a)<0,f"(x)≤0(x>a),则
- 设f(x)在[0,1]上二阶可导,且f(0)=f"(0)=f(1)=f"(1)=0.证明:方程f"(x)=f(x)=0在(0,1)内有根.
- 设函数f(x)在x=0处可导,且f(0)=0,f′(0)=2,则=()。设函数f(x)在x=0处可导,且f(0)=0,f′(0)=2,则=()。
- 设f(x)在[0,1]上二阶连续可导,且f’(0)=f’(1).证明:存在ξ∈(0,1),使得
- 设f(x)连续可导,f(0)=0且f’(0)=b,若在x=0处连续,则C=()。设f(x)连续可导,f(0)=0且f’(0)=b,若在x=0处连续,则C=()。
内容
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