设f″(x)在[0.π]上连续,且f(0)=2,f(π)=1,求∫π0
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设f″(x)在[0.π]上连续,且f(0)=2,f(π)=1,求
∫ | π 0 |
因为
∫π0[f(x)+f″(x)]sinxdx=
∫π0f(x)sinxdx+
∫π0f″(x)sinxdx
又f″(x)在[0.π]上连续,且f(0)=2,f(π)=1,
所
∫π0f″(x)sinxdx
=
∫π0sinxdf′(x)
=f′(x)sin
x|π0−
∫π0f′(x)cosxdx
=-
∫π0cosxdf(x)
=-f(x)cos
x|π0−
∫π0f(x)sinxdx
=f(π)+f(0)-
∫π0f(x)sinxdx
=3-
∫π0f(x)sinxdx
所以
∫π0[f(x)+f″(x)]sinxdx=
∫π0f(x)sinxdx+3−
∫π0f(x)sinxdx=3.
∫π0[f(x)+f″(x)]sinxdx=
∫π0f(x)sinxdx+
∫π0f″(x)sinxdx
又f″(x)在[0.π]上连续,且f(0)=2,f(π)=1,
所
∫π0f″(x)sinxdx
=
∫π0sinxdf′(x)
=f′(x)sin
x|π0−
∫π0f′(x)cosxdx
=-
∫π0cosxdf(x)
=-f(x)cos
x|π0−
∫π0f(x)sinxdx
=f(π)+f(0)-
∫π0f(x)sinxdx
=3-
∫π0f(x)sinxdx
所以
∫π0[f(x)+f″(x)]sinxdx=
∫π0f(x)sinxdx+3−
∫π0f(x)sinxdx=3.
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