设f(x)在(0,π/2(为闭区间)上连续,f(x)=xcosx+∫ f(t)dt 则∫ f(x)dx 等于多少积分都有上限π/2 下限上限是平π/2 下限是0
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![设f(x)在(0,π/2(为闭区间)上连续,f(x)=xcosx+∫ f(t)dt 则∫ f(x)dx 等于多少积分都有上限π/2 下限上限是平π/2 下限是0](/uploads/image/z/1730768-32-8.jpg?t=%E8%AE%BEf%EF%BC%88x%EF%BC%89%E5%9C%A8%EF%BC%880%2C%CF%80%2F2%28%E4%B8%BA%E9%97%AD%E5%8C%BA%E9%97%B4%29%E4%B8%8A%E8%BF%9E%E7%BB%AD%2Cf%28x%29%3Dxcosx%2B%E2%88%AB+f%28t%29dt+%E5%88%99%E2%88%AB+f%EF%BC%88x%EF%BC%89dx+%E7%AD%89%E4%BA%8E%E5%A4%9A%E5%B0%91%E7%A7%AF%E5%88%86%E9%83%BD%E6%9C%89%E4%B8%8A%E9%99%90%CF%80%2F2+%E4%B8%8B%E9%99%90%E4%B8%8A%E9%99%90%E6%98%AF%E5%B9%B3%CF%80%2F2+%E4%B8%8B%E9%99%90%E6%98%AF0)
设f(x)在(0,π/2(为闭区间)上连续,f(x)=xcosx+∫ f(t)dt 则∫ f(x)dx 等于多少积分都有上限π/2 下限上限是平π/2 下限是0
设f(x)在(0,π/2(为闭区间)上连续,f(x)=xcosx+∫ f(t)dt 则∫ f(x)dx 等于多少积分都有上限π/2 下限
上限是平π/2 下限是0
设f(x)在(0,π/2(为闭区间)上连续,f(x)=xcosx+∫ f(t)dt 则∫ f(x)dx 等于多少积分都有上限π/2 下限上限是平π/2 下限是0
记:∫[0,π/2]f(t)dt=k(常数)
则f(x)=xcosx+∫ [0,π/2]f(t)dt可化为
f(x)=xcosx+k
两边在[0,π/2]积分有
∫[0,π/2]f(t)dt=∫[0,π/2]tcostdt+k∫[0,π/2]dt【分部积分】
k=tsint[0,π/2]-∫[0,π/2]sintdt+kπ/2
k=π/2-1+kπ/2
解得k=-1
∫ f(t)dt (上限是π/2 下限是0)是常数,记为t,则
f(x)=xcosx+t
两边积分(上限是π/2 下限是0),得 t=∫xcosxdx+tπ/2
t=[∫xcosxdx]/(1-π/2)(上限是π/2 下限是0).
不定积分:∫xcosxdx=∫xdsinx=xsinx-∫sinxdx=xsinx+cosx+C
定积分∫xcosxdx=π...
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∫ f(t)dt (上限是π/2 下限是0)是常数,记为t,则
f(x)=xcosx+t
两边积分(上限是π/2 下限是0),得 t=∫xcosxdx+tπ/2
t=[∫xcosxdx]/(1-π/2)(上限是π/2 下限是0).
不定积分:∫xcosxdx=∫xdsinx=xsinx-∫sinxdx=xsinx+cosx+C
定积分∫xcosxdx=π/2-1(上限是π/2 下限是0)
故所求t=-1
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