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

题目

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

答案

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/2k=π/...