明顯是0,下面是無(wú)窮大,而上面一定是個(gè)有限值:2>=∫[0->x] sintdt >= -2sint是絕對(duì)值sint，答案不是0，是派/2|sint|是周期為π的函數(shù)∫[0->π] |sint|dt = 2則有∫[0->nπ] |sint|dt = 2n,∫[0->(n+1)π] |sint|dt = 2(n+1)由于f(x)=∫[0->x] |sint|dt是個(gè)增函數(shù)，所以當(dāng)x∈[nπ, (n+1)π]時(shí),f(nπ)<=f(x)<=f((n+1)π),易得:f(nπ)/((n+1)π)<=f(x)/x<=f((n+1)π)/nπ同時(shí)取極限:x->∞時(shí)，n->∞，lim[n->∞] f(nπ)/((n+1)π) = lim[n->∞] 2n/((n+1)π) = 2/πl(wèi)im[n->∞] f((n+1)π)/nπ 亦等于2/π，由夾逼定理得lim[x->∞] ∫[0->x] |sint|dt / x =2/π