高數(shù)大一上練習(xí)題定積分證明題

證明dx(xt)f(t)dtf(x)f dxx(xt)faxa(xt)dfx(xt)f(t)xxf x(ax)f(a)afdx(xt)f(t)dtdxaf(a)ff(xf(a)b設(shè)函數(shù)f(x)在[ab]內(nèi)可導(dǎo)，且f(a)b

f a使f()0baf(x)dxf(1)(ba)f(1)在區(qū)間[a，1]上,應(yīng)用 在一點(diǎn)(a，1)(a,b)使f()0。設(shè)f(x)在[ab]上有定義,且對(duì)[ab]上任意兩點(diǎn)xy

f(x)f(y)

xy則f(x)在[ab]上可積,bf(x)dxbaf(a)1(ba)2證明x(ab)ylimy

f(xx)f(x)f(x)在[ab]上連續(xù),于是f(x)在[ab]上可積f(xf(a)xa(x即f(axaf(xf(ax a[f(a)(xa)]dxaf(x)dxa[f(a)(x(ba)2bf(x)dx(ba)f(a)(b bf(x)dxbaf(a)1(ba)2 設(shè)fx)[ab]上正值，連續(xù)，則在(ab)內(nèi)至少存在一點(diǎn)使f(x)dx f(x)dx1 f(x)dx 2證：令F(xaf(t)dtxf由于x[ab]時(shí)，f(x) F(a)a

f(t)dt0,F(b)a

f(t)dt由根的存在性定理，存在一點(diǎn)(a,b)使F()baf(t)dtbb

f f(x)dx

f(x)dx

f af(x)dx

f(x)dxa

f1

x(0,1)時(shí)，x2 4又x30,4x241 211 dx110 1 1 1 設(shè)函數(shù)f(x)和g(x)在[a,b]上連續(xù)證明:[bf(x)g(x)dx]2bf2x)dxbg2x)dx 考慮以tb[f(xtg(x)]2a顯然f(xtg(x)]20.并由題設(shè)知它在[ab]上連續(xù)[bf(xtg(x)]2dxa 即t2bg2x)dx2tbf(x)g(x)dxbf2x)dx 所以其判別式 即[bf(x)g(x)dx]2[bf2x)dx][bg2x)dx [bf(x)g(x)dx]2[bf2x)dx][bg2x 設(shè)函數(shù)f(x)在[0,1]上連續(xù) 2f(cosx)dx f(cosx)dx 4證顯然fcosx)是以2f(cosx)dx2f(cosx)dx

f(cosx)dx

02

f(cosx在后一積分中令xt f(cosx)dx

f(cos(t))dt2f(cost)dt2f(cosx f(cosx)dx2[2f(cosx)dx 2f(cosx)dx]42f(cosx得證2fcosx)dx12fcosx)dx 4x若函數(shù)f(x)在R連續(xù)，且f(xa

f(t)dt，則f(x0且xR有f(x(xf(t)dt)1faf(x)f(x)考慮函數(shù)p(x)f(x)ex.xp(x)f(x)exf(x)ex[f(x)f(x)]exp(x)c(常數(shù)cfa已知f(a

f(x)cexf(t)dt0f(x) 設(shè)函數(shù)f(x)在[a，b]上連續(xù)且單調(diào)遞增。F(x) xf(t)dt，(axxaF(a)f()， F(x)在[ab]上單調(diào)增.。對(duì)[a,b]內(nèi)的每個(gè)x，由積分中值定理F(x) xf(t)dt f()(x-a)f( (axa x當(dāng)xa時(shí)，a，limF(xlimf(f(aF F(x)在點(diǎn)a連續(xù).從而F(x)在[ab]F(x)f(x) xf(t)dtf(x) [f()(xx

(xa)2

x (xf(x)fx

axf(x)單調(diào)增.且滿足ax，故f()f(x)，從而F(x)0，(axb) F(x)在[a,b]上單調(diào)增。b設(shè)f(x)在[ab]上二階可導(dǎo)且f(x0,證明:f(x)dxbafab)b 2f(x)f(ab)f'(ab)(xab)1f()(xab (介于x與ab之間2由題設(shè)知f(

22b)21f2(ab)(xab baf(x)22b)21f2(ab)(xab ba2bf(x)dx(ba)f(abafab)2設(shè)f(x)[ab][ab]可導(dǎo)，且f(x0F(x xf(t)dt(ab)內(nèi)滿足F(x0axaxF(x)(xa)f(x)ax(x

f(xa)f(x)(xa)f((xf(x)f()(xa)

由已知.x(ab)時(shí)，f(x0，故f(x)(ab)a xf(x)f(又x-aF(x) x(a,b)試證：如果f(x)在[ab]上連續(xù)，且對(duì)于一切x