高数大一上练习题定积分证明题

证明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设函数f(x)在[ab]内可导，且f(a)b

f a使f()0baf(x)dxf(1)(ba)f(1)在区间[a，1]上,应用 在一点(a，1)(a,b)使f()0。设f(x)在[ab]上有定义,且对[ab]上任意两点xy

f(x)f(y)

xy则f(x)在[ab]上可积,bf(x)dxbaf(a)1(ba)2证明x(ab)ylimy

f(xx)f(x)f(x)在[ab]上连续,于是f(x)在[ab]上可积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 设fx)[ab]上正值，连续，则在(ab)内至少存在一点使f(x)dx f(x)dx1 f(x)dx 2证：令F(xaf(t)dtxf由于x[ab]时，f(x) F(a)a

f(t)dt0,F(b)a

f(t)dt由根的存在性定理，存在一点(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)时，x2 4又x30,4x241 211 dx110 1 1 1 设函数f(x)和g(x)在[a,b]上连续证明:[bf(x)g(x)dx]2bf2x)dxbg2x)dx 考虑以tb[f(xtg(x)]2a显然f(xtg(x)]20.并由题设知它在[ab]上连续[bf(xtg(x)]2dxa 即t2bg2x)dx2tbf(x)g(x)dxbf2x)dx 所以其判别式 即[bf(x)g(x)dx]2[bf2x)dx][bg2x)dx [bf(x)g(x)dx]2[bf2x)dx][bg2x 设函数f(x)在[0,1]上连续 2f(cosx)dx f(cosx)dx 4证显然fcosx)是以2f(cosx)dx2f(cosx)dx

f(cosx)dx

02

f(cosx在后一积分中令xt f(cosx)dx

f(cos(t))dt2f(cost)dt2f(cosx f(cosx)dx2[2f(cosx)dx 2f(cosx)dx]42f(cosx得证2fcosx)dx12fcosx)dx 4x若函数f(x)在R连续，且f(xa

f(t)dt，则f(x0且xR有f(x(xf(t)dt)1faf(x)f(x)考虑函数p(x)f(x)ex.xp(x)f(x)exf(x)ex[f(x)f(x)]exp(x)c(常数cfa已知f(a

f(x)cexf(t)dt0f(x) 设函数f(x)在[a，b]上连续且单调递增。F(x) xf(t)dt，(axxaF(a)f()， F(x)在[ab]上单调增.。对[a,b]内的每个x，由积分中值定理F(x) xf(t)dt f()(x-a)f( (axa x当xa时，a，limF(xlimf(f(aF F(x)在点a连续.从而F(x)在[ab]F(x)f(x) xf(t)dtf(x) [f()(xx

(xa)2

x (xf(x)fx

axf(x)单调增.且满足ax，故f()f(x)，从而F(x)0，(axb) F(x)在[a,b]上单调增。b设f(x)在[ab]上二阶可导且f(x0,证明:f(x)dxbafab)b 2f(x)f(ab)f'(ab)(xab)1f()(xab (介于x与ab之间2由题设知f(

22b)21f2(ab)(xab baf(x)22b)21f2(ab)(xab ba2bf(x)dx(ba)f(abafab)2设f(x)[ab][ab]可导，且f(x0F(x xf(t)dt(ab)内满足F(x0axaxF(x)(xa)f(x)ax(x

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

由已知.x(ab)时，f(x0，故f(x)(ab)a xf(x)f(又x-aF(x) x(a,b)试证：如果f(x)在[ab]上连续，且对于一切x