高數(shù)上5五章1節(jié)

第五章知識點1、定積分概念及其應(yīng)用.2、用定積分性質(zhì)求定積分.3、用微積分定理求定積分.4、變限定積分有關(guān)定理及其應(yīng)用.5、用換元法求定積分.6、用分部法求定積分.§5.1

定積分的概念和性質(zhì)一、實例分析：P223---225定積分問題舉例1,21、求如圖曲頂梯形的面積A：曲頂梯形由連續(xù)曲線y

=f

(x)

(f

(x)?0)、x軸與直線x

=a、x

=b所圍成。abxy

=

f

(x)i-1

i ,

x

]：i=1a

=

x0

<

x1

<

x2<<

xn-1

<

xn

=b,記Dxi

=

xi

-

xi-1,

l

=

max(Dxi

),xi

?

[xi-1,

xi

],nlfi

0

i=1i=1

i=1n

n(1)任意分割[a,b]=n

[x任意取點(2)近似作積：DAi

?

f

(xi

)Dxi

,(3)求和：A

=DAi

?

f

(xi

)Dxi

,(4)取極限：A

=lim

f

(xi

)Dxi

.(必存在)2、求變速直線運動的位移S：已知變速直線運動的速度函數(shù)為v(t),求時刻T1到T2

的位移S.1

2i-1

i ,

t

]：i=1nlfi

0

i=1n

ni=1

i=1(1)任意分割[T

,T

]

=

n

[t(3)求和：S

=DSi

?v(ti

)Dti

,(4)取極限：S

=lim

v(ti

)Dti

.(必存在)T1

=t0

<t1<t2

<<tn-1<tn

=T2,記Dti

=

ti

-ti-1,

l

=

max(Dti

),任意取點

ti

?

[ti-1,

ti

],(2)近似作積：DSi

?

v(ti

)Dti

,實例解決方法的共性：(1)分割(化整為零)；(2)近似作積(以不變代變)；(3)求和(積零為整)；(4)取極限(近似到精確).(1)任意分割[a,b]=[xi-1

,xi

]：二、定積分的定義：設(shè)f

(x)在[a,b]上有界,ni=1a

=

x0

<

x1

<

x2

<<

xn-1

<

xn

=b,記Dxi

=

D[xi-1,

xi

]

=

xi

-

xi-1,l

=

max(Dxi

),任意取點xi

?

[xi-1,xi

],(2)作積：DSi

=

f

(xi

)Dxi

,(3)求和：S

=DSi

=

f

(xi

)Dxi

,i=1i=1n

nlfi

0(4)取極限：lim

S

=lim

f

(xi

)Dxi

,lfi

0

i=1n若limn

i

if

(x

)Dx

=A,則lfi

0

i=1若上述極限存在,則此極限稱為f

(x)在區(qū)間[a,b]上的定積分并記為baf

(x)dx,即：baf

(x)dx

=

A.也稱f

(x)在區(qū)間[a,b]上可積.此時稱定積分baf

(x)dx存在,f

(x)dx：f

(x)在[a,b]上的定積分；a其中：

：積分符號；f

(x)：被積函數(shù)；dx中的x：積分變量；f

(x)dx：被積表達(dá)式；[a,b]：積分區(qū)間；a、b：積分下限、上限；bn

f

(xi

)Dxi：積分和.i=1=baf

(x)dxT2T1v(t)dt.如：實例1中所求面積i

iA

=

limf

(x

)Dx

i

iv(S

=

limt

)Dt

=實例2中所求位移：nnlfi

0

i=1lfi

0

i=1常用n等分[a,b],取右端點,i

iin

nl=D(x

)

=b-a,x

=x

=a+i(b-a)(i

=1,2,,n),1、使l

fi

0的分割及取點皆任意,\使l

fi

0的特殊分割及特殊取點也成立.n

fi

￥.且l

fi

0此時有：nanlimnfi

￥i=1b

-

ani

b

-

a

$f

a

+b=

f

(x)dx.從而有：i

iin

nl

=

D(x

)

=

1

,x

=

x

=

i

(i

=1,

2,,

n),nnff1nn1n

nlimnfi

￥

i

i

=

limnfi

￥

i=1i=1若取[a,b]=[0,1],則：n等分[a,b],取右端點,有：10f

(x)dx.$=sinn1nin(1)

limnfi

￥i=1例1、用定積分表示下列極限：(2)

lim

1nnnnfi

￥1i

+i=1

10sin

xdx.=10=(1+

x)dx(3)

lim

1nnnnfi

￥1+cos

ipi=110=1+cos(px)dx注：用定積分表示極限,結(jié)果形式可以多樣.xdx

=

B10(1)n等分[0,1]nnnii

1n

ni=1i=1xi

)Dx

=(3)求和A

=

f

(1

=

B

l

fi

0

2

n

fi

￥

(4)取極限A

fini1ni-1

ii=1=l

=

Dx

=[x

,x

]且i

i

i-1

i(2)取點x

=

x

=

i

?

[x

,

x

][分析]：例2、用定義計算：解：(1)n等分[0,1]：0

=

x0

<

x1

<<

xn

=1ini-1

i?[x

,

x

](2)取xi

=

xi

=21nniiini=1

i=1(3)

A

=f

(x

)Dx

=nfi

￥lfi

0i

in

n

x

=

i

,l

=Dx

=

1(i

=1,2,,n)i

in

nn2

f

(x)Dx

=

i

1

=

i(i

=1,2,,n)2n2n(n+1)=2n2

2nfi

￥(4)B

=

lim

A

=

lim

A

=

lim

n(n

+1)

=

1ba2、f

(x

)dx中x

?

[a

,b

].p02如：=p0sinxdxp2p2sin

xdx=p2p(-sinx)dx0|

sinx

|

dxpsin

x

dx

=|

sinx

|

dx2pp=ba3、f

(x)?

C[a,b]f

(x)dx存在.只有有限個可去間斷點時也可積.反之不一定,如：a

ab

b

ba4、

f

(x)dx

=

f

(t)dt

=

f

(u)du

==相對于積分變量而言的常數(shù).5、定積分由被積函數(shù)和積分區(qū)間共同確定.f

(x)在[a,b]上三、定積分幾何意義：baf

(x)dx

=

A1、"x?[a,b],f

(x)?0

A

=S1

+S2

+2、"x?[a,b],f

(x)￡0

A

=-S1

-S2

-3、x

?[a,b],f

(x)有正有負(fù)

A

=S1

-S2

+S3

-S4

+例3、利用幾何意義計算：(a>0)2-1p-p(2)

sin

xdx

=

04a0a2

-

x2

dx

=

1

pa2(3)S紅=2,S藍(lán)=1

2(1)

xdx

=

3

2y

=

x,

x?[-1,2]四、定積分的性質(zhì)：1、規(guī)定：(1)aaf

(x)dx=02、設(shè)相關(guān)定積分都存在,則：baabf

(x)dxf

(x)dx

=-(2)(1)badx

=

b

-

a12bbaag(x)dx=

kf

(x)dx

+k注意：積分區(qū)間不變.ba(xsint

+

y

cost)dtbaba=

xcostdtsintdt

+

y1

2(2)"k1,

k2

?

R,bag(x)]dx[k f

(x)

+k如：bccabaf

(x)dxf

(x)dx+f

(x)dx

=(3)注意：被積函數(shù)為分段函數(shù)時常用；積分區(qū)間：首尾不變,中間相連.如：03

13011-

x

dx

=

(1-

x)dx

+

(x

-1)dx；50max(1,

x2

,

x3

)dx15301x

dx.=dx

+b

baag(x)dx.f

(x)dx

?Cor：若"x

?

[a,b],f

(x)?g(x),則(4)若"x

?[a,b],f

(x)?0,baf

(

x)dx

?

0.則如：21221(lnx)

dxln

xdx?(5)若f

(x)

?

C[a,b],m

=

min

f

(x),x?[a,b]M

=max

f

(x),則：x?[a,b]baf

(x)dx

￡

M

(b

-a)m(b

-a)

￡(定積分估值定理)31ln

xdx如：?[0,2ln3]130nnxnnfi

￥1+

x例4、若A

=dx,求lim

A

.33n解：0

￡

x

￡

1

0

￡

xn

￡

113nnxn1+x130\

0￡

￡x

￡

11n3n

3n+1\

0

￡

A

￡dx

=fi

0(nfi

￥)(夾逼準(zhǔn)則)130xn\

An

=dx

=

01+

xaf

(x)dx

=

f

(x)(b

-a)(積分中值定理)a2

-x2

在[0,a]上4的平均值為

1

pa

.(6)若f

(x)?

C[a,b]，則$x?

[a,b]使：b其中f

(x)：f

(x)在[a,b]上的平均值.如：f

(x)=2nxen+2

x2nfi￥例5、若An

=

ndx,求lim

A

.解：$x?[n,n

+2]使2ex

exx22x2An

=2

[(n+2)-n]=2且由x?[n,n+2]有：n

fi

￥時,x

fi

￥x2nfi

￥xfi

￥

ex\

lim

An

=

2

lim

=

=

0(洛必達(dá)法則)2nnfi

￥2n+3

2x2

+13x

-2練習(xí)、若An

=2ndx,求lim

A

.解：$x?[2n,2n

+3]使2x2

+1

3(2x2

+1)An

=

3x2

-2[(2n+3)

-2n]=

3x2

-2且x?[2n,2n+3]n

fi

￥時,x

fi

￥2nfi

￥2x2

+1\

lim

An

=3lim=

2xfi

￥

3x

-2(消小法)[a,

b],(7)若f

(x)在[a,b]上可積,[c,d

]則f

(x)在[c,d

]上可積.(8)若f

(x)在[a,b]上可積,則f

(x)在[a,b]上可積.(反之不一定)x

?

Q-1

x

ˇ

Q,則例如：設(shè)f

(x)=

1f

(x)在[0,1]上可積,f

(x)在[0,1]上不可積.小結(jié)：1、定積分的概念及其應(yīng)用.2、定積分的性質(zhì)及其應(yīng)用.作業(yè)5-1：235-236頁10(1,3),13(2,4,5)*(9)

f

(x),

g(x)

?

C[a,b],"x

?

[a,b],g

(x)恒正(負(fù))

$x

?

[a,b]

st：ab

bag(x)dxf

(x)g(x