Download - İntegral 03
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ANA SAYFA
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BELİRSİZ İNTEGRAL TANIM: f:[a,b]→ R tanımlı iki fonksiyon olsun.Eğer F(x) in türevi F(x) veya diferansiyeli f(x).d(x) olan F(x) fonksiyonuna, f(x) fonksiyonunun belirsiz integrali denir ve ∫f(x).d(x)=F(x)+c biçiminde gösterilir.
BELİRSİZ İNTEGRALİN ÖZELLİKLERİ
1) Bir belirsiz integralin türevi, integrali alınan fonksiyona eşittir. Yani,
= = f(x) tir.
2) Bir belirsiz integralin diferansiyeli, integral işaretinin altındaki ifadeye eşittir. Yani,
d(∫f(x).dx) = f(x).dx
.dx))(( 'xf∫ )')(( CxF +
ANA SAYFA
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xexe
3) Bir fonksiyonun diferansiyelinin belirsiz integrali, bu fonksiyon ile bir C sabitinin toplamına eşittir. Yani,
∫d(f(x)) = f(x) + c dir.
iNTEGRAL ALMA KURALLARI
1) ∫ dx= (1/n+1) +c (n≠ -1)
2) ∫(1/x)dx = ln |x| +c
3) ∫ dx = +c
4) ∫ dx = (1/lna). +c
5)∫sinxdx = -cosx+c
6)∫ cosx=sinx+c
7) ∫ tanx.secx.dx = secx+c
nx 1+nx
xaxa
ANA SAYFA
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8) ∫cotx.cosecx.dx= -cosecx+c
9)∫ .dx = ∫(1/ .dx) = (1+ )dx = tanx+c
10)∫ .dx = ∫(1/ ).dx = ∫ (1+ ).dx=-cotx+c
11)∫(1/1+ ).dx =arctanx+ =-arccotx+
12)∫ dx=arcsinx+ =-arccosx+
Örnek: ∫ (2x+1)dx belirsiz integralini bulalım.
Çözüm: ∫ (2x+1).dx= 2∫x.dx+ ∫1.dx=2.( /2)+1.x+c= +x+c bulunur.
Örnek: ∫[-[(2x-3x) / x].dx belirsiz integralini bulalım.
Çözüm: ∫[(2x-3x) / x].dx =∫[(2x/x) -(3x/x)].dx=∫2x.dx-∫3/x.dx =2∫x.dx-3∫(1/x)dx=x-3 ln |x|+c
xsec2
xcos2
xtan2
xcosec2
xsin 2
xcot2
2x 1c2c
2x-1
11c
2c
2x 2x
ANA SAYFA
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İNTEGRAL ALMA METOTLARIİNTEGRAL ALMA METOTLARI
DEĞİŞKEN DEĞİŞTİRME METODU İntegral hesaplarında, uygun bir değişken değiştirmesi yapılarak integrali hesaplanacak ifade ilkeli kolaylıkla bulunabilecek bir ifadeye dönüştürülür. 1) ∫f(x). .dx= ∫f(x).d(f(x)) integralinde fonksiyon ve türevi çarpım şeklinde ise, değişken değiştirme metodu kullanılır. Değişken değiştirme yapılırken hem fonksiyonun hem de diferansiyelin değişimi yapılmalıdır. F(x)=u dönüşümü yapılırsa; d(f(x))=d(u) => .dx= du olur. Bulunan bu değerleri yerlerine yazalım: ∫f(x). dx= ∫u.du=( /2)+c=1/2 (x)+c şeklinde çarpım fonksiyonunun integrali alınmış olur.
(x). f '
(x). f '
(x). f ' 2u 2f
ANA SAYFA
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n[f(x)]
(x)f '
2) ∫ dx =∫ d(f(x)) integralinde genellikle üssü görmeden f(x)=u dönüşümü yapılır. Fakat türev oluşmazsa = u denilmelidir. Burada f(x) = u dönüşümü yapılırsa;
f(x) = u =>d(f(x))=(du) => .dx=du olur. Bulunan bu değerleri yerine yazalım:
.[f(x)] n (x).f ' (x).f n
(x).dxfn ' .[f(x)]∫
Cn
uduu
nn +
+==
+
∫ 1.
1
Cn
++
=+
1
[f(x)] 1n
ANA SAYFA
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(x).f '
3) integralinde,
f(x) dönüşümü yapılırsa ; her iki tarafın diferansiyelini alalım: d(f(x))=d(u) => dx=du olur. Bulunan bu değerleri yerlerine yazalım:
bulunur.
∫ =)(
))((
)(
(x).dxf '
xf
xfd
xf
∫ ∫ +=+== CxfCuu
du
xf
dxxf)(lnln
)(
).('
ANA SAYFA
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{ }∫ −∈ + )1().(. ')( Radxxfa xf
)(' xf
4) integralinde;
f(x) = u dönüşümü yapılırsa ;d(f(x))=d(u) => .dx = du olur. Bulunanları yerlerine yazalım:
bulunur.
5) integralinde, g(x) = u dönüşümü
yapılırsa ;
g(x)=u => d(g(x))= d(u)=> g’(x).dx=du olur.
Ca
aCa
aduadxxfa xfuuxf +=+==∫ ∫ ln
1.
ln
1..).(. )(')(
dxxgxgf ).(.))(( '∫
ANA SAYFA
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Bulunanları yerlerine yazalım:
gibi basit fonksiyon integrali elde edilir.
İNTEGRALDE TRİGONOMETRİK DÖNÜŞÜMLER
İntegrandında bulunan integraller, trigonometrik dönüşümler yardımıyla hesaplanır.
Amaç , yapılacak trigonometrik dönüşümlerle irrasyonel fonksiyon integralini, rasyonel fonksiyon integraline dönüştürmektedir.
1)İntegrandında Varsa(a>0)
olur.
duufdxxgxgf .)().(.))(( ' ∫∫ =
222222 ,, xaaxxa +−−
22 xa −
tattaaxa cos.sin1sin. 222222 =−=−=−
ANA SAYFA
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22 xa +
olur.
1>a
x
2)İntegrandında varsa;
olur.tataataax tan1secsec. 222222 =−=−=−
22 ax −
3)İntegrandında varsa; (a>0)
x = a.tant dönüşümü yapılır.
Buna göre, tatataaxa sec.tan1.tan 222222 =+=+=+
ANA SAYFA
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2)KISMİ İNTEGRASYON METODU
YARDIM:
1)dv’nin integrali kolay olmalı.
2) integrali ilk integral
3) u’yu seçerken genelde aşağıdaki sıra ile seçmek avantajlıdır.
∫ duv.
∫ ∫−= duvvuduu ...
ANA SAYFA
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ÖRNEK: x.cos.dx
u= x ; dv=cosx.dx
du=dx ; v=sinx
=>x.sinx- sinx.dx
=xsinx+cosx+c ÖRNEK2:
lnx/x2
u=lnx dv=1/x2.dx
du=(1/x).dx v=-1/x
=>u.v- v.du
lnx(-1/x)- (1/x).(1/x).dx
∫
∫
∫
∫
∫
ANA SAYFA
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.dx=
=
1+x22 23 +++ xxx
∫ +++
1
22
xxx
= (-lnx/x)-(1/x)+c
= (-lnx-1/x)+c
3) BASİT KESİRLERE AYIRMA METODU
ÖRNEK:
dxx
xxx.
1
22 23
∫ ++++
=x2+x kalan:2
cxxx ++++ 1ln223
23
ANA SAYFA
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B=3 ; C=1 ;A=-3
Örnek: dxxx
xx.
323
2
∫ −++
11)1)(1(
322
++
−+=
+−++
x
C
x
B
x
A
xxx
xx
)1()1()1(32 22 −+++−=++ xCxxBxxAxx
∫
++
−+−
dxxxx
.1
1
1
33
=-3ln|x|+3ln|x-1|+ln|x+1|+c
ANA SAYFA
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1tansec
sin211cos22cos
cossin22sin
1cos2sin
22
22
22
=−−=−=
==+
xx
xxx
xxx
xx
4) TRİGONOMETRİK DÖNÜŞÜMLER YARDIMI İLE
(sina.sinb)= -1/2(cos(a+b)-cos(a-b))
(cosa.cosb)= 1/2(cos(a+b)-cos(a-b))
(sina.cosb)= 1/2(sin(a+b)-sin(a-b))
ANA SAYFA
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ÖRNEKLER:
1)
2)
∫ = ???.tan4 dxx
∫
∫ = ???.2cos.4cos dxxx
3)
4)
sinx2 .dx=???
???x.dxx.cossin 32 =∫
5)
∫ = ???.tan.sec 3 dxxx
ANA SAYFA
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( ) dxxx .2cos6cos2
1∫ +=
( ) ( )
cxx
dxxdxx
+
=+
=
+=
6
2sin
2
1
6
6sin
2
1
.2cos2
1.6cos
2
1
ANA SAYFA
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dxx
dxx
.2
2cos
2
1.
2
2cos1x.dx sin 2 ∫∫
−=−=
cxx +−=
4
2sin
2
ANA SAYFA
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=> u=sinx du=cosx.dx
( ) dxxxx .cos.sin1sin
x.cosx.dxx.cossin
22
22
∫∫
−=
=
( )( )
cxx
cuu
duuu
duuu
+−=
+−=
−=
−=
∫∫
5
sin
3
sin
53
.
.1
53
53
42
22
ANA SAYFA
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( )∫∫
−=
=
dxxxx
dxxxx
.tan.sec1sec
.tan.sec.tan
2
2
=> u=secx du=secx.tanx.dx
( )
cxx
cuu
duu
+−=
+−=
−=∫
sec3
sec
3
.1
3
3
2
ANA SAYFA
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( )∫∫
∫∫
∫
−=
−=
=
=>
dxxdxxx
dxxx
dxxx
dxx
.tan.tan.sec
.tan.1sec
.tan.tan
.tan
222
22
22
4
u=tanx
du=sec2x.dx
( )cxx
x
dxxduu
++−=
−+−=>∫ ∫tan
3
tan
.11tan.
3
22
ANA SAYFA
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5)ÖZEL DÖNÜŞÜMLER
Sadece köklü ifade varsa!!!
ub
axaxb
ub
axxba
ub
axxba
sec*
tan*
sin*
222
222
222
==>−
==>+
==>−
ANA SAYFA
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( )∫ =+−
???664 2x
dx
∫ =++
???1744 2 xx
dx
ÖRNEK1:
ÖRNEK3:
ÖRNEK2:
∫ = ???.sec dxxarc
ANA SAYFA
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( )( )
cx
u
xu
xu
duu
duu
uu
+
+=
+=⇒+=⇒
=
=−=
=
+=
==+
∫∫
8
6arcsin
8
6arcsin
8
6sin
cos8
.cos8
cos8sin18
8sinu-64
6x-64
8cosu.dudx 8sinu 6x
2
2
2
ANA SAYFA
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1744 2 ++ xx =(2x+1)2+42
( )udx
udx
udx
ux
uxxx
dx
2
2
2
2
tan12
tan22
tan442
1tan42
tan4121744
+=+=+=
−==+
++∫ ( )( )
( )u
u
u
u
u
x
xx
2
2
2
2
2
2
2
tan12
1
tan14
tan12
1tan4
16tan4
1612
1744
+=+
++=
+=
++=
++
DEVAMIANA SAYFA
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cuuuu ++=== ∫∫ tansecln2
1sec
2
1sec
2
1 2
tanu=2x+1/4
Secu=1/2 ln| |
uu tantan1 2 ++
cxxx
xxx
+++++=
+++++=
121744ln2
1
4
12
16
17441ln
2
1
2
2
ANA SAYFA
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cxxxarc
cx
xxarc
cx
xxarc
xu
xu
uxux
xx
dxxxarc
dxduxux
dxdu
xarcux
dxxarcxdxxarc
+−
++
+
−−
−===>==>=
−−=
===>−
=
=−
−=
∫
∫∫
lnsec
1lnsec
1lnsec
1lncos
1cos
cos
1sec
1sec
;1
sec1
sec..sec
2
2
2
ANA SAYFA
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( )∫ =b
a
xFdxxf )(.b
a)()( aFbF −= c yok ; c-c=0
ANA SAYFA
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( )∫∏
=+0
???.sincos dxxx
x
dx
1
2
3
2
12−
∫
ANA SAYFA
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663
∏=∏−∏=
2
1
x=
2
3
Arc sinx
dx
1
2
3
2
12−
∫ 2
1sin
2
3sin ArcArc −
ANA SAYFA
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xx cossin −0
∏
2)1()1( =−−+=
ANA SAYFA
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∫ =+=x
dttxF2
???).12()(
BELİRLİ İNTEGRALİN TÜREVİ
)(')).(()(').(()('
).()()(
)(
xgxgfxhxhfxF
duufxFxh
xg
−=
⇒= ∫
ÖRNEK:
ANA SAYFA
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12)('
)0).(5()1).(12()('
+=−+=
xxF
xxF
ANA SAYFA
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???.sin.sin2
2
=∫∏
∏−
dxxx???.cossgn2
2
=∫∏
∏
dxx
ÖZEL FONKSİYONUN İNTEGRALİ
???.2
3
2
=∫−
dxX
ANA SAYFA
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Π−⇒+−+− ∫∫∫Π
Π
Π
Π
Π
Π
2
2
3
2
3
2
.0.1.1 dxdxdx
-1,5
-1
-0,5
0
0,5
1
1,5
Π
2
3Π
Π2
2
Π
ANA SAYFA
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∏
2
∏
2
3∏ ∏2
0
2
∏−
1
1−
0
2
∏−
x
dxxdxx
cos
.sin.0.sin2
0
0
2
⇒
+− ∫∫∏
∏−
1= ANA SAYFA
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1
.1.0.1
2
2111
3
2
2
0
0
2
−⇒
++−
===
∫∫∫−
dxdxdx
ad
ANA SAYFA
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İNTEGRALDE ALAN HESABI1) A)BİR EĞRİNİN ALTINDA KALAN ALAN
y=f(x) , x=a , x=b ve x ekseni
dxxfAb
a
.)(∫=
a
dxxfAb
a
.)(∫=
a b
b
dxxfAb
a
.)(∫−= dxxfdxxfA
AAAc
a
b
a
.)(.)(
21
∫∫ −=
+=
cba
ANA SAYFA
![Page 40: İntegral 03](https://reader034.vdocuments.pub/reader034/viewer/2022051400/55a632281a28ab77488b4583/html5/thumbnails/40.jpg)
B ) x=g(y) , y=c , y=d ve y ekseni
c
∫∫ −=
+=e
c
d
c
dyygdyygA
AAA
).().(
21
∫−=d
c
dyygA ).( ∫=d
c
dyygA ).(
∫=d
c
dyygA .)(
c c
dd
d
e
C
ANA SAYFA
![Page 41: İntegral 03](https://reader034.vdocuments.pub/reader034/viewer/2022051400/55a632281a28ab77488b4583/html5/thumbnails/41.jpg)
2) İKİ EĞRİ ARASINDA KALAN ALAN
y=f(x) , y=g(x) , x=a , x=b
∫ −=d
c
dxxgxfA .)()(
ANA SAYFA
![Page 42: İntegral 03](https://reader034.vdocuments.pub/reader034/viewer/2022051400/55a632281a28ab77488b4583/html5/thumbnails/42.jpg)
y=x2-x-2 x ekseni ve
x=-2 , x=2 doğrusu
y=2-x2 , y=-x arasındaki alan
ANA SAYFA
![Page 43: İntegral 03](https://reader034.vdocuments.pub/reader034/viewer/2022051400/55a632281a28ab77488b4583/html5/thumbnails/43.jpg)
02x2 =−− x 2=x 1; −=x
21−2−
+ +−
dxxxdxxx
AAA
.)2(.)2(2
1
21
2
2
21
∫∫−
−
−
−−−+−−=
+=
2
3
19br= ANA
SAYFA
![Page 44: İntegral 03](https://reader034.vdocuments.pub/reader034/viewer/2022051400/55a632281a28ab77488b4583/html5/thumbnails/44.jpg)
2-x2=-x
x2-x-2=0 x=2 , x=-1
2
2
1
2
2
9
).()2(
br
dxxxA
=
−−−= ∫−
ANA SAYFA
![Page 45: İntegral 03](https://reader034.vdocuments.pub/reader034/viewer/2022051400/55a632281a28ab77488b4583/html5/thumbnails/45.jpg)
DÖNEL CİSİMLERİN HACİMLERİ
a b
İki eğri arasında x ekseni etrafında;
[ ]∫ −∏=b
a
dxxgxfH .)()( 22
X ekseni etrafında;
[ ] dxxfHb
a
.)(2
∫∏=a b
Y ekseni etrafında;
[ ]∫∏=d
c
dyyfH .)( 2
c
d
)(2 xgy =
)(1 xfy =
ANA SAYFA
![Page 46: İntegral 03](https://reader034.vdocuments.pub/reader034/viewer/2022051400/55a632281a28ab77488b4583/html5/thumbnails/46.jpg)
x=a , x=b y=f(x) y ekseni etrafında;
ba
c
a b
∫∏=b
a
dxxxfH ).(2
f(x) ve x=c , x=a ,x=b arası bölge c etrafında
[ ]∫ −∏=b
a
dxcxfH .)( 2
ANA SAYFA
![Page 47: İntegral 03](https://reader034.vdocuments.pub/reader034/viewer/2022051400/55a632281a28ab77488b4583/html5/thumbnails/47.jpg)
Yarıçapı r olan kürenin hacminin olduğunu gösteriniz. 3
4 3r∏
,xey = ,1=x x ekseni arsında kalan bölgenin x ekseni etrafındaki hacmi nedir?
ANA SAYFA
![Page 48: İntegral 03](https://reader034.vdocuments.pub/reader034/viewer/2022051400/55a632281a28ab77488b4583/html5/thumbnails/48.jpg)
)3
(
.)(
.(
32
22
22
22
xxrH
dxxrH
dxxrH
xry
r
r
r
r
−∏=
−∏=
−∏=
−=
∫
∫
−
−
r−
r
3
4 3r∏⇒ ANA SAYFA
![Page 49: İntegral 03](https://reader034.vdocuments.pub/reader034/viewer/2022051400/55a632281a28ab77488b4583/html5/thumbnails/49.jpg)
( ) 322
2
1
0
2
1222
2
)(
.)(
bree
eH
dxeH
x
x
−∏=∏−∏=
∏=
∏= ∫
0
1
ANA SAYFA
![Page 50: İntegral 03](https://reader034.vdocuments.pub/reader034/viewer/2022051400/55a632281a28ab77488b4583/html5/thumbnails/50.jpg)
![Page 51: İntegral 03](https://reader034.vdocuments.pub/reader034/viewer/2022051400/55a632281a28ab77488b4583/html5/thumbnails/51.jpg)
ANA SAYFA
![Page 52: İntegral 03](https://reader034.vdocuments.pub/reader034/viewer/2022051400/55a632281a28ab77488b4583/html5/thumbnails/52.jpg)
ANA SAYFA
![Page 53: İntegral 03](https://reader034.vdocuments.pub/reader034/viewer/2022051400/55a632281a28ab77488b4583/html5/thumbnails/53.jpg)
ANA SAYFA
![Page 54: İntegral 03](https://reader034.vdocuments.pub/reader034/viewer/2022051400/55a632281a28ab77488b4583/html5/thumbnails/54.jpg)
ANA SAYFA
![Page 55: İntegral 03](https://reader034.vdocuments.pub/reader034/viewer/2022051400/55a632281a28ab77488b4583/html5/thumbnails/55.jpg)
ΠANA SAYFA
![Page 56: İntegral 03](https://reader034.vdocuments.pub/reader034/viewer/2022051400/55a632281a28ab77488b4583/html5/thumbnails/56.jpg)
ΠANA SAYFA