[ keisti ] Rymano integralas pasižymi tokiomis savybėmis, kurias gana lengva suprasti, laikant integralą plotu.
∫
a
a
f
(
x
)
d
x
=
0.
{\displaystyle \int _{a}^{a}f(x){\mathsf {d}}x=0.}
Jei
b
<
a
{\displaystyle b<a}
∫
a
b
f
(
x
)
d
x
=
−
∫
b
a
f
(
x
)
d
x
{\displaystyle \int _{a}^{b}f(x){\mathsf {d}}x=-\int _{b}^{a}f(x){\mathsf {d}}x}
Δ
x
i
{\displaystyle \Delta x_{i}}
Jei
c
∈
[
a
;
b
]
{\displaystyle c\in [a;b]}
∫
a
c
f
(
x
)
d
x
+
∫
c
b
f
(
x
)
d
x
=
∫
a
b
f
(
x
)
d
x
{\displaystyle \int _{a}^{c}f(x){\mathsf {d}}x+\int _{c}^{b}f(x){\mathsf {d}}x=\int _{a}^{b}f(x){\mathsf {d}}x}
c
{\displaystyle c}
Jei
f
(
x
)
{\displaystyle f(x)}
g
(
x
)
{\displaystyle g(x)}
f
(
x
)
g
(
x
)
{\displaystyle f(x)g(x)}
∫
a
b
(
f
(
x
)
+
g
(
x
)
)
d
x
=
∫
a
b
f
(
x
)
d
x
+
∫
a
b
g
(
x
)
d
x
.
{\displaystyle \int _{a}^{b}(f(x)+g(x)){\mathsf {d}}x=\int _{a}^{b}f(x){\mathsf {d}}x+\int _{a}^{b}g(x){\mathsf {d}}x.}
Skaičiuoti Rymano integralą naudojantis apibrėžimu ne visada įmanoma, be to, tai yra labai sudėtinga. Dažniausiai praktikoje naudojama Niutono-Leibnico formulė , kuri sieja neapibrėžtinį integralą su apibrėžtiniu, nors iš esmės tai yra visiškai skirtingi dalykai:
∫
a
b
f
(
x
)
d
x
=
F
(
x
)
|
a
b
=
F
(
b
)
−
F
(
a
)
.
{\displaystyle \int _{a}^{b}f(x){\mathsf {d}}x=F(x)\vert _{a}^{b}=F(b)-F(a).}
Čia
F
(
x
)
{\displaystyle F(x)}
f
(
x
)
{\displaystyle f(x)}
pirmykščių funkcijų . Pavyzdžiui, rasime integralą
∫
0
1
x
2
d
x
{\displaystyle \int _{0}^{1}x^{2}{\mathsf {d}}x}
parabolės šaka, apribota parabole
f
(
x
)
=
x
2
{\displaystyle f(x)=x^{2}}
Ox koordinačių ašimi ir tiese, statmena Ox ašiai ir kertančią ją taške
x
=
b
=
1.
{\displaystyle x=b=1.}
Iš pradžių surandame:
∫
x
2
d
x
=
x
2
+
1
2
+
1
+
C
=
x
3
3
+
C
.
{\displaystyle \int x^{2}{\mathsf {d}}x={\frac {x^{2+1}}{2+1}}+C={\frac {x^{3}}{3}}+C.}
Tada į šitą formulę įstatome a ir b reikšmes:
F
(
b
)
=
1
3
3
+
C
=
1
3
+
C
.
{\displaystyle F(b)={\frac {1^{3}}{3}}+C={\frac {1}{3}}+C.}
F
(
a
)
=
0
3
3
+
C
=
C
.
{\displaystyle F(a)={\frac {0^{3}}{3}}+C=C.}
Tada atimame F(a) iš F(b):
F
(
b
)
−
F
(
a
)
=
∫
0
1
x
2
d
x
=
1
3
3
−
0
3
3
=
1
3
.
{\displaystyle F(b)-F(a)=\int _{0}^{1}x^{2}{\mathsf {d}}x={\frac {1^{3}}{3}}-{\frac {0^{3}}{3}}={\frac {1}{3}}.}
Radome plotą esantį po dešine parabolės šaka, kuris apribotas, šiuo atveju, tik viena iš dešinės pusės statmena x ašiai tiese (kurios ilgis yra
1
2
{\displaystyle 1^{2}}
∫
0
1
e
x
d
x
4
e
2
x
+
12
e
x
+
34
=
∫
1
e
d
t
4
t
2
+
12
t
+
34
=
∫
1
e
d
t
(
2
t
+
3
)
2
+
25
=
1
2
∫
5
2
e
+
3
d
u
u
2
+
25
=
{\displaystyle \int _{0}^{1}{e^{x}dx \over 4e^{2x}+12e^{x}+34}=\int _{1}^{e}{dt \over 4t^{2}+12t+34}=\int _{1}^{e}{dt \over (2t+3)^{2}+25}={1 \over 2}\int _{5}^{2e+3}{du \over u^{2}+25}=}
=
1
10
arctan
u
5
|
5
2
e
+
3
=
1
10
arctan
2
e
+
3
5
−
π
40
,
{\displaystyle ={1 \over 10}\arctan {u \over 5}\vert _{5}^{2e+3}={\frac {1}{10}}\arctan {2e+3 \over 5}-{\pi \over 40},}
t
=
e
x
;
{\displaystyle t=e^{x};}
d
t
=
e
x
d
x
;
{\displaystyle dt=e^{x}dx;}
a
=
e
0
=
1
;
{\displaystyle a=e^{0}=1;}
b
=
e
1
=
e
;
{\displaystyle b=e^{1}=e;}
u
=
2
t
+
3
;
{\displaystyle u=2t+3;}
d
u
=
2
d
t
.
{\displaystyle du=2dt.}
∫
a
b
u
d
v
=
u
v
|
a
b
−
∫
a
b
v
d
u
.
{\displaystyle \int _{a}^{b}u\;dv=uv\vert _{a}^{b}-\int _{a}^{b}v\;du.}
∫
0
π
2
x
sin
x
d
x
=
−
x
cos
x
|
0
π
2
+
∫
0
π
2
cos
x
d
x
=
sin
x
|
0
π
2
=
1
,
{\displaystyle \int _{0}^{\pi \over 2}x\sin x\;dx=-x\cos x\vert _{0}^{\pi \over 2}+\int _{0}^{\pi \over 2}\cos x\;dx=\sin x\vert _{0}^{\pi \over 2}=1,}
x
=
u
;
{\displaystyle x=u;}
sin
x
d
x
=
d
v
;
{\displaystyle \sin xdx=dv;}
d
x
=
d
u
;
{\displaystyle dx=du;}
−
cos
x
=
v
.
{\displaystyle -\cos x=v.}
Apskaičiuosime 100 cm ilgio strypo masę, kai jo ilginis tankis
γ
(
x
)
=
2
+
0.001
x
2
{\displaystyle \gamma (x)=2+0.001x^{2}}
(
g
/
c
m
)
.
{\displaystyle (g/cm).}
m
=
∫
0
100
(
2
+
0.001
x
2
)
d
x
=
(
2
x
+
0.001
3
x
3
)
|
0
100
=
200
+
1000
3
=
533
1
3
(
g
)
.
{\displaystyle m=\int _{0}^{100}(2+0.001x^{2})dx=(2x+{0.001 \over 3}x^{3})|_{0}^{100}=200+{1000 \over 3}=533{1 \over 3}\;(g).}
∫
0
1
1
−
x
2
d
x
.
{\displaystyle \int _{0}^{1}{\sqrt {1-x^{2}}}dx.}
x
=
sin
t
,
{\displaystyle x=\sin t,}
d
x
=
cos
t
d
t
.
{\displaystyle dx=\cos tdt.}
sin
t
=
0
{\displaystyle \sin t=0}
t
=
0
{\displaystyle t=0}
sin
t
=
1
,
{\displaystyle \sin t=1,}
t
=
π
2
,
{\displaystyle t={\pi \over 2},}
∫
0
1
1
−
x
2
d
x
=
∫
0
π
2
1
−
sin
2
t
cos
t
d
t
=
∫
0
π
2
cos
2
d
t
=
∫
0
π
2
1
+
cos
(
2
t
)
2
d
t
=
π
4
+
1
4
sin
(
2
t
)
|
0
π
2
=
π
4
.
{\displaystyle \int _{0}^{1}{\sqrt {1-x^{2}}}dx=\int _{0}^{\pi \over 2}{\sqrt {1-\sin ^{2}t}}\cos tdt=\int _{0}^{\pi \over 2}\cos ^{2}dt=\int _{0}^{\pi \over 2}{1+\cos(2t) \over 2}dt={\pi \over 4}+{1 \over 4}\sin(2t)|_{0}^{\pi \over 2}={\pi \over 4}.}
Parabolės. Apskaičiuosime figūros, apribotos kreivių
y
=
x
2
{\displaystyle y=x^{2}}
y
=
x
1
/
2
{\displaystyle y=x^{1/2}}
Pirmiausia turime rasti tų kreivių susikirtimo taškų abscises. Tuo tikslu sprendžiame lygtį
x
2
=
x
1
/
2
{\displaystyle x^{2}=x^{1/2}}
x
1
=
0
,
{\displaystyle x_{1}=0,}
x
2
=
1.
{\displaystyle x_{2}=1.}
S
=
∫
0
1
(
x
−
x
2
)
d
x
=
2
3
x
3
2
|
0
1
−
x
3
3
|
0
1
=
2
3
−
1
3
=
1
3
.
{\displaystyle S=\int _{0}^{1}({\sqrt {x}}-x^{2})dx={2 \over 3}x^{3 \over 2}|_{0}^{1}-{x^{3} \over 3}|_{0}^{1}={2 \over 3}-{1 \over 3}={1 \over 3}.}
Elipsė. Apskaičiuokime figūros, apribotos elipse
x
2
a
2
+
y
2
b
2
=
1
{\displaystyle {x^{2} \over a^{2}}+{y^{2} \over b^{2}}=1}
Apskaičiuokime plotą tos figūros dalies, kuri yra pirmajame ketvirtyje, po to gautą rezultatą padauginsime iš 4. Elipsės kanonine lygtį pakeičiame parametrinėmis lygtimis
x
=
a
cos
t
,
{\displaystyle x=a\cos t,}
y
=
sin
t
.
{\displaystyle y=\sin t.}
x kinta nuo 0 iki a , todėl t kinta nuo
π
2
{\displaystyle {\pi \over 2}}
t reikšmes gavome, įrašę į lygtį
x
=
a
cos
t
{\displaystyle x=a\cos t}
x jo reikšmes 0 ir a ). Į formulę
S
=
∫
a
b
y
d
x
{\displaystyle S=\int _{a}^{b}ydx}
y įrašykime
y
=
b
sin
t
,
{\displaystyle y=b\sin t,}
d
x
{\displaystyle dx}
d
(
a
cos
t
)
=
−
a
sin
t
d
t
,
{\displaystyle d(a\cos t)=-a\sin tdt,}
x
=
a
cos
t
.
{\displaystyle x=a\cos t.}
S
=
−
4
∫
π
/
2
0
b
sin
t
a
sin
t
d
t
=
4
a
b
∫
0
π
/
2
sin
2
t
d
t
=
2
a
b
∫
0
π
/
2
(
1
−
cos
(
2
t
)
)
d
t
=
{\displaystyle S=-4\int _{\pi /2}^{0}b\sin ta\sin tdt=4ab\int _{0}^{\pi /2}\sin ^{2}tdt=2ab\int _{0}^{\pi /2}(1-\cos(2t))dt=}
=
2
a
b
[
π
2
−
∫
0
π
/
2
cos
(
2
t
)
2
d
(
2
t
)
]
=
2
a
b
[
π
2
−
sin
(
2
t
)
2
|
0
π
/
2
]
=
π
a
b
.
{\displaystyle =2ab[{\pi \over 2}-\int _{0}^{\pi /2}{\cos(2t) \over 2}d(2t)]=2ab[{\pi \over 2}-{\sin(2t) \over 2}|_{0}^{\pi /2}]=\pi ab.}
Apskaičiuosime kūno, apriboto paraboloido
z
=
x
2
+
3
2
y
2
{\displaystyle z=x^{2}+{3 \over 2}y^{2}}
z
=
4
{\displaystyle z=4}
Jei paraboloidą kirstume plokštuma
z
=
c
o
n
s
t
,
{\displaystyle z=const,}
x
2
+
3
2
y
2
=
z
,
{\displaystyle x^{2}+{3 \over 2}y^{2}=z,}
x
2
z
+
y
2
2
3
z
=
1.
{\displaystyle {x^{2} \over z}+{y^{2} \over {2 \over 3}z}=1.}
a
=
z
,
b
=
2
z
3
.
{\displaystyle a={\sqrt {z}},\;b={\sqrt {2z \over 3}}.}
Q
(
z
)
=
π
a
b
{\displaystyle Q(z)=\pi ab}
Q
(
z
)
=
π
z
⋅
2
z
3
=
π
z
2
3
.
{\displaystyle Q(z)=\pi {\sqrt {z}}\cdot {\sqrt {2z \over 3}}=\pi z{\sqrt {2 \over 3}}.}
V
=
∫
0
4
π
2
3
z
d
z
=
π
2
3
⋅
z
2
2
|
0
4
=
8
π
6
3
.
{\displaystyle V=\int _{0}^{4}\pi {\sqrt {2 \over 3}}zdz=\pi {\sqrt {2 \over 3}}\cdot {z^{2} \over 2}|_{0}^{4}={8\pi {\sqrt {6}} \over 3}.}
Plotas apribotas parabolės ir tiesės . Apskaičiuosime plotą figūros, apribotos grafikais funkcijų
y
=
f
1
(
x
)
=
x
{\displaystyle y=f_{1}(x)=x}
y
=
f
2
(
x
)
=
2
−
x
2
.
{\displaystyle y=f_{2}(x)=2-x^{2}.}
Rasime abscises taškų susikirtimo tiesės
y
=
x
{\displaystyle y=x}
prabole
y
=
2
−
x
2
.
{\displaystyle y=2-x^{2}.}
x
=
2
−
x
2
,
{\displaystyle x=2-x^{2},}
−
x
2
−
x
+
2
=
0
,
{\displaystyle -x^{2}-x+2=0,}
D
=
b
2
−
4
a
c
=
(
−
1
)
2
−
4
(
−
1
)
2
=
9
,
{\displaystyle D=b^{2}-4ac=(-1)^{2}-4(-1)2=9,}
x
1
,
2
=
−
b
±
D
2
a
=
1
±
3
−
2
=
−
2
;
1
,
{\displaystyle x_{1,2}={-b\pm {\sqrt {D}} \over 2a}={1\pm 3 \over -2}=-2;1,}
x
1
=
−
2
,
{\displaystyle x_{1}=-2,}
x
2
=
1.
{\displaystyle x_{2}=1.}
s
=
∫
−
2
1
[
f
2
(
x
)
−
f
1
(
x
)
]
d
x
=
∫
−
2
1
[
(
2
−
x
2
)
−
x
]
d
x
=
(
2
x
−
x
3
3
−
x
2
2
)
|
−
2
1
=
{\displaystyle s=\int _{-2}^{1}[f_{2}(x)-f_{1}(x)]dx=\int _{-2}^{1}[(2-x^{2})-x]dx=(2x-{x^{3} \over 3}-{x^{2} \over 2})|_{-2}^{1}=}
=
(
2
−
1
3
−
1
2
)
−
(
−
4
+
8
3
−
2
)
=
8
−
3
−
1
2
=
9
2
.
{\displaystyle =(2-{1 \over 3}-{1 \over 2})-(-4+{8 \over 3}-2)=8-3-{1 \over 2}={9 \over 2}.}
Tą patį plotą apribota parabole
y
=
2
−
x
2
{\displaystyle y=2-x^{2}}
y
=
x
{\displaystyle y=x}
x ašimi susikirtimo tašką parabolės
2
−
x
2
=
0
,
{\displaystyle 2-x^{2}=0,}
2
=
x
2
,
{\displaystyle 2=x^{2},}
x
=
±
2
.
{\displaystyle x=\pm {\sqrt {2}}.}
−
2
≤
x
≤
1
:
{\displaystyle -{\sqrt {2}}\leq x\leq 1:}
S
1
=
∫
−
2
1
(
2
−
x
2
)
d
x
=
(
2
x
−
x
3
3
)
|
−
2
1
=
2
−
1
3
−
(
−
2
2
+
2
2
3
)
=
5
3
+
4
2
3
;
{\displaystyle S_{1}=\int _{-{\sqrt {2}}}^{1}(2-x^{2})dx=(2x-{x^{3} \over 3})|_{-{\sqrt {2}}}^{1}=2-{1 \over 3}-(-2{\sqrt {2}}+{2{\sqrt {2}} \over 3})={5 \over 3}+{4{\sqrt {2}} \over 3};}
Dabar surandame plotą po parabole nuo
x
=
−
2
{\displaystyle x=-2}
x
=
−
2
1
/
2
:
{\displaystyle x=-2^{1/2}:}
|
S
2
|
=
|
∫
−
2
−
2
(
2
−
x
2
)
d
x
|
=
|
(
2
x
−
x
3
3
)
|
−
2
−
2
|
=
|
−
2
2
+
2
2
3
−
(
−
4
+
8
3
)
|
=
|
−
4
2
3
+
4
3
|
=
4
2
3
−
4
3
;
{\displaystyle |S_{2}|=|\int _{-2}^{-{\sqrt {2}}}(2-x^{2})dx|=|(2x-{x^{3} \over 3})|_{-2}^{-{\sqrt {2}}}|=|-2{\sqrt {2}}+{2{\sqrt {2}} \over 3}-(-4+{8 \over 3})|=|-{4{\sqrt {2}} \over 3}+{4 \over 3}|={4{\sqrt {2}} \over 3}-{4 \over 3};}
Dabar iš pirmo ploto po parabole atimame trikampio plotą:
S
V
=
S
1
−
S
Δ
1
=
5
3
+
4
2
3
−
1
2
=
7
6
+
4
2
3
;
{\displaystyle S_{V}=S_{1}-S_{\Delta _{1}}={5 \over 3}+{4{\sqrt {2}} \over 3}-{1 \over 2}={7 \over 6}+{4{\sqrt {2}} \over 3};}
Apatinį (ieškomą) plotą trečiajame ketvirtyje gauname atėmę
|
S
2
|
{\displaystyle |S_{2}|}
S
A
=
S
Δ
2
−
|
S
2
|
=
2
2
2
−
(
4
2
3
−
4
3
)
=
10
3
−
4
2
3
;
{\displaystyle S_{A}=S_{\Delta _{2}}-|S_{2}|={2^{2} \over 2}-({4{\sqrt {2}} \over 3}-{4 \over 3})={10 \over 3}-{4{\sqrt {2}} \over 3};}
Susumavę viršutinį (virš ašies Ox ) ieškomą plotą
S
V
{\displaystyle S_{V}}
Ox ) ieškomą plotą
S
A
{\displaystyle S_{A}}
S
=
S
V
+
S
A
=
(
7
6
+
4
2
3
)
+
(
10
3
−
4
2
3
)
=
81
18
=
9
2
.
{\displaystyle S=S_{V}+S_{A}=({7 \over 6}+{4{\sqrt {2}} \over 3})+({10 \over 3}-{4{\sqrt {2}} \over 3})={81 \over 18}={9 \over 2}.}
10.16. Pavyzdis . Skritulys, apribotas apskritimo
x
2
+
(
y
−
a
)
2
=
b
2
{\displaystyle x^{2}+(y-a)^{2}=b^{2}}
Ox (10.16 pav). Apskaičiuokime gauto sukinio, vadinamo toru , tūrį. Čia a yra atstumas nuo koordinačių pradžios taško O iki skritulio centro, kuris sukamas aplink ašį Ox . Šio skritulio spindulys lygus b .Sprendimas . Toro tūris lygus dviejų sukinių tūrių skirtumui: pirmasis sukinys gaunamas sukant kreivinę trapeciją ABCDE , o antrasis - trapeciją ABFDE . Išsprendžiame lygį
x
2
+
(
y
−
a
)
2
=
b
2
{\displaystyle x^{2}+(y-a)^{2}=b^{2}}
y atžvilgiu:
(
y
−
a
)
2
=
b
2
−
x
2
,
{\displaystyle (y-a)^{2}=b^{2}-x^{2},}
y
−
a
=
±
b
2
−
x
2
,
{\displaystyle y-a=\pm {\sqrt {b^{2}-x^{2}}},}
y
=
a
±
b
2
−
x
2
.
{\displaystyle y=a\pm {\sqrt {b^{2}-x^{2}}}.}
Lanko BCD lygtis
y
1
=
a
+
b
2
−
x
2
,
{\displaystyle y_{1}=a+{\sqrt {b^{2}-x^{2}}},}
BFD lygtis
y
2
=
a
−
b
2
−
x
2
.
{\displaystyle y_{2}=a-{\sqrt {b^{2}-x^{2}}}.}
Tuomet toro tūris bus lygus
V
=
2
π
(
∫
0
b
y
1
2
d
x
−
∫
0
b
y
2
2
d
x
)
=
2
π
∫
0
b
(
y
1
2
−
y
2
2
)
d
x
=
2
π
∫
0
b
(
(
a
+
b
2
−
x
2
)
2
−
(
a
−
b
2
−
x
2
)
2
)
d
x
=
{\displaystyle V=2\pi \left(\int _{0}^{b}y_{1}^{2}dx-\int _{0}^{b}y_{2}^{2}dx\right)=2\pi \int _{0}^{b}(y_{1}^{2}-y_{2}^{2})dx=2\pi \int _{0}^{b}\left((a+{\sqrt {b^{2}-x^{2}}})^{2}-(a-{\sqrt {b^{2}-x^{2}}})^{2}\right)dx=}
=
2
π
∫
0
b
(
(
a
2
+
2
a
b
2
−
x
2
+
b
2
−
x
2
)
−
(
a
2
−
2
a
b
2
−
x
2
+
b
2
−
x
2
)
)
d
x
=
2
π
∫
0
b
4
a
b
2
−
x
2
d
x
.
{\displaystyle =2\pi \int _{0}^{b}\left((a^{2}+2a{\sqrt {b^{2}-x^{2}}}+b^{2}-x^{2})-(a^{2}-2a{\sqrt {b^{2}-x^{2}}}+b^{2}-x^{2})\right)dx=2\pi \int _{0}^{b}4a{\sqrt {b^{2}-x^{2}}}dx.}
Pažymėkime:
x
=
b
sin
t
{\displaystyle x=b\sin t}
d
x
=
b
cos
(
t
)
d
t
{\displaystyle dx=b\cos(t)dt}
V
=
8
π
a
∫
0
b
b
2
−
x
2
d
x
=
8
π
a
∫
0
π
2
b
1
−
sin
2
t
b
cos
(
t
)
d
t
=
8
π
a
b
2
∫
0
π
2
cos
2
t
cos
(
t
)
d
t
=
{\displaystyle V=8\pi a\int _{0}^{b}{\sqrt {b^{2}-x^{2}}}dx=8\pi a\int _{0}^{\pi \over 2}b{\sqrt {1-\sin ^{2}t}}b\cos(t)dt=8\pi ab^{2}\int _{0}^{\pi \over 2}{\sqrt {\cos ^{2}t}}\cos(t)dt=}
=
8
π
a
b
2
∫
0
π
2
cos
2
(
t
)
d
t
=
8
π
a
b
2
⋅
1
!
!
2
!
!
⋅
π
2
=
2
π
2
a
b
2
.
{\displaystyle =8\pi ab^{2}\int _{0}^{\pi \over 2}\cos ^{2}(t)dt=8\pi ab^{2}\cdot {\frac {1!!}{2!!}}\cdot {\frac {\pi }{2}}=2\pi ^{2}ab^{2}.}
Čia pasinaudojome dvigubu faktorialu .
Parinkime a =4, b =3. Tuomet
V
=
2
π
2
a
b
2
=
2
π
2
⋅
4
⋅
3
2
=
2
π
2
⋅
36
=
72
π
2
=
710.6115169.
{\displaystyle V=2\pi ^{2}ab^{2}=2\pi ^{2}\cdot 4\cdot 3^{2}=2\pi ^{2}\cdot 36=72\pi ^{2}=710.6115169.}
Pasinaudodami integralų lentele, turime:
∫
a
2
−
x
2
d
x
=
x
2
a
2
−
x
2
+
a
2
2
arctan
x
a
2
−
x
2
+
C
.
{\displaystyle \int {\sqrt {a^{2}-x^{2}}}\;{\mathsf {d}}x={\frac {x}{2}}{\sqrt {a^{2}-x^{2}}}+{\frac {a^{2}}{2}}\arctan {\frac {x}{\sqrt {a^{2}-x^{2}}}}+C.}
x kinta nuo 0 iki b ir rasti toro tūrį:
V
=
8
a
π
∫
0
b
b
2
−
x
2
d
x
=
8
a
π
(
x
2
b
2
−
x
2
+
b
2
2
arctan
x
b
2
−
x
2
)
|
0
b
=
8
a
π
(
b
2
b
2
−
b
2
+
b
2
2
arctan
b
b
2
−
b
2
)
−
8
a
π
(
0
+
b
2
2
arctan
0
)
=
{\displaystyle V=8a\pi \int _{0}^{b}{\sqrt {b^{2}-x^{2}}}\;{\mathsf {d}}x=8a\pi {\Big (}{\frac {x}{2}}{\sqrt {b^{2}-x^{2}}}+{\frac {b^{2}}{2}}\arctan {\frac {x}{\sqrt {b^{2}-x^{2}}}}{\Big )}|_{0}^{b}=8a\pi {\Big (}{\frac {b}{2}}{\sqrt {b^{2}-b^{2}}}+{\frac {b^{2}}{2}}\arctan {\frac {b}{\sqrt {b^{2}-b^{2}}}}{\Big )}-8a\pi (0+{\frac {b^{2}}{2}}\arctan 0)=}
=
8
a
π
(
0
+
b
2
2
arctan
b
0
)
=
8
a
π
⋅
b
2
2
arctan
∞
=
4
a
b
2
π
arctan
∞
=
4
a
b
2
π
⋅
π
2
=
2
a
b
2
π
2
.
{\displaystyle =8a\pi {\Big (}0+{\frac {b^{2}}{2}}\arctan {\frac {b}{0}}{\Big )}=8a\pi \cdot {\frac {b^{2}}{2}}\arctan \infty =4ab^{2}\pi \arctan \infty =4ab^{2}\pi \cdot {\frac {\pi }{2}}=2ab^{2}\pi ^{2}.}
Kadangi
tan
π
2
=
sin
π
2
cos
π
2
=
1
0
=
∞
,
{\displaystyle \tan {\frac {\pi }{2}}={\frac {\sin {\frac {\pi }{2}}}{\cos {\frac {\pi }{2}}}}={\frac {1}{0}}=\infty ,}
arctan
∞
=
π
2
.
{\displaystyle \arctan \infty ={\frac {\pi }{2}}.}
arctan
0
=
0.
{\displaystyle \arctan 0=0.}
Pasinaudodami trigonometrija išintegruokime be dvigubo faktorialo, taigi,
cos
(
2
A
)
=
2
cos
2
A
−
1
;
{\displaystyle \cos(2A)=2\cos ^{2}A-1;}
cos
2
A
=
cos
(
2
A
)
+
1
2
.
{\displaystyle \cos ^{2}A={\frac {\cos(2A)+1}{2}}.}
a =4, b =3, turime:
V
=
8
π
a
b
2
∫
0
π
2
cos
2
(
t
)
d
t
=
8
π
⋅
4
⋅
3
2
∫
0
π
2
cos
(
2
t
)
+
1
2
d
t
=
16
π
⋅
9
∫
0
π
2
(
cos
(
2
t
)
+
1
)
d
t
=
144
π
[
∫
0
π
2
cos
(
2
t
)
d
t
+
∫
0
π
2
d
t
]
=
{\displaystyle V=8\pi ab^{2}\int _{0}^{\pi \over 2}\cos ^{2}(t)dt=8\pi \cdot 4\cdot 3^{2}\int _{0}^{\pi \over 2}{\frac {\cos(2t)+1}{2}}dt=16\pi \cdot 9\int _{0}^{\pi \over 2}(\cos(2t)+1)dt=144\pi [\int _{0}^{\pi \over 2}\cos(2t)dt+\int _{0}^{\pi \over 2}dt]=}
=
144
π
[
∫
0
π
2
cos
(
2
t
)
d
(
2
t
)
2
+
π
2
]
=
144
π
[
1
2
⋅
sin
(
2
t
)
|
0
π
2
+
π
2
]
=
72
π
[
sin
(
2
t
)
|
0
π
2
+
π
]
=
72
π
[
sin
(
2
⋅
π
2
)
−
sin
(
2
⋅
0
)
+
π
]
=
72
π
2
=
710.6115169.
{\displaystyle =144\pi [\int _{0}^{\pi \over 2}\cos(2t){\frac {d(2t)}{2}}+{\pi \over 2}]=144\pi [{\frac {1}{2}}\cdot \sin(2t)|_{0}^{\pi \over 2}+{\pi \over 2}]=72\pi [\sin(2t)|_{0}^{\pi \over 2}+\pi ]=72\pi [\sin(2\cdot {\frac {\pi }{2}})-\sin(2\cdot 0)+\pi ]=72\pi ^{2}=710.6115169.}
![{\displaystyle c\in [a;b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/02b8df5ba00bbc02820be401c0dbc93d73392da8)
![{\displaystyle =2ab[{\pi \over 2}-\int _{0}^{\pi /2}{\cos(2t) \over 2}d(2t)]=2ab[{\pi \over 2}-{\sin(2t) \over 2}|_{0}^{\pi /2}]=\pi ab.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a00ce7bcdb759fdb8d580441c1fb7f1bcb710ee)
![{\displaystyle s=\int _{-2}^{1}[f_{2}(x)-f_{1}(x)]dx=\int _{-2}^{1}[(2-x^{2})-x]dx=(2x-{x^{3} \over 3}-{x^{2} \over 2})|_{-2}^{1}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5364e0c9fbeccecb16c0677a99344edd3db43f1e)
![{\displaystyle V=8\pi ab^{2}\int _{0}^{\pi \over 2}\cos ^{2}(t)dt=8\pi \cdot 4\cdot 3^{2}\int _{0}^{\pi \over 2}{\frac {\cos(2t)+1}{2}}dt=16\pi \cdot 9\int _{0}^{\pi \over 2}(\cos(2t)+1)dt=144\pi [\int _{0}^{\pi \over 2}\cos(2t)dt+\int _{0}^{\pi \over 2}dt]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08c2cbf6a5d138b11dce03f7f04d6848174684f7)
![{\displaystyle =144\pi [\int _{0}^{\pi \over 2}\cos(2t){\frac {d(2t)}{2}}+{\pi \over 2}]=144\pi [{\frac {1}{2}}\cdot \sin(2t)|_{0}^{\pi \over 2}+{\pi \over 2}]=72\pi [\sin(2t)|_{0}^{\pi \over 2}+\pi ]=72\pi [\sin(2\cdot {\frac {\pi }{2}})-\sin(2\cdot 0)+\pi ]=72\pi ^{2}=710.6115169.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3fc2daf2a09eb9eb66178a58ba282a19c0913e9c)