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Integralszamıtas I.
Az eloadashoz kapcsolodo feladatsor megoldokulcsa
Vegezze el a kovetkezo integralasokat:
1. Alapintegralok
(a)∫xα dx =
xα+1
α + 1+ C (α 6= −1)
pl.:∫x2 dx =
x3
3+ C
∫x dx =
x2
2+ C
∫ √x dx =
∫x
12 dx =
x32
32
+ C
(b)
∫1
xdx = ln |x|+ C
(c)∫
sinx dx = − cosx+ C
(d)∫
cosx dx = sinx+ C
(e)∫ex dx = ex + C
(f)∫ax dx =
ax
ln a+ C pl.:
∫2x dx =
2x
ln 2+ C
(g)
∫1
sin2 xdx = − ctg x+ C
(h)
∫1
cos2 xdx = tg x+ C
2. (a)2∫1
x5 dx =[x6
6
]2
1= 26
6− 16
6= 64
6− 1
6= 10, 5
(b)
4∫2
1
xdx =
[ln |x|
]4
2= ln | 4| − ln | 2| = ln 4− ln 2 ≈ 0, 69
(c)
π2∫
0
sinx dx =[− cosx
]π2
0=(− cos π
2
)− (− cos 0) = (0)− (−1) = 1
3.∫f(ax+ b) dx =
F (ax+ b)
a+ C tıpus, ahol a 6= 0 ,
∫f(x) dx alapintegral es
∫f(x) dx = F (x) + C
(a)∫
(3x− 1)2 dx =(3x− 1)3
3 · 3+ C =
1
9(3x− 1)3 + C (itt: f(x) = x2 F (x) = x3
3a = 3 b = −1 )
(b)∫
sin 2x dx =− cos 2x
2+ C (itt: f(x) = sinx F (x) = − cosx a = 2 b = 0 )
(c)∫
cos 3x dx =sin 3x
3+ C
(d)∫ex+3 dx = ex+3 + C
(e)∫
3−2x+1 dx =3−2x+1
(−2) · ln 3+ C
1
(f)
∫3
2x− 5dx = 3
∫1
2x− 5dx = 3 · ln | 2x− 5|
2+ C
(g)∫ −4
3 sin2(1+3x)dx = −4
3
∫1
sin2(3x+1)dx = −4
3· − ctg(3x+1)
3+ C = 4
9ctg(3x+ 1) + C
(h)∫
1−2 cos2(−x+12)
dx = −12
∫1
cos2(−x+12)dx = −1
2· tg(−x+12)
−1+ C = 1
2tg(−x+ 12) + C
(i)
∫3√
2x− 7dx = 3
∫(2x− 7)−
12 dx = 3 · (2x− 7)
12
12· 2
+ C = 3√
2x− 7 + C
4. (a)
4∫0
(2x+ 3)6 dx =
[(2x+ 3)7
7 · 2
]4
0
=117
14− 37
14≈ 1391785
(b)
π4∫
0
cos 2x dx =
[sin 2x
2
]π4
0
=sin π
2
2− sin 0
2=
1
2− 0
2= 0, 5
(c)
1∫−1
5−2x+7 dx =
[5−2x+7
(−2) · ln 5
]1
−1
= − 1
2 ln 5
(55 − 59
)≈ 605802
5.∫fα(x) · f ′(x) dx =
fα+1(x)
α + 1+ C tıpus, ahol α 6= −1
(a)
∫x2(2x3 − 1)12 dx =
1
6
∫6x2︸︷︷︸f ′
(2x3 − 1)12︸ ︷︷ ︸f12
dx =1
6· (2x3 − 1)13
13+ C
(b)
∫−2 sin6 x cosx dx = −2 · sin7 x
7+ C
(c)
∫3 sin 2x cos5 2x dx = −3
2
∫(−2 sin 2x)︸ ︷︷ ︸
f ′
cos5 2x︸ ︷︷ ︸f5
dx = −3
2· cos6 2x
6+ C
msz.: (cos 2x)′ = −2 sin 2x
(d)
∫x3 · 5√−x4 + 11 dx = −1
4
∫(−4x3)︸ ︷︷ ︸
f ′
· (−x4 + 11)15︸ ︷︷ ︸
f15
dx = −1
4· (−x4 + 11)
65
65
+ C
(e)
∫tg6 2x
cos2 2xdx =
1
2
∫2
cos2 2x· tg6 2x dx =
1
2· tg7 2x
7+ C
msz.: (tg 2x)′ =2
cos2 2x
(f)
∫arctg11 2x
4x2 + 1dx =
1
2
∫2
1 + 4x2· arctg11 2x dx =
1
2· arctg12 2x
12+ C
msz.: (arctg 2x)′ =2
1 + (2x)2=
2
1 + 4x2
6. (a)
3∫2
x · (x2 + 3)4 dx =
[(x2 + 3)5
10
]3
2
=125
10− 75
10= 23202, 5
2
msz.:∫x · (x2 + 3)4 dx = 1
2
∫2x︸︷︷︸f ′
· (x2 + 3)4︸ ︷︷ ︸f4
dx = 12· (x2+3)5
5+ C
(b)
π3∫
π4
5 sin2 x cosx dx =5
3
[sin3 x
]π3
π4
=5
3
(sin3 π
3− sin3 π
4
)≈ 0, 49
7.∫f ′(x)
f(x)dx = ln |f(x)|+ C tıpus
(a)
π4∫
0
tg x dx =[− ln | cosx|
]π4
0=(− ln | cos π
4|)− (− ln | cos 0|) = − ln
√2
2≈ 0, 35
msz.:
∫tg x dx =
∫sinx
cosxdx = −
∫− sinx
cosx︸ ︷︷ ︸f ′f
dx = − ln | cosx|+ C
(b)
∫ctg x dx =
∫cosx
sinxdx = ln | sinx|+ C
(c)
1∫0
x
x2 + 2dx =
1
2
1∫0
2x
x2 + 2dx =
[1
2ln |x2 + 2|
]1
0
=1
2ln 3− 1
2ln 2 ≈ 0, 2
(d)
∫sinx
2 + cos xdx = −
∫− sinx
2 + cos xdx = − ln | 2 + cos x|+ C
(e)
∫3
(1 + x2) arctg xdx = 3
∫ 11+x2
arctg xdx = 3 ln |arctg x|+ C
(f)
∫− 8
(arcsinx)√
1− x2dx = −8
∫ 1√1−x2
arcsinxdx = −8 ln |arcsinx|+ C
8.∫g′(x) · f(g(x)) dx = F (g(x)) + C tıpus, ahol
∫f(x) dx = F (x) + C
(a)∫
(x+ 1) · ex2+2x dx = 12
∫(2x+ 2)︸ ︷︷ ︸
g′(x)
· ex2+2x︸ ︷︷ ︸f(g(x))
dx = 12ex
2+2x + C
itt: f(x) = ex g(x) = x2 + 2x g′(x) = 2x+ 2 F (x) = ex
(b)
∫sin(lnx)
xdx =
∫1
x︸︷︷︸g′(x)
· sin(lnx)︸ ︷︷ ︸f(g(x))
dx = − cos(lnx) + C
(c)
1∫0
ex
1 + e2xdx =
[arctg(ex)
]1
0= arctg e− arctg 1 ≈ 0, 43
msz.:
∫ex
1 + e2xdx =
∫ex · 1
1 + (ex)2dx = arctg(ex) + C
(d)∫x2 cos(x3 − 1) dx = 1
3
∫3x2 · cos(x3 − 1) dx = 1
3sin(x3 − 1) + C
3
(e)
∫1
√x√
1− xdx =
∫1√x· 1√
1− (√x)2
dx = 2
∫1
2√x· 1√
1− (√x)2
dx = 2 arcsin√x+C
(f)
∫2tg x
cos2 xdx =
∫1
cos2 x· 2tg x dx =
2tg x
ln 2+ C
(g)∫esinx2(x cosx2) dx = 1
2
∫esinx2(2x cosx2) dx = 1
2esinx2 + C
msz.: (sinx2)′ = 2x cosx2
4
Integralszamıtas II.
Parcialis integralas∫uv′ dx = uv −
∫u′v dx
1. (a)∫xex dx =
∫x︸︷︷︸u
ex︸︷︷︸v′
dx = x︸︷︷︸u
ex︸︷︷︸v
−∫
1︸︷︷︸u′
ex︸︷︷︸v
dx = xex − ex + C
u′ = 1 v = ex
(b)∫x2ex dx = ex(x2 − 2x+ 2) + C
(c)∫
(3x2 − 2x)e2x dx =(
32x2 − 5
2x+ 5
4
)e2x + C
(d)∫
(x+ 1)(e2x − 2e−x) dx =(
12x+ 1
4
)e2x + (2x+ 4)e−x + C
(e)∫x sinx dx = −x cosx+ sinx+ C
(f)∫x cosx dx = x sinx+ cosx+ C
(g)∫x2 cos 2x dx = 1
2x2 sin 2x+ 1
2x cos 2x− 1
4sin 2x+ C
(h)∫
(1− x2)(sin 2x− 2 cos 3x) dx =
=(
12x2 − 3
4
)cos 2x+
(23x2 − 22
27
)sin 3x+
(−1
2x)
sin 2x+(
49x)
cos 3x+ C
(i)∫ex sinx dx = 1
2(sinx− cosx)ex + C
(j)∫e2x cosx dx = 1
5e2x(sinx+ 2 cosx) + C
(k)∫e−x sinx cosx dx = e−x
(−1
5cos 2x− 1
10sin 2x
)+ C
(l)∫
2x(sin 2x− 3 cosx) dx =∫
2x sin 2x dx− 3∫
2x cosx dx =
= − 24+(ln 2)2
2x cos 2x+ ln 24+(ln 2)2
2x sin 2x− 2(ln 2)2
4+(ln 2)2− 3
1+(ln 2)22x sinx− 3 ln 2
1+(ln 2)22x cosx+ C
(m)∫x lnx dx = 1
2x2 lnx− 1
4x2 + C
(n)∫
(2x+ 1) ln2 x dx = (x2 + x) ln2 x− (x2 + 2x) lnx+ 12x2 + 2x+ C
(o)∫
arctg x dx = x arctg x− 12
ln(1 + x2) + C
(p)∫
arcsinx dx = x arcsinx+√
1− x2 + C
2. (a)1∫0
xe3x dx =[
13xe3x − 1
9e3x]1
0≈ 4, 6
(b)2∫−1
(x2 − x+ 1) ex dx = [(x2 − 3x+ 4) ex]2−1 ≈ 11, 835
(c)
π2∫
0
3x cos 2x dx =[
32x sin 2x+ 3
4cos 2x
]π2
0= −3
2
(d)
2π3∫π6
(2x+ 3) sin x2
dx =[(2x+ 3)
(−2 cos x
2
)+ 8 sin x
2
] 2π3π6
≈ 5, 49
(e)2∫0
e−2x cos 3x dx = 913
[e−2x
(13
sin 3x− 29
cos 3x)]2
0≈ 0, 15
(f)2∫1
(x2 + 1) lnx dx =[(
13x3 + x
)lnx− 1
9x3 − x
]21≈ 1, 457
5
(g)e∫
1
(x3 − 2) lnx dx =[(
x4
4− 2x
)lnx− x4
16+ 2x
]e1≈ 8, 3
(h)1∫0
arctg x dx =[x arctg x− 1
2ln (1 + x2)
]10≈ 0, 439
(i)
√3∫
0
x arctg x dx =[x2
2arctg x− x
2+ 1
2arctg x
]√3
0≈ 1, 23
Helyettesıteses integralas
1. (a)∫x√x+ 1 dx =
∫(t2 − 1) · t · 2t dt =
∫(2t4 − 2t2) dt
∗=
t =√x+ 1 x = t2 − 1
dx
dt= 2t dx = 2tdt
∗= 2 · t
5
5− 2 · t
3
3+ C = 2 ·
(√x+ 1
)5
5− 2 ·
(√x+ 1
)3
3+ C
(b)∫
x√x+1
dx = 23
√(x+ 1)3 − 2
√x+ 1 + C
(t =√x+ 1)
(c)∫
x+13√x−2
dx = 35
3√
(x− 2)5 + 92
3√
(x− 2)2 + C
(t = 3√x− 2)
(d)∫ (√
x− 2 + x√x−2
)dx = 4
3
√(x− 2)3 + 4
√x− 2 + C
(t =√x− 2)
(e)∫ 3√x+
√x
6√x5
dx =∫ (
x−12 + x−
13
)dx = 2
√x+ 3
2
3√x2 + C
(Tagonkenti osztassal hatvanyfuggvenyek osszege, nincs szukseg helyettesıtesre!)
(f)∫ √
x4√x−1
dx = 45
4√x5 +
4√x4 + 4
3
4√x3 + 2
4√x2 + 4
√x+ ln | 4
√x− 1|+ C
(t = 4√x)
(g)∫
1+√x+ 3√x+ 6√xx+
6√x7
dx =∫
1+t3+t2+tt6+t7
6t5 dt = 6∫ (
t+ 1t
)dt = 3 3
√x+ 6 ln | 6
√x|+ C
(t = 6√x)
(h)∫
ex
e2x+1dx = arctg ex + C
(t = ex)
(i)∫
e2x
ex+1dx = ex − ln(ex + 1) + C
(t = ex)
(j)∫
cos√x dx = 2
√x sin
√x+ 2 cos
√x+ C
(t =√x)
(k)∫
sin 3√x dx = −3
3√x2 cos 3
√x+ 6 cos 3
√x+ 6 3
√x sin 3
√x+ C
(t = 3√x)
(l)∫e√x dx = 2
√xe√x − 2e
√x + C
(t =√x)
6
(m)∫
1x√x2−1
dx = arctg√x2 − 1 + C
(t =√x2 − 1)
(n)∫x2 · 3√
1− x dx = −3 · (1−x)43
4+ 6 · (1−x)
73
7− 3 · (1−x)
103
10+ C
(t = 3√
1− x = (1− x)13 )
(o)∫x3 · (1− 5x2)10 dx = 1
50· (1−5x2)12
12− 1
50· (1−5x2)11
11+ C
(t = (1− 5x2)10)
2. (a)1∫0
x2
(x+1)√x+1
dx =[
23t3 − 4t− 2
t
]√2
1=[
23
√(x+ 1)3 − 4
√x+ 1− 2√
x+1
]1
0≈ 0, 148
(t =√x+ 1)
(b)1∫0
e2x+ex
ex+2dx = [t− ln(t+ 2)]e1 = [ex − ln(ex + 2)]10 ≈ 1, 265
(t = ex)
(c)
π2
4∫π2
16
sin√x dx = 2 [sin t− t cos t]
π2π4
= 2 [sin√x−√x cos
√x]
π2
4
π2
16
≈ 1, 7
(t =√x)
(d)8∫0
e√
2x dx = [tet − et]40 =[√
2xe√
2x − e√
2x]8
0≈ 164, 8
(t =√
2x)
7
Integralszamıtas III.
Racionalis fuggvenyek integralasa
1. (a)∫
x3−3x2+1
dx = x2
2− 1
2ln(x2 + 1)− 3 arctg x+ C
Nem valodi racionalis tortfuggveny; polinomosztassal elerjuk, hogy a szamlalo alacsonyabbfoku legyen, mint a nevezo.
(x3 − 3) : (x2 + 1) = x−x− 3
Ez alapjan: x3−3x2+1
= x− x+3x2+1∫
x3−3x2+1
dx =∫ (
x− x+3x2+1
)dx =
∫x dx− 1
2
∫2xx2+1
dx− 3∫
11+x2
dx =
= x2
2− 1
2ln(x2 + 1)− 3 arctg x+ C
(b)∫
10x2−1
dx =∫
5x−1
+ −5x+1
dx = 5 ln∣∣x−1x+1
∣∣+ C
utm.: A nevezo masodfoku, diszkriminansa pozitıv: ket darab egyszeres valos gyoke van,ennek megfeleloen bontjuk parcialis tortek osszegere a fuggvenyt:
10x2−1
= 10(x−1)(x+1)
∗= A
x−1+ B
x+1= A(x+1)+B(x−1)
(x−1)(x+1)
10 = A(x+ 1) +B(x− 1)I. modszer: x-hatvanyok egyutthatoinak egyeztetesevel10 = (A+B)x+ (A−B)
0 = A+B10 = A−B
}⇒ A = 5
B = −5II. modszer: A nevezo gyokeinek behelyettesıtesevel
x = −110 = 0 +B(−2)
B = −5
x = 110 = A · 2 + 0
A = 5
Ebbol: 10x2−1
∗= 5
x−1+ −5
x+1
2. (a)∫
1x2+2x+5
dx = 12
arctg(x+1
2
)+ C
(b)∫
74x2+12x+10
dx = 7∫
1(2x+3)2+1
dx = 7 · arctg(2x+3)2
+ C
3. (a)∫
8x2−3x+2
dx = 8 ln∣∣x−2x−1
∣∣+ C
(b)∫
6x−20x2−4x−12
dx = 2 ln |x− 6|+ 4 ln |x+ 2|+ C
utm.: 6x−20x2−4x−12
= 6x−20(x−6)(x+2)
= Ax−6
+ Bx+2
6x− 20 = A(x+ 2) +B(x− 6)
II. modszerrel:x = 6
36− 20 = 8A+ 016 = 8AA = 2
x = −2−12− 20 = 0 +B · (−8)
−32 = −8BB = 4
8
Ebbol: 6x−20x2−4x−12
= 2x−6
+ 4x+2
(c)∫
5x+31x2+2x−3
dx = 9 ln |x− 1| − 4 ln |x+ 3|+ C
(d)∫
13x−1014x2−23x+3
dx = 12
ln |2x− 3|+ 37
ln |7x− 1|+ C
(e)∫
8x+38(x−4)(x+1)(x+3)
dx = 2 ln |x− 4| − 3 ln |x+ 1|+ ln |x+ 3|+ C
4. (a)∫ −5x2+10x−13
(x−3)(x−1)2dx = −7 ln |x− 3|+ 2 ln |x− 1| − 4
(x−1)+ C
utm.: A nevezonek egy egyszeres es egy ketszeres valos gyoke van, ennek megfeleloenbontjuk parcialis tortek osszegere a fuggvenyt:
−5x2+10x−13(x−3)(x−1)2
∗= A
x−3+ B
x−1+ C
(x−1)2
megj.: vigyazzunk, a tortek kozos nevezoje (x−3)(x−1)2, nem pedig (x−3)(x−1)(x−1)2!
−5x2 + 10x− 13 = A(x− 1)2 +B(x− 1)(x− 3) + C(x− 3)
A II. modszerrel kezdunk:
x = 1
−5 + 10− 13 = 0 + 0 + C(−2)−8 = −2CC = 4
x = 3
−45 + 30− 13 = A · 4 + 0 + 0−28 = 4AA = −7
Az I. modszer reszleges alkalmazasaval folytatjuk:
x2 egyutthatoi : −5 = A+B−5 = −7 +B
B = 2
Kaptuk: −5x2+10x−13(x−3)(x−1)2
∗= −7
x−3+ 2
x−1+ 4
(x−1)2
(b)∫
2x2−14x+12(x−2)2(x+2)
dx = − ln |x− 2|+ 2x−2
+ 3 ln |x+ 2|+ C
5. (a)∫
8x2−28x+14(x2+1)(x−9)
dx = 5 ln |x− 9|+ 32
ln |x2 + 1| − arctg x+ C
utm.: 8x2−28x+14(x2+1)(x−9)
∗= A
x−9+ Bx+C
x2+1
8x2 − 28x+ 14 = A(x2 + 1) + (Bx+ C)(x− 9)
A II. modszerrel kezdunk:
x = 9
648− 252 + 14 = 82A+ 0410 = 82AA = 5
Az I. modszer reszleges alkalmazasaval folytatjuk:
x2 egyutthatoi : 8 = A+B8 = 5 +BB = 3
x egyutthatoi : −28 = −9B + C−28 = −27 + C
C = −1
Kaptuk:8x2 − 28x+ 14
(x2 + 1)(x− 9)∗=
5
x− 9+
3x− 1
x2 + 1
9
(b)∫
2x2+3x−10(x2+4)(x−6)
dx = 2 ln |x− 6|+ 32
arctg x2
+ C
utm.: A nevezoben szereplo masodfoku tenyezo diszkriminansa negatıv: a nevezonek nemcsak valos gyokei vannak; ennek megfeleloen bontjuk parcialis tortek osszegere a fuggvenyt:
2x2+3x−10(x2+4)(x−6)
∗= A
x−6+ Bx+C
x2+4
2x2 + 3x− 10 = A(x2 + 4) + (Bx+ C)(x− 6)
A II. modszerrel kezdunk:
x = 672 + 18− 10 = A · 40 + 0
80 = 40AA = 2
Az I. modszer reszleges alkalmazasaval folytatjuk:
x2 egyutthatoi : 2 = A+B2 = 2 +BB = 0
x egyutthatoi : 3 = −6B + C3 = 0 + CC = 3
Kaptuk: 2x2+3x−10(x2+4)(x−6)
∗= 2
x−6+ 0x+3
x2+4= 2
x−6+ 3
x2+4
(c)∫
5x2+15x+37(x2+6x+13)(x−2)
dx =∫ (
2x+1x2+6x+13
+ 3x−2
)dx =
∫2x+6
x2+6x+13dx− 5
∫1
(x+3)2+4+∫
3x−2
dx =
= ln(x2 + 6x+ 13)− 52
arctg(x+3
2
)+ 3 ln |x− 2|+ C
(d)∫ −8x2+8x−4
(9x2−6x+5)(x+1)dx = 1
18ln(9x2 − 6x+ 5) + 2
9arctg
(3x−1
2
)− ln |x+ 1|+ C
6. (a)∫
3x3−7x2−39x−54x2−3x−10
dx = 32x2 + 2x− 7 ln |x− 5|+ 4 ln |x+ 2|+ C
utm.: A szamlalo nem alacsonyabb foku, mint a nevezo: polinomosztas.
(3x3 − 7x2 − 39x− 54) : (x2 − 3x− 10) = 3x+ 22x2 − 9x− 54−3x− 34
Ebbol: 3x3−7x2−39x−54x2−3x−10
= 3x+ 2− 3x+34x2−3x−10
Parcialis tortekre bontas: 3x+34x2−3x−10
= 3x+34(x−5)(x+2)
∗= A
x−5+ B
x+2
3x+ 34 = A(x+ 2) +B(x− 5)
II. modszerrel:
x = −2−6 + 34 = 0 +B(−7)
28 = −7BB = −4
x = 515 + 34 = A · 7 + 0
49 = 7AA = 7
Ebbol:3x+ 34
x2 − 3x− 10∗=
7
x− 5− 4
x+ 2
Osszessegeben tehat: 3x3−7x2−39x−54x2−3x−10
= 3x+ 2− 7x−5
+ 4x+2
(b)2∫0
4x3+22x2+40x+20(x+3)2(x2+1)
dx =[ln |x+ 3|+ 1
x+3+ 3
2ln(x2 + 1) + 2 arctg x
]20≈ 5
10
(c)∫
x5−5x4−x3+21x2−25x+12x3−5x2+4x
dx = x3
3− 5x+ 3 ln |x| − ln |x− 1| − 6 ln |x− 4|+ C
utm.: A szamlalo nem alacsonyabb foku, mint a nevezo: polinomosztas.
(x5 − 5x4 − x3 + 21x2 − 25x+ 12) : (x3 − 5x2 + 4x) = x2 − 5−5x3 + 21x2 − 25x+ 12
−4x2 − 5x+ 12
Ebbol: x5−5x4−x3+21x2−25x+12x3−5x2+4x
= x2 − 5 + −4x2−5x+12x3−5x2+4x
Parcialis tortekre bontas: −4x2−5x+12x3−5x2+4x
= −4x2−5x+12x(x−1)(x−4)
∗= A
x+ B
x−1+ C
x−4
−4x2 − 5x+ 12 = A(x− 1)(x− 4) +Bx(x− 4) + Cx(x− 1)
x = 00− 0 + 12 = A(−1)(−4) + 0 + 0
12 = 4AA = 3
x = 1−4− 5 + 12 = 0 +B · 1 · (−3) + 0
3 = −3BB = −1
x = 4−64− 20 + 12 = 0 + 0 + C · 4 · 3
−72 = 12CC = −6
Ebbol: −4x2−5x+12x3−5x2+4x
∗= 3
x− 1
x−1− 6
x−4
Osszessegeben tehat: x5−5x4−x3+21x2−25x+12x3−5x2+4x
= x2 − 5 + 3x− 1
x−1− 6
x−4
(d)∫
x3+2x2+7x−1x4+5x2+4
dx = ln(x2 + 1)− 12
ln(x2 + 4)− arctg x+ 32
arctg(x2
)+ C
Irracionalis fuggvenyek integralasa
1. (a)∫
(2√x3 − 4 3
√x− 1) dx = 4
5
√x5 − 3 3
√(x− 1)4 + C
(b)∫
(3√x+ 3√x dx =
∫ (3x
12 + x
13
)dx = 2
√x3 + 3
4
3√x4 + C
(c)∫ (
2√
(x+ 2)5 + 4√
2x+ 1− 3 7√
3x+ 11)
dx =
= 47
√(x+ 2)7 + 2
54√
(2x+ 1)5 − 78
7√
(3x+ 11)8 + C
(d)∫ √
1− x2 dx = 12
arcsinx+ 12x√
1− x2 + C
(x = sin t)
(e)∫ √
4− x2 dx = 2 arcsin(x2
)+ x√
1−(x2
)2+ C
(x2
= sin t)
(f)∫ √
1− 3x2 dx = 12√
3arcsin
(√3x)
+ 12x√
1− 3x2 + C
(sin t =√
3x)
(g)∫ √−x2 − 2x dx = 1
2arcsin(x+ 1) + 1
2(x+ 1)
√1− (x+ 1)2 + C
(x+ 1 = sin t)
(h)∫ √−x2 + 6x− 5 dx = 2 arcsin
(x−3
2
)+ (x− 3)
√1−
(x−3
2
)2+ C
(x−32
= sin t)
(i)∫ √
1 + 9x2 dx = 16
arsh 3x+ 12x√
1 + 9x2 + C
(3x = sh t)
(j)∫ √
x2 − 25 dx = 12x√x2 − 25− 25
2arch x
5+ C
(x5
= ch t)
11
2. (a)∫
2x√x2+6
dx =∫
2x · (x2 + 6)−12 dx = (x2+6)
12
12
+ C = 2√x2 + 6 + C
fαf ′ alaku integrandus
(b)∫
3x−6√x2−4x+1
dx = 32
∫2x−4√x2−4x+1
dx = 32
∫(2x− 4)(x2 − 4x+ 1)−
12 dx =
= 32· (x2−4x+1)−
12
12
+ C = 3√x2 − 4x+ 1 + C
fαf ′ alaku integrandus
(c)∫
10√4−x2 dx = 10
2
∫1√
1−(x2
)2dx = 10 arcsin
(x2
)+ C
f(ax+ b) alaku integrandus
(d)∫
6√1+16x2
dx = 6∫
1√1+(4x)2
dx = 6 · arsh 4x4
+ C
f(ax+ b) alaku integrandus
(e)∫
4x−3√9−4x2
dx = −√
9− 4x2 − 32
arcsin(
2x3
)+ C
(2x3
= sin t)
(f)∫
10−x√−x2−6x−5
dx = 13 arcsin(x+3
2
)+ 2√
1−(x+3
2
)2+ C
(x+32
= sin t)
12
Az integralszamıtas alkalmazasai I.
Teruletszamıtas
1. Szamıtsa ki a gorbe es az x-tengely koze zart sıkresz teruletet a megadott intervallumon!
(a) y = x2 − 3x+ 7 [−1, 2 ]
y ≥ 0 ∀x ∈ [−1, 2] eseten, ezert
T =2∫−1
(x2 − 3x+ 7) dx =[x3
3− 3x2
2+ 7x
]2
−1= 117
6≈ 19, 5
(b) y = 12
+ sinx [ 0, π]
y ≥ 0 ∀x ∈ [0, π] eseten, ezert
T =π∫0
(12
+ sinx)
dx =[
12x− cosx
]π0
= 2 + π2≈ 3, 57
(c) y = ex − 1 [−1, 1 ]
ex − 1 = 0, ha x = 0, tovabba ex − 1 < 0, ha x ∈ [−1, 0[ ill. ex − 1 > 0, ha x ∈]0, 1], ezert
T =
∣∣∣∣ 0∫−1
(ex − 1) dx
∣∣∣∣+1∫0
(ex − 1) dx =∣∣[ex − x]0−1
∣∣+ [ex − x]10 =
=∣∣(1− 0)− (1
e+ 1)
∣∣+ (e− 1)− (1− 0) = 1e
+ e− 2 ≈ 1, 086
(d) y = sin2 x− 14
[0, π ]
zerushelyek: sin2 x = 14⇒ sinx = ±1
2
x∈[0,π]⇒ sinx = 12⇒
x1 = π
6
x2 = 5π6
msz.:∫ (
sin2 x− 14
)dx =
∫ (1−cos 2x
2− 1
4
)dx = . . . = 1
4(x− sin 2x) + C
T =∣∣∣[1
4(x− sin 2x)
]π6
0
∣∣∣+[
14
(x− sin 2x)] 5π
6π6
+∣∣∣[1
4(x− sin 2x)
]π5π6
∣∣∣ = . . . ≈ 1, 1278
2. Szamıtsa ki az alabbi, parameteres alakban megadott gorbe es az x-tengely kozotti sıkreszteruletet a megadott intervallumon!
keplet: T = ±β∫α
y(t) · x(t) dt
(− , hax szig. mon. csokk.+ , hax szig. mon. no
)
gyakorlatban: T =
∣∣∣∣∣∣β∫α
y(t) · x(t) dt
∣∣∣∣∣∣(a) x = 2 cos t , y = sin t [ 0, π]
T = −π∫0
sin t(−2 sin t) dt = 2π∫0
sin2 t dt = π
Linearizalo formulaval.
(b) x = t− sin t , y = 1− cos t [ 0, 2π]
T =2π∫0
(1− cos t)(1− cos t) dt =[t− 2 sin t+ t
2+ 1
4sin 2t
]2π0
= 3π
Linearizalo formulaval.
13
(c) x = t2 − 3t , y = et [ 2, 4]
T =4∫2
et · (2t− 3)︸ ︷︷ ︸parc. int.
dt = [et · (2t− 5)]42 ≈ 171, 2
(d) x = t2 − 1 , y = sin t [ 0, π](A peldaban egyebkent adott [2, 4] intervallum nem szerencses, mert ott a sin fv.-nek zerushelye van.)
T =π∫0
2t sin t︸ ︷︷ ︸parc. int.
dt = [−2t cos t+ 2 sin t]π0 = 2π ≈ 6, 28
3. Szamıtsa ki az adott gorbek altal hatarolt korlatos sıkresz teruletet!
(a) y = 6x− x2 − 7 , y = x− 3
6x− x2 − 7 = x− 3 ⇒ x = 1 v. x = 4 ⇒ Az [1, 4] int.-on integralunk.
T =4∫1
((6x− x2 − 7)− (x− 3)) dx = . . . = 4, 5
(b) y = 2x2ex , y = −x3ex
2x2ex = −x3ex ⇒ x = −2 v. x = 0 ⇒ A [−2, 0] int.-on integralunk.
T =0∫−2
((2x2ex)− (−x3ex)) dx =0∫−2
(2x2ex + x3ex)︸ ︷︷ ︸parc. int.
dx = [ex(x3 − x2 + 2x− 2)]0−2 =
= 18e2− 2 ≈ 0, 43
(c) y = x3 , y = 4x
x3 = 4x ⇒ x = −2 v. x = 0 v. x = 2 ⇒ A [−2, 0] ill. [0, 2] int.-on integralunk.
T =0∫−2
(x3 − 4x) dx+2∫0
(4x− x3) dx =[x4
4− 4x2
2
]0
−2+[
4x2
2− x4
4
]2
0= 8
Mindket fv. paratlan, ıgy a szimmetria miatt T = 22∫0
(4x− x3) dx
(d) y = sinx , y =2x
π
sinx = 2xπ⇒ x = −π
2v. x = 0 v. x = π
2
Mivel mindket fv. paratlan, ıgy a szimmetria miatt eleg a [0, π2] int.-on integralnunk.
T = 2
π2∫
0
(sinx− 2
πx)
dx = 2[− cosx− 2
π· x2
2
]π2
0≈ 0, 43
(e) y = x2 , y2 = x
T =1∫0
(√x− x2) dx = . . . = 1
3
4. Hatarozza meg az y = cosx es y = sin 2x gorbek kozotti teruletet a[0; π
2
]intervallumban!
Metszespontok: a cosx = sin 2x egyenletnek a[0; π
2
]intervallumba eso megoldasai.
Ezek: x = π6
ill. x = π2. Erdemes abrazolni a fuggvenygrafikonokat.
14
T =
π6∫
0
(cosx− sin 2x) dx+
π2∫π6
(sin 2x− cosx) dx = . . . = 12
5. Szamıtsa ki a parameteres alakban megadott gorbe altal hatarolt sıkidom teruletet!
keplet: T =
∣∣∣∣∣∣β∫α
y(t) · x(t) dt
∣∣∣∣∣∣(a) x = t2 − 1 , y = sin t , t ∈ [−π, π]
T =
∣∣∣∣ π∫−π
sin t · 2t dt
∣∣∣∣ parcialis int.=
∣∣[−2t cos t+ 2 sin t]π−π∣∣ = |4π| = 4π ≈ 12, 57
(b) x = cos2 t , y = sin 2t , t ∈ [0, π]
T =
∣∣∣∣ π∫0
sin 2t · 2 cos t(− sin t) dt
∣∣∣∣ =
∣∣∣∣− π∫0
sin2 2t dt
∣∣∣∣ linearizalo form.=
∣∣− [12t− 1
8sin 4t
]π0
∣∣ =
=∣∣−π
2
∣∣ = π2≈ 1, 57
(c) x = t2 , y = t3 − 4t , t ∈ [−2, 2]
T =
∣∣∣∣ 2∫−2
(t3 − 4t)2t dt
∣∣∣∣ =
∣∣∣∣ 2∫−2
(2t4 − 8t2 dt
∣∣∣∣ =
∣∣∣∣[2t5
5− 8t3
3
]2
−2
∣∣∣∣ =∣∣−256
15
∣∣ ≈ 17, 07
Testek terfogata
1. Egy negyzet alapu egyenes gula alapele 2m, magassaga 4m. A gula terfogata az ismert V =a2m
3
keplet alapjan16
3m3. Szamıtsuk ki a gula terfogatat integralassal, ıgy ellenorizve a fenti keplet
helyesseget!
2. Egy hengerbol, melynek alaplapja 3 cm sugaru kor, egy eket vagunk le egy az alaplappal 45◦-osszoget bezaro sıkkal, ami atmegy az alaplap kozeppontjan. Mekkora az ek terfogata?
Mo.: V = 18.
3. Egy test alaplapja az y = x2 − 1 es y = 1 − x2 gorbek koze zart terulet az xy sıkon, az x-tengelyre meroleges metszetei szabalyos haromszogek. Szamıtsuk ki a test terfogatat!
Mo.: V = 16√
315≈ 1, 85
4. Szamıtsa ki az adott gorbeıvnek az x-tengely koruli megforgatasaval kapott forgastest terfogatat!
keplet: V = π
b∫a
y2(x) dxkepletgy.
= π
b∫a
f 2(x) dx
(a) y = 4− x2 [−2, 2 ]
V = π2∫−2
(x4 − 8x2 + 16) dx = 512π15≈ 107, 23
15
(b) y =1√
cosx
[0,π
6
]V = π
π6∫
0
1cosx
dx = π[ln∣∣∣1+tg x
2
1−tg x2
∣∣∣]π60
= π2
ln 3 ≈ 1, 726
Az integralas a kepletgyujtemenyben is szereplo t = tg x2
helyettesıtessel vegezheto el.
Utm.:
t = tgx
2x = 2 arctg t
dx
dt=
2
1 + t2dx =
2
1 + t2dt cosx =
1− t2
1 + t2
(c) y =√xe−x [ 0, 1]
V = π1∫0
xe−2x dx = π[e−2x
(−1
2x− 1
4
)]10≈ 0, 466
5. Forgassuk az f(x) =√x fuggveny, az x = 4 egyenes es az y = 1 egyenes altal hatarolt korlatos
sıkidomot az y = 1 egyenes korul! Mekkora a keletkezett forgastest terfogata?
Utm.: toljuk el az abrat a v(−1;−1) vektorral.
V = π3∫0
(√x+ 1− 1
)2dx = . . . = 7π
6
6. Forgassuk meg az y = ex , y = e−x es az x = 1 egyenletu gorbek altal hatarolt veges tartomanytaz x-tengely korul! Mekkora a keletkezett forgastest terfogata?
Mo.: V = π1∫0
e2x dx− π1∫0
e−2x dx = π2
[e2x + e−2x]10 = π
2
(e2 + 1
e2− 2)≈ 8, 67
Mekkora annak a forgastestnek a terfogata, amelyet ugy nyerunk, hogy ugyanezen sıkidomotaz y-tengely korul forgatjuk meg?
Mo.: V = πe∫1e
1 dy −
π 1∫1e
(− ln y)2 dy + πe∫
1
(ln y)2 dy
= πe∫1e
1 dy − πe∫1e
(ln y)2 dy =
= πe∫1e
(1− ln2 y
)dy = π
[−y − y ln2 y + 2y ln y
]e1e
= 4πe≈ 4, 623
(∫
ln2 y dy parcialis integralassal)
7. Szamıtsa ki a kovetkezo, parameteresen megadott gorbeıvek x-tengely koruli megforgatasavalkapott forgastestek terfogatat!
keplet: V = ± πt2∫t1
y2(t) · x(t) dt
(− , hax szig. mon. csokk.+ , hax szig. mon. no
)
gyakorlatban: V =
∣∣∣∣∣∣πt2∫t1
y2(t) · x(t) dt
∣∣∣∣∣∣ kepletgy.=
∣∣∣∣∣∣πt2∫t1
ψ2(t) · ϕ(t) dt
∣∣∣∣∣∣(a) x = cos t , y = sin 2t t ∈
[0,π
2
]V = −π
π2∫
0
sin2 2t(− sin t) dt = 4π
π2∫
0
sin3 t cos2 t dt = 4π
π2∫
0
sin t(1− cos2 t) cos2 t dt =
16
= 4π
π2∫
0
(sin t cos2 t− sin t cos4 t) dt = 4π[− cos3 t
3+ cos5 t
5
]π2
0= 8π
15≈ 1, 67
(b) x = t2 + t , y = et t ∈ [0, 3]
V = π3∫0
e2t(2t+ 1) dt = π [te2t]30 = 3πe6 ≈ 3802
Ivhossz szamıtasa
1. Szamıtsa ki a gorbeıv hosszat a megadott intervallumban!
keplet: s =
b∫a
√1 + (y′(x))2 dx
kepletgy.=
b∫a
√1 + (f ′(x))2 dx
(a) y = 2x32 [ 0; 11]
s =11∫0
√1 + 9x dx = 74
f(ax+ b) alaku integrandus.
(b) y =1
3(x2 + 2)
32 [ 0; 4]
s =4∫0
√x4 + 2x2 + 1 dx =
4∫0
(x2 + 1) dx = . . . = 763
(c) y = x6
√x+ 12 [−11;−3]
s =−3∫−11
x+164√x+12
dx =[
16
√(x+ 12)3 + 2
√x+ 12
]−3
−11= 25
3
Az integralas a t =√x+ 12 helyettesıtessel vegezheto el.
(d) y = ln sin x[π3; 2π
3
]s =
2π3∫π3
√1 + ctg2 x dx =
[ln(tg x
2
)] 2π3π3
= ln 3 ≈ 1, 1
Utm.:∫ √
1 + ctg2 x dx =∫ √
sin2x+cos2 xsin2 x
dx =∫ √
1sin2 x
dx =∫
1sinx
dx
Az integralas a kepletgyujtemenyben is szereplo t = tg x2
helyettesıtessel vegezheto el.
t = tgx
2x = 2 arctg t
dx
dt=
2
1 + t2dx =
2
1 + t2dt sinx =
2t
1 + t2
(e) y = lnx− x2
8[1; 2]
s =2∫1
(1x
+ x4
)dx = . . . = ln 2 + 3
8≈ 1, 07
(f) y =√
1− x2[− 1
2; 1
2
]s =
12∫− 1
2
1√1−x2 dx = . . . = π
3≈ 1, 047
17
(g) y = 3 +ch 2x
2[0; 1]
s =1∫0
ch 2x dx = . . . = e2−e−2
4≈ 1, 81
(h) y = ln(1− x2)[0; 1
2
]s =
12∫
0
1+x2
1−x2 dx =
12∫
0
(−1 + 2
1−x2)
dx . . . = −12
+ ln 3 ≈ 0, 6
2. Szamıtsa ki az alabbi, parameteresen megadott gorbeıv hosszat a megadott intervallumon!
keplet: s =
t2∫t1
√x 2(t) + y 2(t) dt
kepletgy.=
t2∫t1
√ϕ 2(t) + ψ 2(t) dt
(a) x = t2 , y = t(
13− t2
)t ∈[0, 1
3
]s =
13∫
0
√(2t)2 + (1
3− 3t2)2 dt =
13∫
0
√9(t2 + 1
9)2 dt =
13∫
0
3(t2 + 1
9
)dt =
[t3 + 1
3t] 1
3
0= 4
27
(b) x = et sin t , y = et cos t t ∈ [ 0, ln 2]
s =ln 2∫0
√2e2t dt =
ln 2∫0
√2et dt =
[√2et]ln 2
0=√
2
Utm.:x(t) = et(sin t+ cos t)y(t) = et(cos t− sin t)
}⇒ x2(t) = e2t(sin2 t+ 2 sin t cos t+ cos2 t)
y2(t) = e2t(sin2 t− 2 sin t cos t+ cos2 t)
}⇒
x2(t) + y2(t) = e2t(2(sin2 t+ cos2 t)
)= 2e2t
Felszınszamıtas
1. Szamıtsa ki a gorbe x-tengely koruli forgatasaval nyert forgastest palastfelszınet!
keplet: F = 2π
b∫a
y(x)√
1 + (y′(x))2 dxkepletgy.
= 2π
b∫a
f(x)√
1 + (f ′(x))2 dx
(a) y =1
3x3 x ∈ [0; 1]
y′(x) = x2 ⇒ (y′(x))2 = x4
F = 2π1∫0
13x3 ·√
1 + x4 dx = 2π12
1∫0
4x3(1 + x4)12 dx = π
6
[(1+x4)
32
32
]1
0
=
= 2π18
[√(1 + x4)3
]1
0= π
9
(√8−√
1)≈ 0, 64
(b) y =√
9− x2 x ∈ [−3; 3]
y′(x) = 12√
9−x2 · (−2x) ⇒ (y′(x))2 = 4x2
4(9−x2)= x2
9−x2 ⇒
1 + (y′(x))2 = 9−x29−x2 + x2
9−x2 = 99−x2
18
F = 2π3∫−3
√9− x2 ·
√9
9−x2 dx = 2π3∫−3
√9− x2 · 3√
9−x2 dx = 2π3∫−3
3 dx = 2π [3x]3−3 =
2π(9− (−9)) = 36π ≈ 113, 1
(megj.: origo kozeppontu, 3-sugaru gomb felszınerol van szo: F = 4πr2 = 36π)
(c) y =√
3− 2x x ∈ [0; 1]
y′(x) = 12√
3−2x· (−2) ⇒ (y′(x))2 = 4
4(3−2x)= 1
3−2x⇒
1 + (y′(x))2 = 3−2x3−2x
+ 13−2x
= 4−2x3−2x
F = 2π1∫0
√3− 2x ·
√4−2x3−2x
dx = 2π1∫0
√3− 2x ·
√4−2x√3−2x
dx = 2π1∫0
√4− 2x dx =
= 2π
[(−2x+4)
32
32·(−2)
]1
0
= −2π3
(√(−2 + 4)3 −
√(0 + 4)3
)= −2π
3(√
8− 8) ≈ 10, 83
(d) y =√
2x− 4 x ∈ [2; 3]
y′(x) = 12√
2x−4· 2 = 1√
2x−4⇒ (y′(x))2 = 1
2x−4⇒
1 + (y′(x))2 = 2x−42x−4
+ 12x−4
= 2x−32x−4
F = 2π3∫2
√2x− 4 ·
√2x−32x−4
dx = 2π3∫2
√2x− 4 ·
√2x−3√2x−4
dx = 2π3∫2
√2x− 3 dx =
= 2π
[(2x−3)
32
32·2
]3
2
= 2π3
(3
32 − 1
32
)= 2π
3(√
27− 1) ≈ 8, 79
(e) y =x3
9x ∈ [0; 2]
F = 2π2∫0
x3
9·√
1 + x4
9dx ≈ 3, 8 (fα · f ′ alaku integrandus)
(f) y = chx x ∈ [0; ln 2]
F = 2πln 2∫0
chx ·√
1 + sh2 x dx = 2πln 2∫0
ch2 x dx ≈ 5, 13
(ch2 x-re vonatkozo linearizalo formulaval)
2. Forgassa meg az alabbi, parameteres egyenletrendszerrel felırt gorbek megadott darabjat azx-tengely korul, es szamıtsa ki a keletkezo forgastestek palastjanak felszınet!
keplet: F = 2π
b∫a
y(t)√x2(t) + y2(t) dt
kepletgy.= 2π
t2∫t1
ψ(t)
√ϕ2(t) + ψ2(t) dt
(a) x = t2 y = t t ∈ [0, 1]
x(t) = 2t
y(t) = 1
}⇒ x2(t) + y2(t) = 4t2 + 1
19
F = 2π1∫0
t√
4t2 + 1 dt = 2π8
1∫0
8t(4t2 + 1)12︸ ︷︷ ︸
f ′·fα
dt = π4
[(4t2+1)
32
32
]1
0
= 2π12
[√(4t2 + 1)3
]1
0=
= π6(√
125−√
1) ≈ 5, 33
(b) x = a cos2 t y = a sin2 t t ∈ [0, π2]
x(t) = a · 2 cos t(− sin t)
y(t) = a · 2 sin t cos t
}⇒ x2(t) + y2(t) = 8a2 sin2 t cos2 t
F = 2π
π2∫
0
a sin2 t√
8a2 sin2 t cos2 t dt = 2π
π2∫
0
a sin2 t · 2√
2 · a sin t cos t dt =
= 4√
2a2π
π2∫
0
cos t sin3 t︸ ︷︷ ︸f ′·fα
dt = 4√
2a2π[
sin4 t4
]π2
0= 4√
2a2π(
14− 0)
=√
2a2π ≈ 4, 44
(megj.: Olyan forgaskup palastfelszınerol van szo, amely kup alapkorenek sugara a, magassaga a, alkotoja√
2a.)
(c) x = et cos t y = et sin t t ∈ [0, π2]
x(t) = et cos t− et sin t = et(cos t− sin t)
y(t) = et cos t+ et sin t = et(cos t+ sin t)
}⇒ x2(t) + y2(t) =
= e2t(cos2 t−2 cos t sin t+sin2 t)+e2t(cos2 t+2 cos t sin t+sin2 t) = e2t·2(cos2 t+sin2 t) = 2e2t
F = 2π
π2∫
0
et sin t√
2e2t dt = 2√
2π
π2∫
0
et(sin t)et dt = 2√
2π
π2∫
0
e2t sin t︸ ︷︷ ︸parcialis int.
dt = . . . =
= 2√
2π[−1
5e2t cos t+ 2
5e2t sin t
]π2
0= 2√
2π((0 + 2
5eπ)− (−1
5+ 0)
)≈ 84, 02
(d) x = cos t+ ln(tg(t2
))y = sin t t ∈ [π
2, 3π
4]
x(t) = − sin t+ 1tg t
2
· 1cos2 t
2
· 12
= − sin t+ 12 sin t
2cos t
2
= − sin t+ 1sin t
y(t) = cos t
x2(t) + y2(t) = sin2 t− 2 + 1sin2 t
+ cos2 t = 1sin2 t− 1 = 1−sin2 t
sin2 t= cos2 t
sin2 t
F = 2π
3π4∫π2
sin t ·√
cos2 tsin2 t
dt = 2π
3π4∫π2
∣∣ cos t︸︷︷︸≤0
∣∣ dt = 2π
3π4∫π2
(− cos t) dt = 2π [− sin t]3π4π2
=
= 2π(−√
22
+ 1)≈ 1, 84
20
Improprius integralok
Integralas vegtelen intervallumon
1. (a)
∞∫− ln 2
e−2x dx = limω→∞
ω∫− ln 2
e−2x dx = limω→∞
[e−2x
−2
]ω− ln 2
=
= limω→∞
(e−2ω
−2− e−2(− ln 2)
−2
)= 0 +
1
2eln 4 =
1
2· 4 = 2
megj.: a tovabbiakban limω→∞
[F (x)]ωa helyett roviden [F (x)]∞a jeleket ırunk.
(b)∞∫0
x1+x2
dx = 12
∞∫0
2x1+x2
dx = 12
[ln |1 + x2|]∞0 =
= 12
(limx→∞
ln(1 + x2)− ln(1 + 0))
= ”12(∞− 0)” =∞
(Az improprius integral divergens.)
(c)∞∫√
2
x(x2+1)3
dx = 12
∞∫√
2
2x(x2 + 1)−3︸ ︷︷ ︸f ′·fα
dx = 12
[(x2+1)−2
−2
]∞√
2=
= −14
[1
(x2+1)2
]∞√
2= −1
4
(limx→∞
1
(x2 + 1)2− 1
(2 + 1)2
)= −1
4
(0− 1
9
)= 1
36≈ 0, 0277
(d)∞∫e
dxx ln2 x
=∞∫e
1x ln2 x
dx =∞∫e
1x· ln−2 x dx =
[ln−1 x−1
]∞e
= −[
1lnx
]∞e
=
= −(
limx→∞
1
lnx− 1
ln e
)= −(0− 1) = 1
(e)∞∫0
1x2+2x+2
dx =∞∫0
1(x+1)2+1
dx = [arctg(x+ 1)]∞0 = limx→∞
arctg(x+ 1)− arctg(1) =
= π2− π
4= π
4≈ 0, 785
(f)∞∫1
(2x+ 3)e1−x︸ ︷︷ ︸parcialis int.
dx = . . . = [(−2x− 5)e1−x]∞1 =
(limx→∞
(−2x− 5)e1−x)− (−7) =
=
(limx→∞
−2x− 5
ex−1
)+ 7
L’H=
(limx→∞
−2
ex−1
)+ 7 = 0 + 7 = 7
(g)∞∫0
dxex+√ex∗=
msz.: t =√ex helyettesıtessel:
t =√ex ⇒ t2 = ex ⇒ x = ln t2 ⇒ dx
dt=
1
t2· 2t =
2
t⇒ dx =
2
tdt∫
1
ex +√ex
dx =
∫1
t2 + t· 2
tdt =
∫2
t3 + t2dt =
∫2
t2(t+ 1)dt =
21
=
∫ (A
t+B
t2+
C
t+ 1
)︸ ︷︷ ︸
parc. tortekre bontas
dtHF=
∫ (−2
t+
2
t2+
2
t+ 1
)dt =
= −2 ln |t| − 2
t+ 2 ln |t+ 1|+ C = 2 ln
∣∣∣∣t+ 1
t
∣∣∣∣− 2
t+ C = 2 ln
∣∣∣∣√ex + 1√ex
∣∣∣∣− 2√ex
+ C
∗= lim
x→∞
(2 ln
∣∣∣∣√ex + 1√ex
∣∣∣∣− 2√ex
)−
(2 ln
∣∣∣∣∣√e0 + 1√e0
∣∣∣∣∣− 2√e0
)=
= limx→∞
(2 ln
∣∣∣∣1 +1√ex
∣∣∣∣− 2√ex
)−
(2 ln
∣∣∣∣∣√e0 + 1√e0
∣∣∣∣∣− 2√e0
)=
= (0− 0)− (2 ln 2− 2) = 2− 2 ln 2 ≈ 0, 61
(h)0∫−∞
ex+1 dx = [ex+1]0−∞ = e0+1 − lim
x→−∞ex+1 = e− 0 = e ≈ 2, 72
(i)−1∫−∞
x2e2x︸ ︷︷ ︸parc. int.
dx =[(
x2
2− x
2+ 1
4
)e2x]−1
−∞=(
12
+ 12
+ 14
)1e2− lim
x→−∞
(x2
2− x
2+
1
4
)e2x︸ ︷︷ ︸
∞·0
=
=5
4e2− lim
x→−∞
x2
2− x
2+ 1
4
e−2x
L’H=
5
4e2− lim
x→−∞
x− 12
−2e−2x
L’H=
5
4e2− lim
x→−∞
1
4e−2x=
= 54e2− 0 = 5
4e2≈ 0, 17
(j)∞∫−∞
dx1+4x2
=∞∫−∞
11+(2x)2
dx =[
arctg 2x2
]∞−∞ = lim
x→∞
arctg 2x
2− limx→−∞
arctg 2x
2=π
4−(−π
4
)=π
2
(k)∞∫−∞
xe−x2
2 dx = −∞∫−∞
(−x)e−x2
2︸ ︷︷ ︸g′(x)·f(g(x))
dx = −[e−
x2
2
]∞−∞
= −(
limx→∞
e−x2
2 − limx→−∞
e−x2
2
)=
= −(0− 0) = 0
megj.: Ez ugy lehetseges, hogy az xe−x2
2 integrandus paratlan fuggveny.
Adott intervallumon nem korlatos fuggveny integralasa
1. (a)1∫0
dx√1−x
∗=
[(1−x)
12
12·(−1)
]1
0
= −2[√
1− x]1
0= −2
(limx→1−
√1− x−
√1− 0
)= −2(0− 1) = 2
(∗) Az f(x) = 1√1−x fuggveny nincs ertelmezve az x = 1 helyen, tovabba lim
x→1−
1√
1− x=
1
0+= ∞, ezert improprius
integralrol van szo.
(b)1∫0
dx
1− x2︸ ︷︷ ︸kepletgy.
∗=[
12
ln∣∣1+x
1−x
∣∣]10
=1
2limx→1−
ln
∣∣∣∣1 + x
1− x
∣∣∣∣− 1
2ln
∣∣∣∣1 + 0
1− 0
∣∣∣∣ =1
2· ∞ − 1
2· 0 =∞
Az improprius integral divergens.(∗) Az f(x) = 1
1−x2 fuggveny nincs ertelmezve az x = 1 helyen, tovabba limx→1−
1
1− x2=
1
0+= ∞, ezert improprius
integralrol van szo.
22
(c)
π2∫
0
cosx√sinx
dx∗=
π2∫
0
cosx(sinx)−12︸ ︷︷ ︸
f ′·fα
dx =[2√
sinx]π
2
0= 2√
sin π2−2 lim
x→0+
√sinx = 2 ·1−2 ·0 = 2
(∗) Az f(x) = cos x√sin x
fuggveny nincs ertelmezve az x = 0 helyen, tovabba limx→0+
cosx√
sinx=
1
0+= ∞, ezert improprius
integralrol van szo.
(d)4∫0
dxx+√x
∗= [2 ln |
√x+ 1|]40 = 2 ln 3− lim
x→0+ln∣∣√x+ 1
∣∣ = 2 ln 3 ≈ 2, 2
t =√x helyettesıtessel
∫1
x+√x
dx =∫
2tt2+t
dt =∫
2t+1
dt = 2 ln |t+ 1|+ C = 2 ln∣∣√x+ 1
∣∣+ C
(∗) Az f(x) = 1x+√x
fuggveny nincs ertelmezve az x = 0 helyen, tovabba limx→0+
1
x+√x
=1
0+= ∞, ezert improprius
integralrol van szo.
(e)1∫0
lnx︸︷︷︸parc. int.
dx∗= [x lnx− x]10 = (0− 1)− lim
x→0+(x lnx− x) = (0− 1)− 0 = −1
felh.: limx→0+
x lnx = limx→0+
lnx1x
L’H= lim
x→0+
1x
− 1x2
= limx→0+
(−x) = 0−
(∗) Az f(x) = lnx fuggveny nincs ertelmezve az x = 0 helyen, tovabba limx→0+
lnx = −∞, ezert improprius integralrol van
szo.
(f)1∫−1
dx1−x2
A (b) feladat alapjan1∫0
dx1−x2 =∞, ıgy ez az improprius integral is divergens:
1∫−1
dx1−x2 =∞
(g)1∫0
dxx ln2 x
∗=
1∫0
1
x· (lnx)−2︸ ︷︷ ︸f ′·fα
dx =[− 1
lnx
]10
=
(− lim
x→1−
1
lnx
)−(− lim
x→0+
1
lnx
)=∞− 0 =∞
(∗) Az f(x) = 1x ln2 x
fuggveny nincs ertelmezve sem az x = 0, sem az x = 1 helyen, tovabba limx→0+
1
x ln2 x= ∞ ill.
limx→1−
1
x ln2 x=∞, ezert improprius integralrol van szo.
(h)π∫0
tg x dx∗=
π2∫
0
tg x dx+π∫π2
tg x dx
(∗) Az f(x) = tg x fuggveny az x = π2
helyen, az integralasi tartomany ([0, π]) egy belso pontjaban nincs ertelmezve.
Mivel
π2∫
0
tg x dx = −π2∫
0
− sinxcosx
dx = [− ln |cosx|]π20 =
(− lim
x→π2−
ln |cosx|
)− (− ln |cos 0|) =
∞+ 0 =∞, ezert aπ∫0
tg x dx integral is divergens.
23
