Integrace Subst
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Substitucˇnı´ metoda Lenka Bara´kova´ 16. srpna 2005
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..
c
Lenka Bara´kova´, 2005 ×
Obsah Z
Z
Z
//
/
e x +7 dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
3
xe1− x dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
1 √ dx . . . . . . . . . . . . . . . . . . . . . . . . . . . (2 + x ) 1 + x
18
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
e x +7 dx Z
e x +7 dx =
x+7 = t dx = dt
=
Z
et dt
= e t + c = e x +7 + c
//
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
e x +7 dx Z
e x +7 dx =
x+7 = t dx = dt
=
Z
et dt
= e t + c = e x +7 + c
Vnitrˇnı´ slozˇka je x + 7. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
e x +7 dx Z
e x +7 dx =
x+7 = t dx = dt
=
Z
et dt
= e t + c = e x +7 + c
Zavedeme substituci x + 7 = t. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
e x +7 dx Z
e x +7 dx =
x+7 = t dx = dt
=
Z
et dt
= e t + c = e x +7 + c
Nalezneme vztah mezi dx a dt. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
e x +7 dx Z
e x +7 dx =
x+7 = t dx = dt
=
Z
et dt
= e t + c = e x +7 + c
Dosadı´me substituci. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
e x +7 dx Z
e x +7 dx =
x+7 = t dx = dt
=
Z
et dt
= e t + c = e x +7 + c
Integrujeme. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
e x +7 dx Z
e x +7 dx =
x+7 = t dx = dt
=
Z
et dt
= e t + c = e x +7 + c
Pouzˇijeme substituci k na´vratu k promeˇnne´ x. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
2
xe1− x dx.
Z
//
/
.
..
2
xe1− x dx
1 − x2 = t
−2x dx = dt 1 x dx = − dt 2 Z 1 =− et dt 2 1 = − et + c 2 2 1 = − e1− x + c 2
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
2
xe1− x dx.
Z
2
xe1− x dx
1 − x2 = t
−2x dx = dt 1 x dx = − dt 2 Z 1 =− et dt 2 1 = − et + c 2 2 1 = − e1− x + c 2
Vy´raz je soucˇinem polynomu a slozˇene´ exponencia´lnı´ funkce. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
2
xe1− x dx.
Z
2
xe1− x dx
1 − x2 = t
−2x dx = dt 1 x dx = − dt 2 Z 1 =− et dt 2 1 = − et + c 2 2 1 = − e1− x + c 2
2
Zkusı´me substituovat za vnitrˇnı´ slozˇku slozˇene´ funkce e 1− x . //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
2
xe1− x dx.
Z
2
xe1− x dx
1 − x2 = t
−2x dx = dt 1 x dx = − dt 2 Z 1 =− et dt 2 1 = − et + c 2 2 1 = − e1− x + c 2
Hleda´me vztah mezi diferencia´ly. Derivujeme obeˇ strany substituce. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
2
xe1− x dx.
Z
2
xe1− x dx
1 − x2 = t
−2x dx = dt 1 x dx = − dt 2 Z 1 =− et dt 2 1 = − et + c 2 2 1 = − e1− x + c 2
Vyja´drˇ´ıme odsud vy´raz x dx, ktery´ figuruje uvnitrˇ integra´lu. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
2
xe1− x dx.
Z
2
xe1− x dx
1 − x2 = t
−2x dx = dt 1 x dx = − dt 2 Z 1 =− et dt 2 1 = − et + c 2 1 1− x 2 +c =− e 2
Dosadı´me. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
2
xe1− x dx.
Z
2
xe1− x dx
1 − x2 = t
−2x dx = dt 1 x dx = − dt 2 Z 1 =− et dt 2 1 = − et + c 2 1 1− x 2 +c =− e 2
Vypocˇteˇte integra´l pomocı´ vzorce. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
2
xe1− x dx.
Z
2
xe1− x dx
1 − x2 = t
−2x dx = dt 1 x dx = − dt 2 Z 1 =− et dt 2 1 = − et + c 2 2 1 = − e1− x + c 2
Pouzˇijeme substituci pro na´vrat k pu˚vodnı´ promeˇnne´. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
//
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Prˇ´ıklady s odmocninou z linea´rnı´ho cˇlenu rˇesˇ´ıme vzˇdy druhou substitucˇnı´ metodou. Zbavujeme se tak neprˇ´ıjemne´ odmocniny. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Zavedeme proto substituci t = //
/
.
..
√
1 + x. c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Odmocninu vzˇdy prˇevedeme umocneˇnı´m na tvar bez odmocniny, prˇecha´zı´me takto vlastneˇ k inverznı´ funkci. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Inverznı´ funkce bude v prˇepisu take´ trˇeba. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Hleda´me vztah mezi diferencia´ly. Derivujeme obeˇ strany inverznı´ substituce. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Vsˇechny vy´razy s x zameˇnı´me pomocı´ substituce za ekvivalentnı´ vy´razy s t. Nejdrˇ´ıve pouzˇijeme za x inverznı´ substituce. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Odmocnina odpovı´da´ t. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Diferencia´l take´ substituujeme. Vsˇechny cˇleny s x jsme nahradili. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Zkra´tı´me a konstantu prˇevedeme prˇed integra´l. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Integrujeme. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Navra´tı´me se k pu˚vodnı´ promeˇnne´. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
//
/
.
..
c
Lenka Bara´kova´, 2005 ×
Obsah Z
Z
Z
//
/
e x +7 dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
3
xe1− x dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
1 √ dx . . . . . . . . . . . . . . . . . . . . . . . . . . . (2 + x ) 1 + x
18
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
e x +7 dx Z
e x +7 dx =
x+7 = t dx = dt
=
Z
et dt
= e t + c = e x +7 + c
//
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
e x +7 dx Z
e x +7 dx =
x+7 = t dx = dt
=
Z
et dt
= e t + c = e x +7 + c
Vnitrˇnı´ slozˇka je x + 7. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
e x +7 dx Z
e x +7 dx =
x+7 = t dx = dt
=
Z
et dt
= e t + c = e x +7 + c
Zavedeme substituci x + 7 = t. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
e x +7 dx Z
e x +7 dx =
x+7 = t dx = dt
=
Z
et dt
= e t + c = e x +7 + c
Nalezneme vztah mezi dx a dt. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
e x +7 dx Z
e x +7 dx =
x+7 = t dx = dt
=
Z
et dt
= e t + c = e x +7 + c
Dosadı´me substituci. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
e x +7 dx Z
e x +7 dx =
x+7 = t dx = dt
=
Z
et dt
= e t + c = e x +7 + c
Integrujeme. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
e x +7 dx Z
e x +7 dx =
x+7 = t dx = dt
=
Z
et dt
= e t + c = e x +7 + c
Pouzˇijeme substituci k na´vratu k promeˇnne´ x. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
2
xe1− x dx.
Z
//
/
.
..
2
xe1− x dx
1 − x2 = t
−2x dx = dt 1 x dx = − dt 2 Z 1 =− et dt 2 1 = − et + c 2 2 1 = − e1− x + c 2
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
2
xe1− x dx.
Z
2
xe1− x dx
1 − x2 = t
−2x dx = dt 1 x dx = − dt 2 Z 1 =− et dt 2 1 = − et + c 2 2 1 = − e1− x + c 2
Vy´raz je soucˇinem polynomu a slozˇene´ exponencia´lnı´ funkce. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
2
xe1− x dx.
Z
2
xe1− x dx
1 − x2 = t
−2x dx = dt 1 x dx = − dt 2 Z 1 =− et dt 2 1 = − et + c 2 2 1 = − e1− x + c 2
2
Zkusı´me substituovat za vnitrˇnı´ slozˇku slozˇene´ funkce e 1− x . //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
2
xe1− x dx.
Z
2
xe1− x dx
1 − x2 = t
−2x dx = dt 1 x dx = − dt 2 Z 1 =− et dt 2 1 = − et + c 2 2 1 = − e1− x + c 2
Hleda´me vztah mezi diferencia´ly. Derivujeme obeˇ strany substituce. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
2
xe1− x dx.
Z
2
xe1− x dx
1 − x2 = t
−2x dx = dt 1 x dx = − dt 2 Z 1 =− et dt 2 1 = − et + c 2 2 1 = − e1− x + c 2
Vyja´drˇ´ıme odsud vy´raz x dx, ktery´ figuruje uvnitrˇ integra´lu. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
2
xe1− x dx.
Z
2
xe1− x dx
1 − x2 = t
−2x dx = dt 1 x dx = − dt 2 Z 1 =− et dt 2 1 = − et + c 2 1 1− x 2 +c =− e 2
Dosadı´me. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
2
xe1− x dx.
Z
2
xe1− x dx
1 − x2 = t
−2x dx = dt 1 x dx = − dt 2 Z 1 =− et dt 2 1 = − et + c 2 1 1− x 2 +c =− e 2
Vypocˇteˇte integra´l pomocı´ vzorce. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
2
xe1− x dx.
Z
2
xe1− x dx
1 − x2 = t
−2x dx = dt 1 x dx = − dt 2 Z 1 =− et dt 2 1 = − et + c 2 2 1 = − e1− x + c 2
Pouzˇijeme substituci pro na´vrat k pu˚vodnı´ promeˇnne´. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
//
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Prˇ´ıklady s odmocninou z linea´rnı´ho cˇlenu rˇesˇ´ıme vzˇdy druhou substitucˇnı´ metodou. Zbavujeme se tak neprˇ´ıjemne´ odmocniny. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Zavedeme proto substituci t = //
/
.
..
√
1 + x. c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Odmocninu vzˇdy prˇevedeme umocneˇnı´m na tvar bez odmocniny, prˇecha´zı´me takto vlastneˇ k inverznı´ funkci. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Inverznı´ funkce bude v prˇepisu take´ trˇeba. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Hleda´me vztah mezi diferencia´ly. Derivujeme obeˇ strany inverznı´ substituce. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Vsˇechny vy´razy s x zameˇnı´me pomocı´ substituce za ekvivalentnı´ vy´razy s t. Nejdrˇ´ıve pouzˇijeme za x inverznı´ substituce. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Odmocnina odpovı´da´ t. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Diferencia´l take´ substituujeme. Vsˇechny cˇleny s x jsme nahradili. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Zkra´tı´me a konstantu prˇevedeme prˇed integra´l. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Integrujeme. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
Vypocˇteˇte
Z
1 √ dx (2 + x ) 1 + x
√
Z
1+x = t
1 + x = t2 x = t2 − 1
1 √ dx = (2 + x ) 1 + x
dx = 2t dt Z 1 1 = · 2t dt = 2 dt (2 + t2 − 1) t 1 + t2 √ = 2 arctg t + c = 2 arctg 1 + x + c Z
Navra´tı´me se k pu˚vodnı´ promeˇnne´. //
/
.
..
c
Lenka Bara´kova´, 2005 ×
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