ЛЕКЦІЇ ² ВПРАВИ

â³äíîñò³ äî ôîðìóëè äèôåðåíö³þâàííÿ ñêëàäåíî¿ ôóíêö³¿
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( ⎡ϕ () t ⎤) = x
⎡ϕ() t ⎤⋅ϕ ′() t = f ⎡ϕ() t ⎤ϕ′
( x)
⎣ ⎦ ⎣ ⎦ ⎣ ⎦ . Îòæå, ìàòèìåìî:
β
t
() ′() () ( ) () ( ) ∫ ( ) .
∫f⎡⎣ϕ t ⎤⎦ϕ t dt = ⎡⎣ϕ β ⎤⎦−⎡⎣ϕ α ⎤⎦
= b − a = f x dx
α
Òåîðåìó äîâåäåíî.
Ðîçãëÿíåìî ïðèêëàäè âèêîðèñòàííÿ ö³º¿ òåîðåìè.
Ïðèêëàä 9.5.1. Çíàéòè ïëîùó ô³ãóðè, îáìåæåíî¿ åë³ïñîì.
Âðàõîâóþ÷è ñèìåòðè÷í³ñòü åë³ïñà, ÷åòâåðòèíà øóêàíî¿ ïëîù³
S îá÷èñëþºòüñÿ çà ôîðìóëîþ:
1 a
b 2 2
S = ∫ a −x dx .
4 0 a
Çðîáèìî çàì³íó çì³ííî¿: x = a sint. Òîä³ áóäåìî ìàòè
2 2 2 2 2
1 b
a ⎡dx = a cos tdt, a − x = a − a sin t = a cost⎤
2 2
S = ∫ a − x dx = ⎢
⎥ =
4 a 0
⎢ x = 0⇒ t = 0, x = a⇒ t = π ⎥
⎣
2 ⎦
π
π
2 2
2 ab
( )
= ba ∫ cos tdt = ∫ 1+ cos 2t dt =
0 2 0
π
2
ab ⎛ 1 ⎞ πab
= ⎜t+ sin 2 t⎟
= .
2 ⎝ 2 ⎠ 4
Òàêèì ÷èíîì, S = πab. Çîêðåìà, ÿêùî à = b = R, òî îòðèìàºìî:
S = πR 2 (ôîðìóëà äëÿ îá÷èñëåííÿ ïëîù³ êðóãà).
Çàóâàæåííÿ. Íà â³äì³íó â³ä ìåòîäó ï³äñòàíîâêè ó íåâèçíà÷åíîìó
³íòåãðàë³ íåìà ïîòðåáè ïîâåðòàòèñÿ äî ñòàðî¿
çì³ííî¿, îñê³ëüêè âèçíà÷åíèé ³íòåãðàë äîð³âíþº ñòàë³é âåëè-
÷èí³.
Òåîðåòè÷í³ ïðèêëàäè. Âñòàíîâèìî çà äîïîìîãîþ çàì³íè
çì³ííî¿ òàê³ òâåðäæåííÿ.
1. ßêùî f(x) º íåïàðíîþ, òîáòî f(–x) =–f(x), òî ∀à∈R:
a
( ) 0
0
I = ∫ f x dx = . (9.5.2)
−a
b
a
Ñïî÷àòêó ðîç³á’ºìî öåé ³íòåãðàë íà äâà
0
−a
a
( ) ( )
I = ∫ f xdx+
∫ f x dx. (9.5.3)
Ïåðøèé ³íòåãðàë ïåðåòâîðèìî òàêèì ÷èíîì:
0 ⎡ x =− t,
dx =−dt
⎤
∫ f( x)
dx = ⎢ =
−a
x =−a ⇒ t = a, x = 0 ⇒ t = 0
⎥
⎣
⎦
0
a
a
() () ( )
0
0
a
= ∫f tdt = − ∫ f t dt =−∫ f x dx. (9.5.4)
Ó ëàíöþæêó ð³âíîñòåé (9.5.4) ìè ïîñë³äîâíî âèêîðèñòàëè
âëàñòèâîñò³ 2 ³ 1 âèçíà÷åíîãî ³íòåãðàëà (ï. 9.3).
ßêùî òåïåð ï³äñòàâèìî (9.5.4) â (9.5.3), òî ä³éñíî îòðèìàºìî
(9.5.2).
2. ßêùî f(õ) ïàðíà, òîáòî f(–x) =f(x), òî ∀à∈R:
a
−a
( ) = 2 ( )
∫ f x dx ∫ f x dx .
Äîâîäèòüñÿ àíàëîã³÷íî 1. ×èòà÷åâ³ ïðîïîíóºìî çðîáèòè
öå ñàìîñò³éíî.
Ïðèêëàäè 9.5.2 – 9.5.4
5 3
xdx 2
4
2 2⎛t
⎞ 4 2⎛64 1 ⎞
9.5.2. ∫ = ∫( t − 1) dt = ⎜ − t⎟ = −4− + 1 = 4
0 1 3 9 1 9 3 1
⎜ ⎟
+ ⎝ ⎠ 9⎝ 3 3
.
x
⎠
2
t −1 2
Òóò ââåäåíà íîâà çì³ííà t = 1+ 3 x ⇒ x = , dx = tdt .
3 3
Ïðè öüîìó íîâ³ ìåæ³ ³íòåãðóâàííÿ çíàõîäÿòüñÿ òàêèì ÷èíîì:
t 1
= 1+ 3⋅ 0 = 1, t 2
= 1+ 3⋅ 5 = 4 .
9.5.3.
ln 3 3 3
x −x
−1 2
ln 2 2 2
a
0
dx dt dt 1 t −1 3 1⎛ 2 1⎞
1
∫ = ∫ = ∫ = ln = ⎜ln − ln ⎟ = ln1,5.
e −e t( t−t ) t −1
2 t + 1 2 2⎝ 4 3⎠
2
x
ln 2 ln 3
Ïîêëàäåíî t = e ⇒ x = ln t, dx = dt / t, t = e = 2, t = e = 3 .
0
1 2
304 305