Transcript
CAPÍTULO 6 6.2 – EXERCÍCIOS – pg. 246 Nos exercícios de 1 a 10, calcular a integral e, em seguida, derivar as respostas para conferir os resultados. 1.
dx
∫x
3
x −2 −1 +c= 2 +c −2 2x d −1 1 = dx 2 x 2 x 3
−3 ∫ x dx =
2.
∫ 9 t
2
+
1 t3
d −1
t3 t 2 2 ( 9 t + t ) dt = 9 . + + c = 3t 3 − + c. ∫ 3 −1 t 2 −3 2
2
d dt
3.
1 −3 1 3 2 + c = 9t 2 − 2. − . t 2 = 9t 2 + . 3t − 2 t t3
∫ (ax
4
+ bx 3 + 3c)dx
x5 x4 = a + b + 3cx + C. 5 4 d a 5 b 4 a 4 b 3 4 3 x + x + 3cx + C = 5 x + 4 x + 3c = ax + bx + 3c. dx 5 4 4 5
4.
1
∫
x
+
x x dx 3
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1
5
x 2 1 x2 2 5 −1 1 3 = ∫ x 2 + x 2 dx = + . + c = 2 x + x 2 + c. 1 3 3 5 15 2 2 d 2 5 1 −1 2 5 3 1 x x + 2 x + x2 + c = 2 x 2 + . .x2 = dx 15 2 15 2 3 x
5.
∫ (2 x
∫ (4 x
2
− 3) 2 dx
4
− 12 x 2 + 9 dx = 4
)
x5 x3 4 − 12 + 9 x + c = x 5 − 4 x 3 + 9 x + c. 5 3 5
d 4 5 4 4 3 2 4 2 x − 4 x + 9 x + c = 5 x − 12 x + 9 + c = 4 x − 12 x + 9 + c. dx 5 5
6.
dx
∫ sen
2
x
= ∫ sen −2 x dx = ∫ cos sec 2 x dx = − cot g x + c. d (− cot g x + c ) = cos sec2 x = 1 2 . dx sen x
7.
2 y − 1 dy ∫ 2 y
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1 1 −21 = ∫ 2 . y2 − . y dy = 2
3
1
2 y2 1 y2 − . +c 3 2 1 2 2 2 2 32 2 12 2 = y − . y + c = 2 y y − 1 + c. 3 2 3 2 2 32 2 2 3 12 1 2 12 2 1 − 12 1 − 12 2 = − + − = − y . y c . y . y 2 y y 3 3 2 2 2 2 2 1 = 2y − . 2y d dx
8.
∫ 3t
2 dt 2 = arc tg t + c +3 3
2
d 2 2 1 2 = arc tg t + c . = 2 . 2 dt 3 3t + 3 3 1+ t
9.
∫x
3
x dx 9
x2 x dx = +c ∫ 9 2 d 2 92 2 9 7 3 x + c = . x2 = x x dx 9 9 2 7 2
10.
x5 + 2x 2 − 1 ∫ x 4 dx
(
= ∫ x + 2x
d dx
−2
−x
−4
x2 x −1 x −3 x2 2 1 dx = +2 − +c= − + 3 +c 2 −1 − 3 2 x 3x
)
x2 2 2 x − 2 − 1 . 9 x2 1 2 1 x5 + 2 x 2 − 1 − + 3 + c = − 2 + = x + − = . x x2 x4 x4 9 x6 2 x 3x 2
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Nos exercícios de 11 a 31, calcular as integrais indefinidas. 11.
x2 ∫ x 2 + 1 dx 1 = ∫ 1 − 2 dx = x − arc tg x + c. x +1
12.
x2 +1 ∫ x 2 dx −1
x 1 ∫ (1 + x )dx = x + − 1 + c = x − x + c. −2
13.
sen x dx 2 x
∫ cos =∫
14.
∫
9 dx 1− x2
=∫
15.
3 1 − x2
dx = 3arc sen x + c.
4 dx x − x2
∫
4
∫x 16.
sen x 1 . dx = ∫ tg x . sec x dx = sec x + c cos x cos x
2
x2 − 1
dx = 2arc sec x + c.
8 x 4 − 9 x3 + 6 x2 − 2 x + 1 dx ∫ x2
(
)
= ∫ 8 x 2 − 9 x + 6 − 2 x −1 + x −2 dx =
8 x3 9 x2 1 − + 6 x − 2 ln x − + c 3 2 x
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17.
et 1 ∫ 2 + t + t dt 3
1 t2 1 2 3 = et + + ln t + c = et + t 2 + ln t + c. 3 2 2 3 2 18.
∫ cos θ . tg θ dθ = ∫ cos θ .
19.
∫ (e
senθ dθ = ∫ senθdθ = − cosθ + c. cosθ
− e − x )dx
x
= ∫ 2 sen h x dx = 2 cosh x + c. 20.
∫ (t +
t + 3 t + 4 t + 5 t )dt 3
4
5
6
t2 t 2 t 3 t 4 t 5 = + + + + +c 2 3 4 5 6 2 3 4 5 5 2 t 2 3 3 4 4 5 6 = + t2 + t3 + t4 + t5 + c 2 3 4 5 6 21.
x −1 / 3 − 5 ∫ x dx
−4 5 = ∫ x 3 − dx x =
22.
− 13
1 − 3
∫2 =
23.
x
t
− 5 ln | x | +c = −3 x
− 13
− 5 ln | x | + c.
− 2e t + cosh t )dt
2t − 2 et + senh t + c. ln 2
∫ sec
2
x(cos 3 x + 1)dx
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1 = ∫ . cos 3 x + sec 2 x dx = senx + tgx + c. 2 cos x 24.
dx , a ≠ 0, constante 2 + a2
∫ (ax)
dx dx 1 =∫ 2 2 = 2 arc tg x + c 2 a x +a a ( x + 1) a
=∫
25.
2 2
x2 − 1 ∫ x 2 + 1 dx 2 = ∫ 1 − 2 dx = x − 2arc tg x + c. x +1 3
26.
∫
3
1 8(t − 2) t + dt 2 6
1 1 = ∫ 2 (t − 2) 2 t + dt = ∫ 2 t 2 − 4t + 4 t + dt 2 2 1 7 = 2 ∫ t 3 + t 2 − 4t 2 − 2t + 4t + 2 dt = 2 ∫ t 3 − t 2 + 2t + 2 dt 2 2
(
)
t4 7 t3 t2 t 4 7t 3 = 2 − . + 2 + 2t + c = − + 2 t 2 + 4t + c. 4 2 3 2 2 3 27.
∫ e
t
− 4 16t +
3 dt t3
5
t4 t −2 8 5 3 =e −2 +3 + c = et − t 4 − t − 2 + c. 5 −2 5 2 4 t
28.
29.
ln x
∫ x ln x
2
dx
=∫
ln x 1 dx dx = ∫ = ln | x | + c. x 2 ln x 2 x
∫ tg
x cos ec 2 x dx
2
=∫
sen 2 x 1 dx = ∫ sec 2 x dx = tg x + c. 2 2 cos x sen x
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30.
∫ ( x − 1) ( x + 1) 2
( = ∫ (x
2
dx
)(
)
= ∫ x 2 − 2 x + 1 x 2 + 2 x + 1 dx 4
)
+ 2 x 3 + x 2 − 2 x 3 − 4 x 2 − 2 x + x 2 + 2 x + 1 dx
(
)
= ∫ x 4 − 2 x 2 + 1 dx = 31.
dt , onde n ∈ z 1 n n − t 2
∫
∫ − 2dt = −2t + c
Se n = 0,
dt 2 dt =∫ = 2 ln | t | +c 1 n t n − t 2
∫
Se n = 1,
1
Se n ≠ 1,
32.
x5 x3 − 2 + x + c. 5 3
1 n − 2
−n ∫ t dt =
t 1− n +c 1 (1 − n ) n − 2 1
.
Encontrar uma primitiva F , da função f ( x) = x 2 / 3 + x, que satisfaça F (1) = 1. F ( x) = ∫
(
)
5
x 3 x2 x + x dx = + +c 5 2 3 2 3
3 53 x 2 x + +c 5 2 3 1 F (1) = + + c = 1 5 2 F ( x) =
3 1 10 − 6 − 5 − 1 c =1− − = = 5 2 10 10 F ( x) =
33.
3 53 x 2 1 x + − . 5 2 10
Determinar a função f (x) tal que
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1 + cos 2 x + c 2 d 2 1 1 x + cos 2 x + c = 2 x + (− sen 2 x).2 = 2 x − sen 2 x. dx 2 2
∫ f ( x)dx = x
34.
2
Encontrar uma primitiva da função f ( x) =
1 + 1 que se anule no ponto x = 2. x2
x −1 1 1 −2 F ( x) = ∫ 2 + 1dx = ∫ x + 1 dx = +x+c=− +x+c −1 x x 1 F ( 2) = − + 2 + c 2 1 1− 4 − 3 c= −2= = 2 2 2 1 3 F ( x) = − + x − x 2
(
35.
)
Sabendo que a função f (x) satisfaz a igualdade. 1
∫ f ( x)dx = sen x − x cos x − 2 x
2
+ c, determinar f (π / 4).
d 1 2 1 sen x − x cos x − x + c = cos x − ( x(− sen x) + cos x) − 2 x dx 2 2 = cos x + x sen x − cos x − x = x sen x − x = x ( sen x − 1)
π π 2 π 2 − 1 π π π f = sen − 1 = − 1 = . = 4 4 2 2 4 4 4 36.
(
)
2 −2 . 8
Encontrar uma função f tal que f ′( x) + sen x = 0 e f (0) = 2. f ′( x) + sen x = 0 f ′( x) = − senx
∫ − sen x dx = + cos x + c f ( x) = + cos x + c f (0) = cos 0 + c = 2 c = 2 −1 = 1 ∴ f ( x) = cos x + 1.
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