1 1) (Total: 1.0) Calcule ∫ tan x dx. 2)

1
Cálculo 2
PURO-UFF - 2016.1
P1 - 4/jul/2016 - Eduardo Ochs
Links importantes:
http://angg.twu.net/2016.1-C2.html (página do curso)
http://angg.twu.net/2016.1-C2/2016.1-C2.pdf (quadros)
http://angg.twu.net/LATEX/2016-1-C2-material.pdf
[email protected] (meu e-mail)
1) (Total: 1.0) Calcule
R
2) (Total: 1.0) Calcule
R x=2
3) (Total: 1.5) Calcule
R
cos4 x dx.
4) (Total: 1.5) Calcule
R
x2
x2 +x−2
5) (Total: 1.5) Calcule
R
x ex cos x dx.
tan x dx.
x=−1
x−4 dx.
dx.
6) (Total: 2.0)
R x=1 √
6a) (1.5 pts) Calcule x=0 4 − x2 dx.
R x=1 √
6b) (0.1 pts) Represente x=0 4 − x2 dx graficamente.
R x=1 √
6c) (0.4 pts) Mostre como calcular x=0 4 − x2 dx pelo gráfico.
7) (Total: 2.5) Calcule
R x=2
x=−1
|ex − 1| dx.
2016-1-C2-P1 July 11, 2016 13:24
2
Método de Heaviside:
β
α
Se f (x) = x−a
+ x−b
+
γ
x−c
=
então limx→a f (x)(x − a) = α
p(x)
(x−a)(x−b)(x−c) ,
p(a)
= (a−b)(a−c)
.
Substituição:
R x=b
x=b
g(h(x))|x=a = x=a g 0 (h(x)) d h(x)
dx dx
||
R u=h(b)
u=h(b)
g(u))|u=h(a) = u=h(a) g 0 (u) du
Fórmulas:
R x=b
f (g(x)) d g(x)
dx dx
x=a
R x=b
= x=a f (u) du
dx
R u=g(b) dx
= u=g(a) f (u) du
fR(g(x)) d g(x)
dx dx
du
= R f (u) dx dx
= f (u) du
R
[ u=g(x) ]
[ u=g(x) ]
Substituição inversa:
R x=h−1 (β)
x=h−1 (β)
g(h(x))|x=h−1 (α) = x=h−1 (α) g 0 (h(x)) d h(x)
dx dx
||
R u=h(h−1 (β))
u=h(h−1 (β))
g(u))|u=h(h−1 (α)) = u=h(h−1 (α)) g 0 (u) du
||
R u=β
u=β
g(u))|u=α = u=α g 0 (u) du
Fórmulas:
R u=β
f (u) du
u=α
R x=g−1 (β)
= x=g−1 (α) f (u) du
dx dx
R x=g−1 (β)
= x=g−1 (α) f (g(x)) d g(x)
dx dx
R
f (u) du
R
= f (u) du
dx dx
R
= f (g(x)) d g(x)
dx dx
Substituição trigonométrica:
√
R s=b
F (s, 1 − s2 ) ds
s=a
√
R θ=arcsen b
θ
= θ=arcsen a F (sen θ, 1 − sen2 θ) d sen
dθ dθ
R θ=arcsen b
= θ=arcsen a F (sen θ, cos θ) cos θ dθ
√
R z=b
F (z, z 2 − 1) dz
z=a
√
R θ=arcsec b
θ
= θ=arcsec a F (sec θ, sec2 θ − 1) d sec
dθ dθ
R θ=arcsec b
= θ=arcsec a F (sec θ, tan θ) sec θ tan θ dθ
√
R t=b
F (t, 1 + t2 ) dt
t=a
√
R θ=arctan b
θ
= θ=arctan a F (tan θ, 1 + tan2 θ) d tan
dθ
dθ
R θ=arctan b
2
= θ=arctan a F (tan θ, sec θ) sec θ dθ
h
u=g(x)
x=g −1 (u)
i
[ x=g−1 (u) ]
√
FR (s, 1√− s2 ) ds
ds
= F (s, 1 − s2 ) dθ
dθ
R
= F (s, c)c dθ
√
R
2 − 1) dz
FR (z, z√
= F (z, z 2 − 1) dz
dθ dθ
R
= F (z, t)zt dθ
√
R
FR (t, 1√+ t2 ) dt
dt
= F (t, 1 + t2 ) dθ
dθ
R
2
= F (t, z)z dθ
R
2016-1-C2-P1 July 11, 2016 13:24
s=sen θ θi
h θ=arcsen
s=sen θ
c=cos θ
θ=arcsen θ
z=sec θ zi
h θ=arcsec
z=sec θ
θ=arcsec z
t=tan θ
t=tan θ ti
h θ=arctan
t=tan θ
θ=arctan t
z=sec θ
3
Gabarito (não revisado, contém erros de vários tipos!):
1)
2)
R
s
c dθ
1
Rc s dθ
− 1c dc
=
=
=
=
=
R
R
R −4
x dx
R x=2 −4
x dx
x=−1
=
=
=
tan θ dθ
− ln |c|
− ln | cos θ|
=
=
=
x−3
R−3
x=0
R x=2
x−4 dx + x=0 x−4 dx
R x=a
R x=2
lima→0− x=−1 x−4 dx
+ limb→0+ x=b x−4 dx
x=a
x=2 x−3 x−3 −
+
lima→0
+ limb→0
−3 −3 −3 x=−1 −3 −3x=b −3 lima→0− a−3 − (−1)
+ limb→0+ 2−3 − b−3
−3
1/8
−∞
−1
+∞
−
+
−
−3
−3
−3
−3
x=−1
3) Seja E = eiθ . Então
−1
cos θ = E+E
,
2
2
−2
(cos θ)2 = E +1+E
,
4
E 4 +4E 2 +6+4E −2 +E −4
4
(cos θ) =
16
4
−4
2
−2
= 18 E +E
+ 12 E +E
+ 38
2
2
1
1
3
= R8 cos 4θ + 2 cos 2θ + 8
R
1
1
3
(cos θ)4 dθ =
8 cos 4θ + 2 cos 2θ + 8 dθ
1 cos4θ
1 cos 2θ
3
= 8 4 + 2 2 + 8θ
2
2
x
4) x2 +x−2
= (x +x−2)−x+2
=1+
x2 +x−2
−x+2
A
B
Se (x+2)(x−1) = x+2 + x−1 então
4
limx→−2 −x+2
x−1 = A = −3 e
1
limx→1 −x+2
x+2 = B = 3 .
Conferindo:
1/3
A
B
= −4/3
x+2 + x−1
x+2 + x−1
=
=
Daı́:
R
x2
x2 +x−2
dx
−x+2
x2 +x−2
=1+
−x+2
(x+2)(x−1)
− 43 (x−1)+ 13 (x+2)
(x+2)(x−1)
−x+2
(x+2)(x−1)
R
1/3
=
1 + −4/3x + 2 + x−1
dx
= x − 43 ln |x + 2| + 13 ln |x − 1|
2016-1-C2-P1 July 11, 2016 13:24
4
5)
c = cos x, s = sen x.
R Sejam
R Então
x
x
e
c
dx
=
e
s
−
ex |{z}
s dx
|{z} |{z}
|{z} |{z}
|{z}
0
0
g
g
f
f
f
R x
R x g
R
x
e
s
dx
=
e
(−c)
−
e (−c) dx = −ex c + ex c dx
|{z} |{z}
|{z} |{z}
|{z}
|{z}
0
f
g
f
g
f0
g
R x
R x
x
x
x
x
x
x
x
Re cxdx = e sx − ex s dx = e s − (−e c + e c dx) = e s + e c − e c dx
2
R xe c dx = ex s + ex c
e c dx = (e s + e c)/2
R x
R
e s dx = −ex c + ex c dx = −ex c + (ex s + ex c)/2 = (ex s − ex c)/2
R
R
x |{z}
ex c dx = |{z}
x (ex s + ex c)/2 − |{z}
1 (ex s + ex c)/2 dx
|{z}
|
{z
}
|
{z
}
g0
f
f
f0
g
g
R
R
= 12 (xex s + xex c) − 12 ex s dx − 12 ex c dx
= 12 (xex s + xex c) − 14 (ex s − ex c) − 14 (ex s + ex c)
= 12 (xex s + xex c) − 12 ex s
= 12 (xex s + xex c − ex s)
R
R
√
6)
R√
4 − x2
=
=
=
4 − x2 dx
=
=
=
=
=
=
=
=
7)
R x=2
x=−1
|ex − 1| dx
p
2
p4 − 4(x/2)
4(1 − (x/2)2 )
p
2 1 − (x/2)2
R p
2 1 − (x/2)2 dx
R √
2 1 − s2 · 2 ds
R√
4
1 − s2 ds
Rp
4 R 1 − (sen θ)2 cos θ dθ
4 R (cos θ)2 dθ
2θ
4 1+cos
dθ
2
θ
sen 2θ
4( 2 + 4 )
2θ + sen 2θ
h
s=x/2
x=2s
dx=2ds
h
s=sen θ
θ=arcsen s
ds=cos θdθ
i
i
R x=0
R x=2
= x=−1 |ex − 1| dx + x=0 |ex − 1| dx
R x=0
R
x=2
= x=−1 1 − ex dx + x=0 ex − 1 dx
x=0
x=2
= (x − ex )|x=−1 + (ex − x)|x=0
= (0 − e0 ) − (−1 − e−1 ) + (e2 − 2) − (e0 − 0)
= −1 + 1 + e−1 + e2 − 2 − 1
= −3 + e−1 + e2
2016-1-C2-P1 July 11, 2016 13:24