Uploaded by Leen Alajarmeh

Revisión de Examen de Cálculo Integral: Integrales y Volúmenes

advertisement
INTEGRAL
MIDTERM
#
REVIEW
Evaluate
(1
2
=>
xi
&
Ef
+
64x
de
U
·
3/2
(1 + 64x2)
192
=
+
25131
-
192
1
:
:
(8) de
let y
2
=
CALCULUS FINAL REVIEW
du
=
=
1 + 64
1282 dx
Israe
#2)
x
=
dx
dx
,
is
sec -I
Seco
=
Sec xtan
~
Fa
do
=
I
f
S
Sec· di (Secx tanx)
tans
sex
tan
Gose
I
Stosc
I t
9
tanx
ilt
let U
du
=
Sinc doc
Cossed
=
du
=
=
-
I
3sino
↓?
=
50(3
Jice-s
#3
do
O
INTEGRATION
By PARTS
f(x)g(x
ff(x) g(x)
f(x)
=
x
F()
=
1
-
:
9(x) =
g(x)
-
S
=
5
-
-
j
f-este
-
5
*
-Easye
O
-sco -Esco
=
e
1+ u
-
-
=
du
=
5xdx
da
#4)
[Cos Sin dx
Evaluate
let U
Sinc
=
du
cose da
=
Scosic
cosoc
.
.
Since,
S(1-sin2x) Sincer
((
4/22
-
42)y
*
124
-
I
-
Sinlc-Sinks
e
+ C
#5) Evaluate
&
di
f
+ 72
+
x(x
1)
A
t
x
A(x +1)
A(1)
A
=
+ 0
Px + C
+
(x
1)
+
+
(x + C(x)
0
= 1
1
↳!
((x) +
In ke
-
f -x
:
H
+Ink +1) +C
=
1
x= 0
x
=
1
x
=
-
1
#6) Prove
Sostcdx
[costco I costosine
=
=
Coghosinx + (n 1)/ Cost sin doc
=
Cosh" c sinx +
-
/costcoc
=
Scostco
=
(n-1)] Cost-Ec /I-cos2] do
cosicsinat (n-1
+ (n-1)
/cossdx
n
h[cos" Tick t
Cos" sinc -S-(n-1) Cost-E sinx. Since
=
-
Soscdic
+
=
/cost3dx- (n 1) Costecda
=
Cost sinc + (n-1) Cost-d
Cost Sinc + (n-1
Costco
+costin sinc (n-1) /cosick
+
#A)
↑
n
Y
=
T
↳
k
T
↓ '
:=
y
>y
=
1/5
=
x
=
!
+
x 2
=
y
x =3
Y-bounds
V
b
v
=
O
(πR(U)
2
-
(h) dh
1/3
>
y
=
=
112
1/2
:
7/3
->
112
a
i
(3)2
12
-
(2) dy
+
πf(6
2)2 16 1/4 e
-
-
-
#8)
I
renos a
ado
E
R y
=
z
V
L
·
=af Rugbye
·,
2π/y - y3 dy
G
↓=
T
fl
#a)
M
·
17
E
-
&
Coco)
v
·
y
A (x) dx
Of
CO
y
a
2
y
f W(y)2
/Ex 1)
Y
-
=
:
tasino
=
Sin
O
=
Y
(t
y
=
recoil
y
=
x+
b
o
=
1)
=
=
Es
=
1/2
qx
-
+b
1
O
y
E
=
Ex
-
1
Download