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4 - MTH101 Final Term Solved
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4 - MTH101 Final Term Solved
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Post Term
makalah post term
automotive term
Automotive
!"#
6 7 8
#57 8
1 $
d dx
f ( x) = x
n
n
[ x n ] =
x n −1
nx
n −1
nx
n +1
( n − 1) x
n +1
6 7
#57 8
1 $
! "
#
6 7 9
$ #57 8
1 $
y = f ( g ( h ( x ) ) )
u = g (h ( x)) dy v = h ( x)
= __________
dx
dy du dv . . du dv dx
dy du dv du dv dx
dv du dy . . du dv dx
6 7 : #57 8 1 $ %%% & '( & '( ) * 6 7 ;
#57 8 x
1 $
2
t
2 dt a
+
t
a
-
x
x
6 7 <
,
a
#57 8
dx = ________ cf ( x) dx
1 $
.
c
f (cx)dx
c f ( x ) dx
6 7 = #57 8 1 $ / 7
y = x + 6
y = x 2
x = 2
x = 0
6 7 >
#57 8
&,
x = 0 and x = 2
x = 0 and x = 3
x = 2 and x = 3
x = −2 and x = 3
1 $
y = x 2 and y = x + 6
0
6 7 #57 8 1 $ 1 0
1+ 2 + 3 + 4 + _ _ _ _ _ _+ n n +1
2
(n + 1)( n + 2) 2
n( n + 2)
2 n(n + 1) 2
6 7 8
If b > 0 then
d dx
#57 8
1 $
[b x ] = ___________
. x −
xb 1 ln b
x
b ln b
6 7 88 #57 8 1 $ 2 # / 3 3 '( # & 0
b
V = A( y ) dx a
b
V = A( x) dx a
A( x )
[b − a]dx
V =
0
A ( x )
[b + a]dx
V =
0
6 7 8 #57 8 1 $ 2 /
y =
x
;
x = 1, x = 4
&& / 1 4 & & / / 0 4
V = 2π x x dx 1
4
V = 2 x x dx 1
4
V = 2 x x dx 0
4
V=
2x
x dx
−4
6 7 89 #57 8 1 $ & 35 π
4
π
2
π
5
6 7 8:
#57 8
1 $
+ 6+789# %%%%%%% :4
6 7 8;
#57 8
1 $
4/ / 4/ / & / & / / / %%%%%%
6 7 8<
#57 8
1 $
2
+ 4
y = x − 4 x + 5
#
+ :
6 7 8=
#57 8
1 $
lim f ( x) = ........where f ( x) = k x →a
+
;
;$< ;$5 ;
6 7 8>
#57 8
x
1 $ 1 2
+2
dx
2
2
t = x +2
0
x
1 2
+2
dx =
1
t dt
6 *
6 7 8
log b
1 t
#57 8
1 $
#57 8
1 $
= ________
log b t 1 − log b t 1 + log b t − log b t
6 7 7
k =5
1 − 3 + 5 − 7 + 9 − 11 0
(−1) (2k + 1) k
k =0
k =5
(−1) (2k + 1) k
k =1
k =5
(2k + 1) k =1
k =5
(2k + 1) k =1
6 7 8
#57 8
1 $
n
*
f ( x k ) ∆ xk
k =1
+
; ,
# 8 # # 8 # 6 7
#57 8
1 $ n
f ( x
* k
) ∆ xk
k =1
1 == # * * #& * 2 * #4
0
6 7 9 #57 8 1 $ & / & ' (
w( y ) ≥ v( y) for c ≤ y ≤ d
/ , 7
6 >
- > 6
6 7 :
#57 8
1 $
7 0
2(1) + 2( 2) + 2(3) + 2(4) + 2(5)
5
2k
2
k =0
5
2k
2
k =1
5
2k k = 0
5
2k k =1
6 7 ; n
k =1
#57 8
1 $
3
k
2
= ___________
n(n + 1)
4
[n( n + 1)]2 8
n(n + 1)( 2n + 1)
12
(n + 1)(2n + 1) 6
6 7 <
#57 8
a1 < a2 < a3 < ..... < an < .... * )
1 $
4
{an }
*
6 7 =
#57 8
a1 ≥ a2 ≥ a3 ≥ ..... ≥ an ≥ ....
1 $
4
{an }
* ) *
6 7 >
#57 8
1 $
&
{an }
an+1 − an > 0
4
; 7 7 ) * *
6 7
#57 8
1 $
an+1 ? 4 ; 7 7
{an }
an
>1
&
4
) * *
6 7 9
#57 8
1 $
an+1 ? 4 ; 7
{an }
an &
) *
≥1
4
*
6 7 98
f ( n) = an
f '( n) > 0
#57 8
1 $
4
f ( n)
4 7 7
) * *
6 7 9
#57 8
1 $ 2
3
a + ar + ar + ar + ... + ar
k −1
+ ... where (a ≠ 0)
r < 1 & 0 & )& 8& 6 7 99
#57 8
1 $
a + ar + ar + ar + ... + ar k − + ... where ( a ≠ 0) 2
3
& 0 & )& 8& 6 7 9:
#57 8
∞
| u
k
k =1
&
k
k =1
&
/
)& &
k
|
k =1
∞
u
1 $
∞
u
r ≥ 1
1
/
&
6 7 9;
u
#57 8
1 $
k
2
@
ρ = lim k →∞
| uk +1 | | uk |
>1
0
u
+
u
k
&
k
+
& * 6 7 9<
#57 8
1 $
# f g '' ( b
c [ f ( x) + g (x )] dx = ______________ a
b
b
a
a
b
b
a
a
f (cx)dx + g (cx)dx
f ( x) dx + g ( x)] dx b
b
a
a
c f ( x )dx + c g ( x )dx .
6 7 9= #57 8 1 $ 1 # / && & / 3 3 < 0 0 3 .
2
S = 2π f ( x) 1 + [ f ( x)]dx 0
2
2
S = 2π f ( x ) 1 + [ f '( x )] )] dx 0
2
S = 2π f ( x ) 1 + [ f '( x ))]]dx 0
2
S = 2 1 + [ f '( x)]dx 0
6 7 9> #57 8 1 $ 1 / && & y = 4 x ; 1 ≤ x ≤ 4 4
2π ( 4 x ) 1
0 0 2
1 + ( 4 x ) dx
4
(
2π 4 x
)
1
′
(
)
1 + 4 x dx
4
2π + 1
2
′ 1 + ( 4 x ) dx
4
2π ( 4 x ) 1
′ 1 + ( 4 x )
2
dx
6 7 9 #57 8 1 $ 1 ; 1 A & & A ? & & '.B(0
3
W = 3 xdx 2
3
W = 3 xdx 0
3
W = F ( x) dx 0
0
W = F ( x) dx 3
6 7 :
#57 8
1 $
5
0
f ( x) dx = 1
f ( x) dx = 2
0
1
&
5
f ( x) dx 1
0 B 5 5 B 6 7 :8
#57 1 2 x
)&& 6 7 :
?
#57 ρ =
uk
lim
k → + ∞
k
u k
& & & 6 7 :9
#57
C
:& , 3
1
x
2
dx
1
6 7 :: #57 9 D & , 2 x ; 0 ≤ x < 1 f ( x) = 3 x ; 1 ≤ x ≤ 2
2
f ( x) dx
where
0
6 7 :;
#57 9 ∞
(−1) k = 2
#
k −1
2k −1 (k − 1) !
& /
6 7 :< #57 9 : # ) & π
2
Cos x dx Cos −π
2
6 7 := #57 ; :
6 7 :> #57 ; 7 ; 4 / 5.. 5E " /0
6 7 : #57 ; :& , 2
1
2 x +
x
2
x
dx
6 7 ; #57 8 D 2F7F &
lim (1 + sin x)cot x
x →0
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