Интегральное исчисление функции одной переменной. - 40 стр.

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40 §2. ïÐÒÅÄÅÌÅÎÎÙÊ ÉÎÔÅÇÒÁÌ, ÏÓÎÏ×ÎÙÅ Ó×ÏÊÓÔ×Á. . .
ÒÁÓÓÍÏÔÒÉÍ ÐÏÄÓÔÁÎÏ×ËÕ x =
1
cos t
, ÔÁË ËÁË x ∈ [1, 2], ÔÏ t ∈
0;
π
3
.
2
Z
1
√
x
2
− 1
x
4
dx =
π/3
Z
0
q
1
cos
2
t
− 1
1/ cos
4
t
·
sin t
cos
2
t
dt =
π/3
Z
0
sin
2
t ·cos t dt =
sin
3
t
3
π/3
0
=
√
3
8
.
ðÒÉÍÅÒ 5. (ÓÍ. Ó×ÏÊÓÔ×Ï 5).
1
Z
0
arctg x dx = x · arctg x
1
0
−
1
Z
0
x dx
1 + x
2
= x arctg x
1
0
−
−
1
2
ln |1 + x
2
|
1
0
=
π
4
−
1
2
(ln 2 − ln 1) =
π
4
−
ln 2
2
.
2.2. çÅÏÍÅÔÒÉÞÅÓËÉÅ ÐÒÉÌÏÖÅÎÉÑ ÏÐÒÅÄÅÌÅÎÎÏÇÏ ÉÎÔÅÇÒÁÌÁ
I. äÌÉÎÁ ËÒÉ×ÏÊ.
I.1. ëÒÉ×ÁÑ ÚÁÄÁÎÁ Ñ×ÎÙÍ ÕÒÁ×ÎÅÎÉÅÍ y = f(x), x ∈ [a, b], ÔÏÇÄÁ ÄÌÉÎÁ
ËÒÉ×ÏÊ ×ÙÞÉÓÌÑÅÔÓÑ ÐÏ ÆÏÒÍÕÌÅ:
l =
b
Z
a
p
1 + [f
0
(x)]
2
dx.
ðÒÉÍÅÒ 6. îÁÊÔÉ ÄÌÉÎÕ ËÒÉ×ÏÊ y =
x
2
6
ÎÁ ÕÞÁÓÔËÅ x ∈ [0, 4].
y
0
=
x
2
6
0
=
x
3
l =
4
Z
0
r
1 +
x
3
2
dx =
1
3
4
Z
0
p
3
2
+ x
2
dx =
=
1
3
1
2
x
p
3
2
+ x
2
+
9
2
ln(x +
p
3
2
+ x
2
)
4
0
=
=
1
3
1
2
4 ·
p
3
2
+ 4
2
+
9
2
ln(4 +
p
3
2
+ 4
2
)−
−
1
2
· 0 ·
p
3
2
+ 0 −
9
2
ln(0 +
p
3
2
+ 0)
=
1
3
·
10 +
9
2
ln 9−
−
9
2
ln 3
=
1
3
10 +
9
2
ln 3
=
10
3
+
3
2
ln 3.
40                      §2. ïÐÒÅÄÅÌÅÎÎÙÊ ÉÎÔÅÇÒÁÌ, ÏÓÎÏ×ÎÙÅ Ó×ÏÊÓÔ×Á. . .

ÒÁÓÓÍÏÔÒÉÍ ÐÏÄÓÔÁÎÏ×ËÕ x = cos1 t , ÔÁË ËÁË x ∈ [1, 2], ÔÏ t ∈ 0; π3 .
                                                                       

Z2 √ 2                                                                            √
                    q
               Zπ/3     1
                            − 1              Zπ/3                           π/3
    x −1                 2
                      cos t       sin t             2                sin3 t         3
       4
         dx =               4
                                ·    2
                                        dt =     sin t · cos t dt =             =     .
     x               1/ cos t cos t                                    3 0         8
1                    0                                0

     ðÒÉÍÅÒ 5. (ÓÍ. Ó×ÏÊÓÔ×Ï 5).
    Z1                         1      Z1                    1
                                            x dx
         arctg x dx = x · arctg x −               = x arctg x −
                               0           1 + x2            0
    0                                 0
                                                  1
                                1             π 1               π ln 2
                               − ln |1 + x2| = − (ln 2 − ln 1) = −     .
                                2           0 4 2               4  2

2.2. çÅÏÍÅÔÒÉÞÅÓËÉÅ ÐÒÉÌÏÖÅÎÉÑ ÏÐÒÅÄÅÌÅÎÎÏÇÏ ÉÎÔÅÇÒÁÌÁ

  I. äÌÉÎÁ ËÒÉ×ÏÊ.
  I.1. ëÒÉ×ÁÑ ÚÁÄÁÎÁ Ñ×ÎÙÍ ÕÒÁ×ÎÅÎÉÅÍ y = f (x), x ∈ [a, b], ÔÏÇÄÁ ÄÌÉÎÁ
ËÒÉ×ÏÊ ×ÙÞÉÓÌÑÅÔÓÑ ÐÏ ÆÏÒÍÕÌÅ:
                                  Zb p
                               l=     1 + [f 0(x)]2 dx.
                                      a
                                                      2
     ðÒÉÍÅÒ 6. îÁÊÔÉ ÄÌÉÎÕ ËÒÉ×ÏÊ y = x6 ÎÁ ÕÞÁÓÔËÅ x ∈ [0, 4].
                                      2 0
                                      x          x
                               y0 =           =
                                       6         3
                    Z4 r       x 2          Z4
                                            1 p 2
                l=       1+          dx =            3 + x2 dx =
                                3           3
                    0                         0
                    p                                       4
                 1 1                  9           p
               =       x 32 + x2 + ln(x + 32 + x2)              =
                 3 2                  2                       0
                    
                  1 1      p              9          p
                =       4 · 32 + 42 + ln(4 + 32 + 42)−
                  3 2                     2
                                                         
           1    p           9          p                1        9
         − · 0 · 32 + 0 − ln(0 + 32 + 0) = · 10 + ln 9−
           2                2                           3        2
                                             
                  9          1          9            10 3
                − ln 3 =          10 + ln 3 =           + ln 3.
                  2          3          2             3   2

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