北京邮电大学高等数学ppt
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0 0 a 0 a
Let x t
f ( x ) f ( x ) f ( x ) f ( x )
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Example: Find Solution:
1
1
2 x 2 x cos x dx . 2 1 1 x
1
1
1
1 x cos x 2x2 dx dx 1 2 2 1 1 x 1 1 x
Example: Prove Proof
2 0
sin xdx 2 cos n xdx ( n is any positive integer).
n 0
2 0
sin xdx sin n t d t 2 2 2
n 0
cos n tdt
2
18
2(e 2 1).
Integration by Parts
Example: Evaluate
1 2
0
arcsin xdx .
dx 1 x
2
b
a
udv uv a vdu
b a
b
Solution Let u arcsin x , dv dx , then du
0
T
a
0
f ( x )dx f (a x )dx
0
a
(let t a x )
2 0
f (sin x )dx 2 f (cos x )dx
0
0
f (sin x )dx 2 2 f (sin x )dx
0
(7)
0
xf (sin x )dx f (sin x )dx
2 0
2
0
f (sin x )dx
Integration by Parts
Integration by parts for indefinite integrals
When u and v are differentiable functions of x, the Product Rule for differentiation tell us that
偶函数
奇函数
40
2 2 x 1 x (1 1 x ) dx 4 dx 2 2 0 1 1 x 1 (1 x )
2
40 (1 1 x )dx 4 40
2
1
1
1 x 2 dx
4 .
单位圆的面积
(t ) (t )
必需注意换元必换限 , 原函数中的变量不必代回 .
b
a
udv uv a vdu
b a
b
4
0
e x dx .
Solution Let
2 x t , then x t , dx 2tdt . Thus
4
0
e dx e 2tdt 2 tde 2 te
x t t 0 0
2
2
t 2 0
e t dt
0
2
0
2 cos n tdt
0
2 cos n xdx .
0
6
Integration by Substitutions for definite integrals
1 Example: Find lim 2 x 0 x sin xt 0 t dt du u . t 0 u 0; t x u x 2 . Solution Let xt u, then t , dt x x Then x 2 sin u du x sin xt x 2 sin u 0 t dt 0 u x 0 u du x So 0 2 x sin u 0 sin x 2 0 u du x 2 2 x 1 x sin xt lim 2 dt lim 1. lim 2 0 x 0 x x 0 x 0 t x 2x
b
Since
a
f ( x )dx F (b) F (a )
d F ( t ) f ( t ) ( t ), dt
we have
f ( t ) ( t )dt F ( t )
F ( ) F ( )
F (b) F (a )
, v x,
1 2
0
arcsin xdx x arcsin x 0
1 2
1 2
xdx 1 x2
0
1 1 1 1 2 d (1 x 2 ) 2 6 2 0 1 x2
a a
f ( x )dx 0
a a
(2) If f ( x )=f ( x )， then
f ( x )dx 2 f ( x )dx
0
a
偶倍奇零
(3) If f ( x T ) f ( x ) , then
(4)
(5) (6)
a T
a
f ( x )dx f ( x )dx
例 若 f ( x ) 在[0,1]上连续，证明
0
2
f (sin x )dx f (cos x )dx .
0
2
Proof （1）设 x t 2 x 0 t , 2
dx dt,
0
2
x t 0, 2 0 f (sin x )dx f sin t dt 2 2
School of Science, BUPT
Integration by substitution and by parts in definite integrals
2
Integration by Substitutions for definite integrals
When we want to evaluate the value of a definite integral can find the corresponding indefinite integral
0
0
xf (sin x )dx f (sin x )dx. 2 0
0
Example: Find
Solution
0
x sin x dx 2 1 cos x
0
x sin x sin x dx dx 2 2 2 0 1 cos x 1 cos x
, I . Then
b
a
f ( x )dx f ( t ) ( t )dt .
3
Integration by Substitutions for definite integrals
Proof Let F be an antiderivative of f on the interval I, then
a
2 0
a cos t dt 2 2 a sin t a (1 sin t )
cos t sin t 1 dt sin t cos t 1 1 2 . ln sin t cos t 0 4 2 2 2
2 0
cos t 1 2 dt sin t cos t 2 0
2arcsin( ln x )
e
. 6
1 Example: Find dx . (a 0) 2 2 0 x a x Solution: Let x a sin t , dx a cos tdt , x a t , x 0 t 0, 2
2
when x 1 ; thus by the formula, we have
1
0
1 x 2 dx 2 cos 2 tdt
0
Βιβλιοθήκη Baidu
1 1 2 t sin 2t 2 2 0
.
4
5
Integration by Substitutions for definite integrals
Therefore, the formula holds.
4
Integration by Substitutions for definite integrals
Example: Evaluate
1
0
1 x 2 dx .
Solution Let x sin t ,
dx cos tdt . Then t 0 when x 0; t
x
7
Example: Find
3 e4
e
3 e4
dx . x ln x(1 ln x )
Solution
e
d (ln x ) ln x(1 ln x ) d (ln x ) 2 ln x (1 ln x )
3 e4
3 e4
3 e4
e
e
d ln x 1 ( ln x )2
d dv du ( uv ) u v dx dx dx
Integrating both sides with respect to x and rearranging leads to the integral equation
dv d du u dx ( uv ) dx v dx dx dx dx du uv v dx . dx
0
x t 0,
0 xf (sin x )dx ( t ) f [sin( t )]dt ( t ) f (sin t )dt f (sin t )dt tf (sin t )dt 0 0
0
f (sin x )dx xf (sin x )dx,
17
Integration by Parts
Integration by parts for definite integrals
When u and v are differentiable functions of x on the interval [a , b]. Then
Example: Evaluate
2 2
f (cos t )dt f (cos x )dx;
0 0
例 若 f ( x ) 在[0,1]上连续，证明
0
xf (sin x )dx f (sin x )dx . 2 0
x t dx dt ,
Proof: Let
x 0 t ,
1 d (cos x ) arctan(cos x )0 2 2 0 1 cos x 2 2 ( ) . 4 2 4 4
xf (sin x )dx f (sin x )dx . 2 0
0
(1) If f ( x )=f ( x )， then
b
a
f ( x )dx , if we
f ( x )dx , then we can use
the Newton-Leibniz formula to obtain the value immediately.
Theorem (Integration by substitution for definite integrals) Suppose that the function x ( t ) is continuously differentiable on the interval [ , ] , ( ) a , ( ) b , and that f is continuous in the range of
偶倍奇零
(1) If (2) If Proof
a 0
then
a
a
f ( x )dx 2 f ( x )d x
0
a
then
a
a
a
f ( x )d x 0
a
a f ( x) dx a f ( x) dx 0 f ( x) dx
f (t ) d t f ( x) dx [ f ( x ) f ( x ) ] dx
Let x t
f ( x ) f ( x ) f ( x ) f ( x )
机动 目录 上页 下页 返回 结束
Example: Find Solution:
1
1
2 x 2 x cos x dx . 2 1 1 x
1
1
1
1 x cos x 2x2 dx dx 1 2 2 1 1 x 1 1 x
Example: Prove Proof
2 0
sin xdx 2 cos n xdx ( n is any positive integer).
n 0
2 0
sin xdx sin n t d t 2 2 2
n 0
cos n tdt
2
18
2(e 2 1).
Integration by Parts
Example: Evaluate
1 2
0
arcsin xdx .
dx 1 x
2
b
a
udv uv a vdu
b a
b
Solution Let u arcsin x , dv dx , then du
0
T
a
0
f ( x )dx f (a x )dx
0
a
(let t a x )
2 0
f (sin x )dx 2 f (cos x )dx
0
0
f (sin x )dx 2 2 f (sin x )dx
0
(7)
0
xf (sin x )dx f (sin x )dx
2 0
2
0
f (sin x )dx
Integration by Parts
Integration by parts for indefinite integrals
When u and v are differentiable functions of x, the Product Rule for differentiation tell us that
偶函数
奇函数
40
2 2 x 1 x (1 1 x ) dx 4 dx 2 2 0 1 1 x 1 (1 x )
2
40 (1 1 x )dx 4 40
2
1
1
1 x 2 dx
4 .
单位圆的面积
(t ) (t )
必需注意换元必换限 , 原函数中的变量不必代回 .
b
a
udv uv a vdu
b a
b
4
0
e x dx .
Solution Let
2 x t , then x t , dx 2tdt . Thus
4
0
e dx e 2tdt 2 tde 2 te
x t t 0 0
2
2
t 2 0
e t dt
0
2
0
2 cos n tdt
0
2 cos n xdx .
0
6
Integration by Substitutions for definite integrals
1 Example: Find lim 2 x 0 x sin xt 0 t dt du u . t 0 u 0; t x u x 2 . Solution Let xt u, then t , dt x x Then x 2 sin u du x sin xt x 2 sin u 0 t dt 0 u x 0 u du x So 0 2 x sin u 0 sin x 2 0 u du x 2 2 x 1 x sin xt lim 2 dt lim 1. lim 2 0 x 0 x x 0 x 0 t x 2x
b
Since
a
f ( x )dx F (b) F (a )
d F ( t ) f ( t ) ( t ), dt
we have
f ( t ) ( t )dt F ( t )
F ( ) F ( )
F (b) F (a )
, v x,
1 2
0
arcsin xdx x arcsin x 0
1 2
1 2
xdx 1 x2
0
1 1 1 1 2 d (1 x 2 ) 2 6 2 0 1 x2
a a
f ( x )dx 0
a a
(2) If f ( x )=f ( x )， then
f ( x )dx 2 f ( x )dx
0
a
偶倍奇零
(3) If f ( x T ) f ( x ) , then
(4)
(5) (6)
a T
a
f ( x )dx f ( x )dx
例 若 f ( x ) 在[0,1]上连续，证明
0
2
f (sin x )dx f (cos x )dx .
0
2
Proof （1）设 x t 2 x 0 t , 2
dx dt,
0
2
x t 0, 2 0 f (sin x )dx f sin t dt 2 2
School of Science, BUPT
Integration by substitution and by parts in definite integrals
2
Integration by Substitutions for definite integrals
When we want to evaluate the value of a definite integral can find the corresponding indefinite integral
0
0
xf (sin x )dx f (sin x )dx. 2 0
0
Example: Find
Solution
0
x sin x dx 2 1 cos x
0
x sin x sin x dx dx 2 2 2 0 1 cos x 1 cos x
, I . Then
b
a
f ( x )dx f ( t ) ( t )dt .
3
Integration by Substitutions for definite integrals
Proof Let F be an antiderivative of f on the interval I, then
a
2 0
a cos t dt 2 2 a sin t a (1 sin t )
cos t sin t 1 dt sin t cos t 1 1 2 . ln sin t cos t 0 4 2 2 2
2 0
cos t 1 2 dt sin t cos t 2 0
2arcsin( ln x )
e
. 6
1 Example: Find dx . (a 0) 2 2 0 x a x Solution: Let x a sin t , dx a cos tdt , x a t , x 0 t 0, 2
2
when x 1 ; thus by the formula, we have
1
0
1 x 2 dx 2 cos 2 tdt
0
Βιβλιοθήκη Baidu
1 1 2 t sin 2t 2 2 0
.
4
5
Integration by Substitutions for definite integrals
Therefore, the formula holds.
4
Integration by Substitutions for definite integrals
Example: Evaluate
1
0
1 x 2 dx .
Solution Let x sin t ,
dx cos tdt . Then t 0 when x 0; t
x
7
Example: Find
3 e4
e
3 e4
dx . x ln x(1 ln x )
Solution
e
d (ln x ) ln x(1 ln x ) d (ln x ) 2 ln x (1 ln x )
3 e4
3 e4
3 e4
e
e
d ln x 1 ( ln x )2
d dv du ( uv ) u v dx dx dx
Integrating both sides with respect to x and rearranging leads to the integral equation
dv d du u dx ( uv ) dx v dx dx dx dx du uv v dx . dx
0
x t 0,
0 xf (sin x )dx ( t ) f [sin( t )]dt ( t ) f (sin t )dt f (sin t )dt tf (sin t )dt 0 0
0
f (sin x )dx xf (sin x )dx,
17
Integration by Parts
Integration by parts for definite integrals
When u and v are differentiable functions of x on the interval [a , b]. Then
Example: Evaluate
2 2
f (cos t )dt f (cos x )dx;
0 0
例 若 f ( x ) 在[0,1]上连续，证明
0
xf (sin x )dx f (sin x )dx . 2 0
x t dx dt ,
Proof: Let
x 0 t ,
1 d (cos x ) arctan(cos x )0 2 2 0 1 cos x 2 2 ( ) . 4 2 4 4
xf (sin x )dx f (sin x )dx . 2 0
0
(1) If f ( x )=f ( x )， then
b
a
f ( x )dx , if we
f ( x )dx , then we can use
the Newton-Leibniz formula to obtain the value immediately.
Theorem (Integration by substitution for definite integrals) Suppose that the function x ( t ) is continuously differentiable on the interval [ , ] , ( ) a , ( ) b , and that f is continuous in the range of
偶倍奇零
(1) If (2) If Proof
a 0
then
a
a
f ( x )dx 2 f ( x )d x
0
a
then
a
a
a
f ( x )d x 0
a
a f ( x) dx a f ( x) dx 0 f ( x) dx
f (t ) d t f ( x) dx [ f ( x ) f ( x ) ] dx