df(x) = f'(x).dx
df(u) = f'(u).u'.dx = f'(u).du
example:
d sin(sqrt(x)) = cos(sqrt(x)).d sqrt(x)
df(x) = f'(x).dx and df(u) = f'(u).u'.dx = f'(u).du |
So, if F(x) is primitive function of f(x), then F(x) + an arbitrary constant is also a primitive function of f(x).
All these primitive functions of f(x) are written as F(x) + C
|
F(x) is a root function or primitive function of f(x)
if and only if F'(x) = f(x). |
/
| f(x)dx = F(x) + C
/
The integral sign is clumsy due to the html file, but that is not essential.
/
| sin(x)dx = -cos(x) + C because d(-cos(x)) = sin(x)dx
/
/
| f(x)dx = F(x) + C <=> d F(x) = f(x)dx
/
|
/ r ur+1
| u du = ------- + C (for all real r except r = -1)
/ r+1
|
Proof:
r+1
u 1
d ------ = ----. (r+1).ur .u'.dx = ur .du
r+1 r+1
In the same way you can prove the following formulas.
/ 1 ___
| -------du = 2.V u + C
| ___
/ V u
/
| eu du = eu + C
/
/ u au
| a du = ----- + C
/ ln(a)
/
| sin(u)du = -cos(u) + C
/
/
| cos(u)du = sin(u) + C
/
/ 1
| ---------du = -cot(u) + C
| sin2 u
/
/ 1
| ---------du = tan(u) + C
| cos2 u
/
/ 1
| ---------du = arctan(u) + C
| 1+u2
/
/ 1
| --------------du = arcsin(u) + C
| ________
| | 2
/ \| 1 - u
/ 1
| ---du = ln|u| + C
/ u
/ /
| a.u.dx = a.| u.dx (with a constant)
/ /
/ / /
| (u+v)dx = | udx + | vdx
/ / /
|
/
| x6 x = x7 /7 + C
/
/ dx
| ------ = x-2/(-2) + C
/ x3
/
| (x+3)dx = x2 /2 + 3x + C
/
/
| 3x dx = 3x /ln(3) + C
/
/ 7 /
| (x+2) dx = | (x+2)7 d(x+2) = (x+2)8 /8 + C (since d(x+2)=dx)
/ /
in the same way
/ x+2 /
| 3 dx= | 3x+2 d(x+2) = 3x+2/ln(3) + C
/ /
/
| 5.cos(x)dx = 5 sin(x) + C
/
Since d(5x) = 5.dx
/ 1 / 1
| sin(5x)dx = -. | sin(5x)d(5x) = -.(-cos(5x)) + C
/ 5 / 5
Since d(5x+3) = 5.dx
/ 7 1 / 7 1 8
| (5x+3) dx = -. | (5x+3) d(5x+3) = --. (5x+3) +C
/ 5 / 40
|
If there is a part g(x) in the integral such that g'(x).dx is in that integral too, then do the substitution g(x) = u. |
/ 1 1
| ------- dx Let ln(x) = u => ---dx = du
/ x.ln(x) x
/ 1
= | ---- du = ln|u| + C = ln|ln(x)| + C
/ u
/ 2x
| -------- dx Let x2 + 7 = u => 2xdx = du
| x2 + 7
/
/ du
= | ------ = ln|u| + C = ln|x2 + 7| + C
/ u
/
| sin2 (2x).cos(2x)dx Let sin(2x) = u => 2.cos(2x)dx = du
/
/ 2 u3 sin3 (2x)
=1/2 . | u du = ----- + C = ------------- + C
/ 6 6
/ 1
| ---------du (with k > 0 )
| k + u2
/
1 / 1
= --- . | -------------------du
k | 1 + (u/sqrt(k))2
/
1 / sqrt(k)
= --- . | ----------------- d(u/sqrt(k))
k | 1 + (u/sqrt(k))2
/
1 / 1
= ------ | ------------------- d((u/sqrt(k))
___ | 1 + (u/sqrt(k))2
V k /
(think (u/sqrt(k)) as v )
1 u
= ------- .arctan(--------) + C
___ ___
V k V k
for k > 0
/ 1
| ---------du
| k + u2
/
1 u
= ------- .arctan(--------) + C
___ ___
V k V k
|
/ 1 u
| --------------du = arcsin(------- ) + C (with k > 0)
| 2 ___
/ sqrt(k - u ) V k
|
/ 1 2
| ------------- du = ln(|u + sqrt(k + u ) |)
| 2
/ sqrt(k + u )
|
Proof :
2 1 2u
d ln(|u + sqrt(k + u ) |) = -----------------( 1 + ---------------)u'dx
2 2
u + sqrt(k + u ) 2.sqrt(k + u )
1 u + sqrt(k + u2)
= -----------------( -------------------) du
u + sqrt(k + u2) sqrt(k + u2)
1
= ------------- du
________
| 2
\| k + u
Example:
/ dx / dx / d(x + 1 )
| ------------- = | --------------- = | -------------
/ x.x + 2x + 3 | 2 | 2
/ (x + 1 ) + 2 / (x + 1 ) + 2
1 x + 1
= ---- * arctan(-----) + C
___ ___
V 2 V 2
/ /
| u dv = u.v - | v du
/ /
|
Proof:
/
d ( u.v - | v du ) = ... = u dv
/
Examples:
/
I = | x2 . sin(2x) dx
/
Let x2 = u and sin(2x) dx = dv
cos(2 x)
Then v = - --------
2
So,
2 cos(2 x) / cos(2 x)
I =-x . -------- - | -------- .2xdx
2 / 2
2 cos(2 x) /
I =-x . -------- + | x.cos(2x)dx
2 /
Again, let x = u and cos(2x) dx = dv
Then v = sin(2x) / 2
2 /
I =-x .cos(2x)/2 + x.sin(2x)/2 -| sin(2x)dx
/
2
I =-x .cos(2x)/2 + x.sin(2x)/2+(1/4).cos(2x) + C
/ x
I = | e .sin(2x) dx
/
x
Let e = u and sin(2x) dx = dv
Then v = -cos(2x) / 2
So,
x /
I = - e .cos(2x) / 2 + | ex .cos(2x).(1/2) dx
/
Again let ex = u and cos(2x) dx = dv
Then v = sin(2x) / 2
So,
x x / x
I = - e .cos(2x)/2 + e .sin(2x)/4 - (1/4) | e .sin(2x). dx
/
I = - ex.cos(2x)/2 + ex.sin(2x)/4 - (1/4) . I
(5/4)I = - ex.cos(2x)/2 + ex.sin(2x)/4 + C
I = (4/5).( - ex.cos(2x)/2 + ex.sin(2x)/4)
I = (2/5).ex. (sin(2x)/2 - cos(2x) ) + C
Integration by parts is useful to calculate the integrals
/ | xn .sin(rx) dx Let xn = u / / | xn .cos(rx) dx Let xn = u / / | xn.erx dx Let xn = u / / | xn .ln(rx) dx Let ln(rx) = u / / | erx.sin(sx) dx Let erx= u / / | erx.cos(sx) dx Let e rxu / / | arcsin(x) dx Let arcsin(x) = u / / | arccos(x) dx Let arccos(x) = u / / | arctan(x) dx Let arctan(x) = u /
N(x) rest(x)
---- = Quotient(x) + --------
D(x) D(x)
Example
3x2 + 4x 7
---------- = (3x + 7) + ---------
x - 1 x - 1
A A
------- ; --------- ;
(x - a) (x - a)n
Ax + B Ax + B
-------------- ; ----------------- with b2 - 4ac < 0
ax2 + bx + c (ax2 + bx + c)n
(x - a)n and (ax2 + bx + c)n
Each factor (x - a)n in the denominator causes the sum of
n elementary fractions
A B C L
------- + ------- + ------- + ... + -------
2 3 n
(x - a) (x - a) (x - a) (x - a)
Each factor (ax2 + bx + c)n in the denominator causes the sum of
n elementary fractions
Ax + B Cx + D Lx + M
--------------- + --------------- + ...+ ---------------
2 2 2 2 n
(ax + bx + c) (ax + bx + c) (ax + bx + c)
2x + 5 A B C
---------------- = ------- + ------- + -------
(x-3)(x+2)(x-2) (x - 3) (x + 2) (x - 2)
(2x + 5) = A(x + 2)(x - 2) + B(x - 3)(x - 2) + C(x - 3)(x + 2)
(2x + 5) = (A + B + C)x.x + (-5B - C)x + (-4A +6B -6C)
<=>
/ A + B + C = 0
| -5B - C = 2
\ -4A +6B -6C=5
<=>
A = 2.2 B = 0.05 C = -2.25
7x A B C
-------- = ------- + ------- + -------
3 2 3
(x - 2) (x - 2) (x - 2) (x - 2)
with the same method as in previous example we have
A = 0 ; B = 7; C = 14
2 2
5x + 4x + 1 5x + 4x + 1
----------------- = -----------------
3 2 2 2 3
(x + x )(x + 1) (x )(x + 1)
A B C D E
= --- + ---- + ------- + ------- + -------
2 2 3
x x (x + 1) (x + 1) (x + 1)
with the same method as in previous example we have
A = 1 ; B = 1; C = -1 ; D = -2 ; E = 2
(2x + 1) Ax + B Cx + D
---------------------- = --------------- + -----------------
2 2 2 2
( x + 1)( x + x + 1) ( x + 1) ( x + x + 1)
...
A = -1 ; B = 2 ; C = 1; D = -1
/ A
| -------dx = A.ln|x-a| + C
/ (x - a)
- n+1
/ A (x - a)
| ------- dx = A . -------------- + C
| n - n+1
/ (x - a)
/ Ax + B
| -------------- dx
| 2
/ ax + bx + c
is shown by means of an example. The method is general.
/ 2x + 3
| -------------- dx =
| 2
/ 3x + x + 5
write the derivative of the denominator in the numerator and adjust.
/ (1/3)(6x + 1) + 8/3
| --------------------- dx
| 2
/ 3x + x + 5
split the integral in two parts
/ (1/3)(6x + 1) / 8/3
= | ---------------- dx + | ---------------- dx
| 2 | 2
/ 3x + x + 5 / 3x + x + 5
Let I1 = the first integral and I2 the second one.
2
For I1 we write u = 3x + x + 5 => du =(6x + 1)dx then
1 / du 2
I1 = --- | ---- = (1/3) ln|3x + x + 5 | + C
3 / u
For I2 we have
/ 1
I2 = (8/3)(1/3) | --------------------- dx
| 2
/ x + (1/3)x + (5/3)
/ 1
I2 = (8/9) . | --------------------- dx
| 2
/ (x + (1/6)) + (59/36)
x + (1/6)
I2 = (8/9) .sqrt(36/59). arctan(------------)
sqrt(59/36)
/ Ax + B
| ----------------- dx
| 2 n
/ (ax + bx + c)
is beyond the scope of this tutorial.
|
Each rational fraction can be written as the sum of a polynomial and a
sum of elementary fractions. Each polynomial and each elementary fraction can be integrated. Therefore, each rational fraction can be integrated in the way as shown above. |
/ / / | cos(rx).cos(sx) dx ; | sin(rx).sin(sx) dx ; | sin(rx).cos(sx) dx / / / |
cos((r + s)x) + cos((r - s)x) = 2cos(rx)cos(sx)
cos((r + s)x) - cos((r - s)x = - 2sin(rx)sin(sx)
sin((r + s)x) + cos((r - s)x) = 2sin(rx)cos(sx)
Example :
/ /
| sin(x).sin(7x) dx = (-1/2). | (cos(8x) - cos(-6x))dx
/ /
/ /
= (-1/2). | (cos(8x) dx + (1/2).| (cos(6x) dx
/ /
= (-1/16).sin(8x) +(1/12).sin(6x) + C
/ | sinm (u).cosn (u) dx / |
/
I = | sin3 (x).cos2 (x) dx
/
Let cos(x) = t => sin(x)dx = -dt
sin2 (x) = 1 - t2
/
So, I= - | (1 - t2 ) t2 dt = (-1/3)t3 + (1/5)t5 + C = ...
/
/
I = | sin2 (x).cos2 (x) dx
/
We know that sin2 (4x) = 4.sin2 (2x)cos2 (2x)
/
I = (1/4) . | sin2 (4x) dx
/
With Carnot we know 1 - cos(8x) = 2.sin2 (4x)
/
I = (1/8) . | (1 - cos(8x))dx
/
I = (1/8) . (x - (1/8)sin(8x)) + C
| To integrate a rational function of sin(u) and cos(u) use the t-formules. |
Let t = tan(u/2) , then
2
1 - t 2t 2t
cos(u) = --------- ; sin(u) = -------- ; tan(u) = -------
2 2 2
1 + t 1 + t 1 - t
and dt
u/2 = arctan(t) => du = 2 ---------
2
1 + t
Examples :
/ dx
I = | ------------- Let t = tan(x) , then
/ (2 + cos(2x))
/ 1 dt
I = | ------------ . --------
| 2 2
/ 1 - t 1 + t
2 + --------
2
1 + t
...
/ 1
I = | --------- dt
/ 3 + t2
t
I = sqrt(1/3) . arctan(--------) + C
sqrt(3)
tan(x)
I = sqrt(1/3) . arctan(--------) + C
sqrt(3)
/ dx
I = | -------- Let t = tan(x/2) , then
/ cos(x)
/ 1 2 dt
I = | ------------ . --------
| 2 2
/ 1 - t 1 + t
--------
2
1 + t
...
/ - 2 dt
I = | --------
/ t2-1
using partial fractions we have
- 2 1 1
------- = -------- - --------
t2 - 1 t + 1 t - 1
/ 1 / 1
I = | -------dx - | ------dx = ln|t + 1| -ln|t - 1| + C
/ t + 1 / t - 1
|t + 1| |tan(x/2) + tan(pi/4)|
I = ln -------- = ln ------------------------ + C
|1 - t| |1 - tan(x/2).tan(pi/4)|
I = ln|tan(x/2 - pi/4)| + C
| Sometimes the integration of an irrational function is possible with the help of a suitable substitution. |
/ 1
I = | --------------- dx
| _______
| | 2
/ x \| x - 1
Let sqrt(x2 -1 ) = t => (x2 -1 ) = t2
xdx = tdt and dx/x = (tdt)/(t2 +1)
/ t dt / dt
I = |------------ = |------------ arctan(t) + C
| 2 | 2
/ t.( t + 1) / ( t + 1)
I = Arctan(sqrt(x2 -1 )) + C
/ sqrt(x + 1)
I = | ------------- dx
/ x
2
Let sqrt(x + 1) = t => (x + 1) = t => dx = 2tdt
2 2
/ t dt / ( t - 1 + 1)
I =2|---------- = I =2|-------------- dt
| 2 | 2
/ ( t - 1) / ( t - 1)
/ / dt
I =2| dt + 2 | ---------
/ /(t-1)(t+1)
using partial fractions we find
I = 2t + ln|t - 1| -ln|t + 1| + C with sqrt(x + 1) = t
Now we'll show how to integrate the forms
/ _____________
| | 2
| \| x + b x + c dx
/
and
/ _____________
| | 2
| \| c + b x - x dx
/
|
First consider
/ ________
| | 2
I = | \| t + k dt with k real and constant.
/
To integrate this we use integration by parts
________
| 2
Let u = \| t + k and dt = dv then
/
________ |
| 2 | t
I = \| t + k .t - |t. ---------------dt
| ________
| | 2
/ \| t + k
/
________ | 2
| 2 | t + k - k
I = \| t + k .t - |---------------dt
| ________
| | 2
/ \| t + k
/ /
________ | 2 |
| 2 | t + k | 1
I = \| t + k .t - |-------------- dt + k |---------------- dt
| ________ | ________
| | 2 | | 2
/ \| t + k / \| t + k
/ /
________ | ________ |
| 2 | | 2 | 1
I = \| t + k .t - | \| t + k dt + k |---------------- dt
| | ________
| | | 2
/ / \| t + k
/
________ |
| 2 | 1
2.I = \| t + k .t + k |---------------- dt
| ________
| | 2
/ \| t + k
________ ________
| 2 | 2
2.I = \| t + k .t + k ln |t + \| t + k | + C
________ ________
| 2 | 2
\| t + k .t + k ln |t + \| t + k | + C
I = --------------------------------------------
2
With this formula we can integrate
/ _____________
| | 2
| \| x + b x + c dx
/
Example :
/ _____________ / _____________
| | 2 | | 2
I = |\| x + 4 x + 6 dx = | \| (x + 2) + 2 dx
| |
/ /
Now let x+2 = t , then we have
/ ________
| | 2
I = | \| t + 2 dt
/
________ ________
| 2 | 2
\| t + 2 .t + 2 ln |t + \| t + 2 | + C
I = --------------------------------------------
2
In the same way we can integrate
/ ________
| | 2
I = | \| k - t dt with k real and constant.
/
and so all integrals of the form
/ _____________
| | 2
| \| c + b x - x dx
/
|
If the integrand contains sqrt(a2 - u2 ) or sqrt(a2 + u2 ) or sqrt(u2 - a2 ) you can use a trigonometric substitution.
For sqrt(a2 - u2 ) use the substitution u = a.sin(t)
with t in [-pi/2,pi/2]
For sqrt(a2 + u2 ) use the substitution u = a.tan(t)
with t in [-pi/2,pi/2]
For sqrt(u2 - a2 ) use the substitution u = a.sec(t)
with t in [0, pi/2[ if u > 0 and
with t in [ pi, 3.pi/2] if u < 0
|
/
I = | sqrt(9 - x2 ) dx
/
x = 3.sin(t) with t in [-pi/2,pi/2]
=> 9 - x2 = ... = 9.cos2 (t) and dx = 3.cos(t) dt
/
I = |3.cos(t). 3.cos(t) dt
/
/
I =9 |cos2 (t) dt
/
and with Carnot formula
/
I = (9/2). |(1 + cos (2t)) dt
/
I = (9/2)t + (9/4)sin(2t) + C with t = arcsin (x/3)
/ dt
I = | ---------------
| sqrt( 9 + t2 )
/
x = 3.tan(t) with t in [-pi/2,pi/2]
=> 9 + x2 = ... = 9/cos2(t) and dx = 3/cos2(t) .dt
/ 3.cos(t)
I = | -------------
| 2
/ 3. cos (t)
/ 1
I = | --------- dt
/ cos(t)
We have calculated this integral as an example of a rational
function of sin(t) and cos(t).
Appealing on that result, we have
I = ln|tan(t/2 - pi/4)| + C with t = arctan(x/3)
/ dx
I = | --------------- with x > 0
| sqrt( x2 - 9)
/
Let t = 3/cos(t) with t in [0, pi/2[
3.sin(t)dt
=> dx = ----------- and x2 - 9 = ... = 9 tan2 (t)
cos2 (t)
Then
/ 3.sin(t)
I = | ----------------dt
| 3.tan(t).cos2t)
/
/ 1
I = | --------- dt
/ cos(t)
As in previous example we have
I = ln|tan(t/2 - pi/4)| + C with t = arccos(3/x)
s = sum mi.(xi - xi-1)
i
and the 'upper sum'
S = sum Mi.(xi - xi-1)
i
Now we choose in each interval [xi-1, xi] in new value. In that way,
we create 2n intervals in [a,b]. With these intervals corresponds a new
lower sum and a new upper sum.
/b
| f(x).dx = I
/a
n lower sum upper sum
10 0.317 0.385
20 0.325 0.358
40 0.329 0.346
80 0.331 0.340
160 0.332 0.336
...
2560 0.33327 0.3335
Both sums converge to the same limit I = 1/3.
/1
| x2dx = 1/3
/0
/b /a
| f(x).dx = - | f(x).dx
/a /b
|
/b /c /b
| f(x).dx = | f(x).dx + | f(x).dx
/a /a /c
|
/b /a /b
| f(x).dx = | f(x).dx + | f(x).dx
/c /c /a
/b /a /b
<=> | f(x).dx - | f(x).dx = | f(x).dx
/c /c /a
/b /c /b
<=> | f(x).dx + | f(x).dx = | f(x).dx
/c /a /a
/c /b /b
<=> | f(x).dx + | f(x).dx = | f(x).dx
/a /c /a
Q.E.D.
/a
| f(x).dx = 0
/a
|
xi-1 =< si =< xi
=> mi =< f(si) =< Mi
=> mi.(xi - xi-1) =< f(si).(xi - xi-1) =< Mi.(xi - xi-1)
We make the sum of these expressions for i from 1 to n.
sum mi.(xi - xi-1) =< sum f(si).(xi - xi-1) =< sum Mi.(xi - xi-1)The sum f(si).(xi - xi-1) is called a Rieman sum.
We take the limit for n --> infinity
I =< lim sum f(si).(xi - xi-1) =< I
Therefore
/b
I = | f(x).dx = lim sum f(si).(xi - xi-1)
/a
Each definite integral is the limit of an appropriate Rieman sum and each
limit of an appropriate Rieman sum is a definite integral.
/b
| f(x).dx = (b-a).f(c)
/a
Proof :
sum m.(xi - xi-1) =< sum mi.(xi - xi-1)
<=> m.sum (xi - xi-1) =< sum mi.(xi - xi-1)
and
sum Mi.(xi - xi-1) >= sum M.(xi - xi-1)
sum Mi.(xi - xi-1) >= M.sum (xi - xi-1)
Taking the limit, we have
m.(b-a) =< I =< M.(b-a)
Hence I/(b-a) is a number between the smallest
and the biggest image in [a,b].
/b
| f(x).dx = (b-a).f(c)
/a
1 /b
The mean value of f(x) in [a,b] = ------- | f(x).dx
(b - a) /a
|
/b
| f(x).dx = (b-a).f(c) = number independent of x
/a
So, the name x is not important. Hence,
/b /b /b
| f(x).dx = | f(u).du = | f(t).dt = ...
/a /a /a
/x
| f(t).dt
/a
is a number depending on the upper limit x. It is a function of x.
We call that function g(x). Hence
/x
g(x) = | f(t).dt
/a
The derivative of the function
/x
| f(t).dt
/a
is f(x) .
|
Let
/x
g(x) = | f(t).dt
/a
We calculate the derivative of g(x) appealing on the definition of
derivative.
d (g(x+h)-g(x))
-- g(x) = lim ---------------
dx h->0 h
/x+h /x
| f(t).dt - | f(t).dt
/a /a
= lim -----------------------------
h->0 h
/x+h /a
| f(t).dt + | f(t).dt
/a /x
= lim -----------------------------
h->0 h
/x+h
| f(t).dt
/x
= lim -------------
h->0 h
The theorem of the mean says that there is at least one c
in [x,x+h] such that
h.f(c)
= lim -------------
h->0 h
= lim f(c)
h->0
If h -> 0 , c -> x
= lim f(c)
c->x
= f(x)
d /x /x --- | f(t).dt = f(x) , | f(t).dt is a primitive function of f(x) dx /a /a
If F(x) is a primitive function of f(x), then
/b
| f(t).dt = F(b) - F(a)
/a
|
/x
| f(t).dt is a primitive function of f(x) too.
/a
Thus,
/x
| f(t).dt = F(x) + C FOR ALL x in [a,b] (1)
/a
For x = a this gives:
/a
0 = | f(t).dt = F(a) + C FOR ALL x in [a,b]
/a
=> C = - F(a)
/x
(1) => | f(t).dt = F(x) - F(a) FOR ALL x in [a,b]
/a
For x = a this gives:
/b
| f(t).dt = F(b) - F(a)
/a
Since the name t is not important,
/b
| f(x).dx = F(b) - F(a)
/a
The last expression is noted as
/b b
| f(x).dx = [ F(x) ]
/a a
/1
| x(x2 + 7)dx
/0
We calculate a primitive function of x(x2 + 7).
/
| x(x2 + 7)dx = ... = (1/4)(x2 + 7)2 + C
/
Now,
/1 2 2 2 1
| x(x + 7)dx = [ (1/4)(x + 7) ] = 64/4 - 49/4 = 15/4
/0 0
The area enclosed between two continuous curves y = f(x) and y = g(x)
in the interval [a,b] with f(x) >= g(x) in [a,b] is
/b | (f(x) - g(x))dx /a |
Proof:
In interval [a,b], we choose values
x1,x2,x3,...,xn-1. Take a = x0 and b = xn.
We choose in each interval [xi-1,xi] a value = si.
Consider the area of the rectangle in [xi-1,xi]
( f(si) - g(si) ).(xi - xi-1)
Now, take the sum of the area of such rectangles in each subinterval.
sum ( f(si) - g(si) ).(xi - xi-1)
This is a Rieman sum. The limit a this sum , for n --> infinity,
is the definite integral
/b | (f(x) - g(x))dx /aThis is the area enclosed between the continuous curves y = f(x) and y = g(x).
You see that the area is the sum of an infinity number of elementary rectangles. This procedure is used to solve a lot of problems in physics.
If the condition f(x) >= g(x) is not satisfied in [a,b], you can divide the interval in a suitable way such that the condition is satisfied in each subinterval.
Example:
We calculate the area between y = cos(x) and y = sin(x) in [0,pi].
In [0,pi/4], we have cos(x) >= sin(x) and in [pi/4,pi] we have sin(x) >= cos(x).
The area A =
/pi/4 /pi
| (cos(x) - sin(x)) dx + | (sin(x) - cos(x)) dx
/0 /pi/4
|pi/4 |pi
= (sin(x) + cos(x))| + (-cos(x) - sin(x))|
|0 |pi/4
= ... = 2.8284
We'll find a formula to calculate this volume V.
In interval [a,b], we choose values
x1,x2,x3,...,xn-1. Take a = x0 and b = xn.
We choose in each interval [xi-1,xi] a value = si.
Consider the rectangle with base (xi - xi-1) and height si.
If this rectangle revolves around the x axis, the volume of the
cylinder is
pi. si2 . (xi - xi-1)The sum of all such cylinders is a Rieman sum
sum (pi. si2 . (xi - xi-1))
The limit a this sum, for n --> infinity, is the definite integral
/b
pi. | (f(x))2 dx
/a
A continuous function f(x) with positive images in interval [a,b],
revolving around the x-axis, defines a volume
/b
V = pi. | (f(x))2 dx
/a
|
Example:
When the curve of y = sqrt(r2-x2) revolves around the x-axis,
it generates a sphere with radius r. The volume =
/r
V = pi . | (r2-x2) dx = ... = (4/3). pi . r3
/-r
In interval [a,b], we choose values x1,x2,x3,...,xn-1. Take a = x0 and b = xn.
In interval [xi-1,xi], we take the points P(xi-1,f(xi-1)) and Q(xi,f(xi)). We identify the curve in that interval, with the segment PQ.
(length PQ)2 = (xi - xi-1)2 + (f(xi) - f(xi-1))2
(f(xi) - f(xi-1))2
= ( 1 + ------------------------) . (xi - xi-1)2
(xi - xi-1)2
According to Lagrange's theorem
there is an si in [xi-1,xi], such that
f(xi) - f(xi-1)
--------------------- = f'(si)
(xi - xi-1)
Now, we have
(length PQ)2 = (1 + (f'(si))2 ) . (xi - xi-1)2
<=>
________________
|
(length PQ) = \| 1 + (f'(si))2 . (xi - xi-1)
The sum of all the distances from each subinterval is a Rieman sum
and the limit of this sum is a definite integral.
The integral is the sum of an infinity number of elementary parts
of the curve. Each elementary part has length
_____________
|
\| 1 + (f'(x))2 dx
_________
|
= \| 1 + y'2 dx
Therefore we
can write
The length of the curve y = f(x) in an interval [a,b] is
/b _________
| |
L = | \| 1 + y'2 dx
|
/a
|
Example:
Calculute the length of the curve y = (ex + e-x)/2 in interval [-1,1].
y' = (1/2).(ex - e-x)
1 + y'2 = ... = (1/4).(ex + e-x)2
/1
|
L = | (1/2).(ex + e-x) dx = ... = 2.35
|
/-1
You can find solved problems about
integration
here