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 u^{r+1} | u du = ------- + C (for all real r except r = -1) / r+1 |
Proof: r+1 u 1 d ------ = ----. (r+1).u^{r} .u'.dx = u^{r} .du r+1 r+1In a similar way you can prove the following formulas.
/ 1 ___ | -------du = 2.V u + C | ___ / V u / | e^{u} du = e^{u} + C / / u a^{u} | a du = ----- + C / ln(a) / | sin(u)du = -cos(u) + C / / | cos(u)du = sin(u) + C / / 1 | ---------du = -cot(u) + C | sin^{2} u / / 1 | ---------du = tan(u) + C | cos^{2} u / / 1 | ---------du = arctan(u) + C | 1+u^{2} / / 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 / / / |
/ | x^{6} x = x^{7} /7 + C / / dx | ------ = x^{-2}/(-2) + C / x^{3} / | (x+3)dx = x^{2} /2 + 3x + C / / | 3^{x} dx = 3^{x} /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= | 3^{x+2} d(x+2) = 3^{x+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 x^{2} + 7 = u => 2xdx = du | x^{2} + 7 / / du = | ------ = ln|u| + C = ln|x^{2} + 7| + C / u
/ | sin^{2} (2x).cos(2x)dx Let sin(2x) = u => 2.cos(2x)dx = du / / 2 u^{3} sin^{3} (2x) =1/2 . | u du = ----- + C = ------------- + C / 6 6
/ 1 | ---------du (with k > 0 ) | k + u^{2} / 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 + u^{2} / 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 + u^{2}) = -----------------( -------------------) du u + sqrt(k + u^{2}) sqrt(k + u^{2}) 1 = ------------- du ________ | 2 \| k + uExample:
/ 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 = | x^{2} . sin(2x) dx / Let x^{2} = 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 + | e^{x} .cos(2x).(1/2) dx / Again let e^{x} = 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 = - e^{x}.cos(2x)/2 + e^{x}.sin(2x)/4 - (1/4) . I (5/4)I = - e^{x}.cos(2x)/2 + e^{x}.sin(2x)/4 + C I = (4/5).( - e^{x}.cos(2x)/2 + e^{x}.sin(2x)/4) I = (2/5).e^{x}. (sin(2x)/2 - cos(2x) ) + CIntegration by parts is useful to calculate the integrals
/ | x^{n} .sin(rx) dx Let x^{n} = u / / | x^{n} .cos(rx) dx Let x^{n} = u / / | x^{n}.e^{rx} dx Let x^{n} = u / / | x^{n} .ln(rx) dx Let ln(rx) = u / / | e^{rx}.sin(sx) dx Let e^{rx}= u / / | e^{rx}.cos(sx) dx Let e^{rx}= u / / | arcsin(x) dx Let arcsin(x) = u / / | arccos(x) dx Let arccos(x) = u / / | arctan(x) dx Let arctan(x) = u /
Each rational fraction can be written as the sum of a polynomial and partial fractions. This is called the partial fraction decomposition. Partial fraction decomposition is essential to integrate rational fractions.
The procedure is explained here as a separate subject.
/ 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 + cis 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) / x / 1 = A.| ----------------- dx + B.| ----------------- dx | 2 n | 2 n / (ax + bx + c) / (ax + bx + c) and then you can use the recursion formulas andThe proof of these formulas is beyond the frame of this course
Each rational fraction can be written as the sum of a polynomial and a
sum of partial fractions. Each polynomial and each partial 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
/ | sin^{m} (u).cos^{n} (u) dx / |
/ I = | sin^{3} (x).cos^{2} (x) dx / Let cos(x) = t => sin(x)dx = -dt sin^{2} (x) = 1 - t^{2} / So, I= - | (1 - t^{2} ) t^{2} dt = (-1/3)t^{3} + (1/5)t^{5} + C = ... /
/ I = | sin^{2} (x).cos^{2} (x) dx / We know that sin^{2} (4x) = 4.sin^{2} (2x)cos^{2} (2x) / I = (1/4) . | sin^{2} (4x) dx / With Carnot we know 1 - cos(8x) = 2.sin^{2} (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-formulas. |
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 + tExamples :
/ 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 + t^{2} 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 = | -------- / t^{2}-1 using partial fractions we have - 2 1 1 ------- = -------- - -------- t^{2} - 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(x^{2} -1 ) = t => (x^{2} -1 ) = t^{2} xdx = tdt and dx/x = (tdt)/(t^{2} +1) / t dt / dt I = |------------ = |------------ = arctan(t) + C | 2 | 2 / t.( t + 1) / ( t + 1) I = Arctan(sqrt(x^{2} -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 a similar 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(a^{2} - u^{2} ) or sqrt(a^{2} + u^{2} ) or sqrt(u^{2} - a^{2} ) you can use a trigonometric substitution. For sqrt(a^{2} - u^{2} ) use the substitution u = a.sin(t) with t in [-pi/2,pi/2] For sqrt(a^{2} + u^{2} ) use the substitution u = a.tan(t) with t in [-pi/2,pi/2] For sqrt(u^{2} - a^{2} ) 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 - x^{2} ) dx / x = 3.sin(t) with t in [-pi/2,pi/2] => 9 - x^{2} = ... = 9.cos^{2} (t) and dx = 3.cos(t) dt / I = |3.cos(t). 3.cos(t) dt / / I =9 |cos^{2} (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 + t^{2} ) / x = 3.tan(t) with t in [-pi/2,pi/2] => 9 + x^{2} = ... = 9/cos^{2}(t) and dx = 3/cos^{2}(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( x^{2} - 9) / Let t = 3/cos(t) with t in [0, pi/2[ 3.sin(t)dt => dx = ----------- and x^{2} - 9 = ... = 9 tan^{2} (t) cos^{2} (t) Then / 3.sin(t) I = | ----------------dt | 3.tan(t).cos^{2}t) / / 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 m_{i}.(x_{i} - x_{i-1}) iand the 'upper sum'
S = sum M_{i}.(x_{i} - x_{i-1}) iNow we choose in each interval [x_{i-1}, x_{i}] 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 0.3335Both sums converge to the same limit I = 1/3.
/1 | x^{2}dx = 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 /aQ.E.D.
/a | f(x).dx = 0 /a |
x_{i-1} =< s_{i} =< x_{i} => m_{i} =< f(s_{i}) =< M_{i} => m_{i}.(x_{i} - x_{i-1}) =< f(s_{i}).(x_{i} - x_{i-1}) =< M_{i}.(x_{i} - x_{i-1})We make the sum of these expressions for i from 1 to n.
sum m_{i}.(x_{i} - x_{i-1}) =< sum f(s_{i}).(x_{i} - x_{i-1}) =< sum M_{i}.(x_{i} - x_{i-1})The sum f(s_{i}).(x_{i} - x_{i-1}) is called a Riemann sum.
We take the limit for n --> infinity
I =< lim sum f(s_{i}).(x_{i} - x_{i-1}) =< ITherefore
/b I = | f(x).dx = lim sum f(s_{i}).(x_{i} - x_{i-1}) /aEach definite integral is the limit of an appropriate Riemann sum and each limit of an appropriate Riemann sum is a definite integral.
/b | f(x).dx = (b-a).f(c) /aProof :
sum m.(x_{i} - x_{i-1}) =< sum m_{i}.(x_{i} - x_{i-1}) <=> m.sum (x_{i} - x_{i-1}) =< sum m_{i}.(x_{i} - x_{i-1}) and sum M_{i}.(x_{i} - x_{i-1}) >= sum M.(x_{i} - x_{i-1}) sum M_{i}.(x_{i} - x_{i-1}) >= M.sum (x_{i} - x_{i-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 /aSo, 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 /ais 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 /ais f(x) . |
Let /x g(x) = | f(t).dt /aWe 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(x^{2} + 7)dx /0 We calculate a primitive function of x(x^{2} + 7). / | x(x^{2} + 7)dx = ... = (1/4)(x^{2} + 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
x_{1},x_{2},x_{3},...,x_{n-1}. Take a = x_{0} and b = x_{n}.
We choose in each interval [x_{i-1},x_{i}] a value = s_{i}.
Consider the area of the rectangle in [x_{i-1},x_{i}]
( f(s_{i}) - g(s_{i}) ).(x_{i} - x_{i-1})Now, take the sum of the area of such rectangles in each subinterval.
sum ( f(s_{i}) - g(s_{i}) ).(x_{i} - x_{i-1})This is a Riemann 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
x_{1},x_{2},x_{3},...,x_{n-1}. Take a = x_{0} and b = x_{n}.
We choose in each interval [x_{i-1},x_{i}] a value = s_{i}.
Consider the rectangle with base (x_{i} - x_{i-1}) and height s_{i}.
If this rectangle revolves around the x axis, the volume of the
cylinder is
pi. s_{i}^{2} . (x_{i} - x_{i-1})The sum of all such cylinders is a Riemann sum
sum (pi. s_{i}^{2} . (x_{i} - x_{i-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(r^{2}-x^{2}) revolves around the x-axis,
it generates a sphere with radius r. The volume =
/r V = pi . | (r^{2}-x^{2}) dx = ... = (4/3). pi . r^{3} /-r
In interval [a,b], we choose values x_{1},x_{2},x_{3},...,x_{n-1}. Take a = x_{0} and b = x_{n}.
In interval [x_{i-1},x_{i}], we take the points P(x_{i-1},f(x_{i-1})) and Q(x_{i},f(x_{i})). We identify the curve in that interval, with the segment PQ.
(length PQ)^{2} = (x_{i} - x_{i-1})^{2} + (f(x_{i}) - f(x_{i-1}))^{2} (f(x_{i}) - f(x_{i-1}))^{2} = ( 1 + ------------------------) . (x_{i} - x_{i-1})^{2} (x_{i} - x_{i-1})^{2}According to Lagrange's theorem there is an s_{i} in [x_{i-1},x_{i}], such that
f(x_{i}) - f(x_{i-1}) --------------------- = f'(s_{i}) (x_{i} - x_{i-1})Now, we have
(length PQ)^{2} = (1 + (f'(s_{i}))^{2} ) . (x_{i} - x_{i-1})^{2} <=> ________________ | (length PQ) = \| 1 + (f'(s_{i}))^{2} . (x_{i} - x_{i-1})The sum of all the distances from each subinterval is a Riemann 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} dxTherefore 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:
Calculate the length of the curve y = (e^{x} + e^{-x})/2 in interval [-1,1].
y' = (1/2).(e^{x} - e^{-x}) 1 + y'^{2} = ... = (1/4).(e^{x} + e^{-x})^{2} /1 | L = | (1/2).(e^{x} + e^{-x}) dx = ... = 2.35 | /-1