| Course ID | Course | Professor | Time | Location |
| Introduction to integral calculus | - |
Introduction to integral calculus
Repeat the chapter Basic integral calculus
Definition
For linear functions, such as f (x) = ax + b or f (x) = x / c + dx etc., the area of the integral to be determined can be calculated geometrically below the curve.
For functions that have curvatures, the subintervals, Δx, must be chosen to be infinitely small and the infinitely narrow areas under the curve must be added together.
For the function f (x) = x, Δx = 1, see below. The function is linear. the integral ∫ (5,1) i. of f (x) = x from x = 1 to x = 5
according to integral calculus ∫f (x) = x = x² / 2. [5² / 2 – 1² / 2] = 12 Geometric, if you add the blue and the red squares together and divide the sum by 2:
(uppersum + lowersum) / 2, 14 + 10 = 24, 24/2 = 12
Another example of a linear function:
f (x) = x / 2
we again determine the integral according to the principles of integral calculus (see above)
∫ (x / 2) dx = x² / 4
The integration limits are between 10 and 50.
What is the area?
∫50.10 (x / 2) dx = [50² / 4 – 10² / 4] = 625 – 25 = 600
The integral ∫ (50,10) f (x) dx = ∫50,10 (x / 2) dx, which means that the area between x = 10 and x = 50 is 600.
The same result can be determined graphically. It is recommended to use a millimeter paper, if no other Tools (including special computer programs) are available.
Here Δx is 10:
The result is (Σ uppersums + Σ lowersums) / 2
See picture below:
Non-linear functions
Before we finally derive a definition, we want to determine computationally and graphically an integral of a non-linear function.
f (x) = x²
We calculate the area under the graph of f (x) from x = 1 to x = 4.
First, we integrate ∫f (x) dx = ∫x²dx according to the basics of integral calculus (see above):
∫x²dx = x³ / 3
Calculation of the area from x = 1 to x = 4
∫4.1 (x²) dx = [4³ / 3 – 1³ / 3] = 64/3 -1/3 = 21
The calculation of the area under the graph again can be done by means of upper and lower sum, but the Δx now must be Chosen small as far as possible.
This is the only way to get closer to the result = 21.
See pictures below:
In order to calculate the surface area between x = 1 and x = 4 of f (x) = x² as accurately as possible, Δx must be chosen as small as possible in this way.
The subintervals of Δx must be as close as possible to 0. It is therefore sufficient to work alone with the upper or lower sum. From this the following rule can be formulated:
Limes Δxi -> 0 (i = 1) Σf (xi) * Δxi = ∫b, a f (x) dx
In our example f (x) = x² b = 4 and a = 1 and the area is in the selection of smallest possible Δx between x = 1 and x = 4. 21.
∫4,1 (x²) dx = 21
The sum of the surfaces of arbitrarily small subintervals Δx, which tend to become more numerous, is called the Riemann sum.
Definite and indefinite integrals
Definite integrals
As the subintervals Δx become smaller and smaller – but this does not have to be the same length – the number of subareas n become more and more numerous. (n goes against ∞)
The above-mentioned Riemann sum is approaching a limit. This limit value is called the definite integral of f(x), with the limits a and b. In the above example, f (x) = x², the limits are 1 and 4. The rule can be defined is as follows:
∫b, a f (x) dx = limit Δxi -> 0 (i = 1) Σf (xi) * Δxi
Indefinite integrals
In our example, the so-called root function is f (x) = x².
If we integrate this parent function, this gives ∫x²dx = x³ / 3.
If we derive this integrated function again we get x².
If we derive
x³ / 3 + 2,
x³ / 3 + 4,
x³ / 3 + 1599 or
x³ / 3 + A
from dx,
we also get x².
This means that the basic functions are not explicitly determined. That’s why a C is added to the integrated function:
∫f (x) dx = F (x) + C in our example ∫x²dx = x³ / 3 + C
The indefinite integral is therefore a amount of functions.
Further details won’t be given here, because not having any relevance to the practice now.
More complicated integrals
There are functions that can not be integrated with the integration rules that will be discussed in the near future.
With the function f (x) = x * e high -x² or x * exp (-x²)
It is not easy to integrate this function.
You can create a lookup table and plot the function values graphically. This is done here now.
This function, (x) = x * e high -x², is to be integrated from 0 to infinity,.
the areas under the graph will be calculated.
The value table: see below
The graph
Calculate the area of a rectangle:
0.1* .025 = 0.0025
Now add the little rectangles under the curve: The number of complete rectangles is 185. If you include the incomplete rectangles just below the curve, you get 200 rectangles.
We multiply 200 * 0.0025 and get 0.5 = 1/2.
Comment:
It is crucial to know :The smaller the intervals chosen, the more accurate you get the result of this integral, because the complete rectangles become smaller and more numerous. In the natural sciences, it is not uncommon for measurement results to be evaluated graphically for which there is no mathematical function or, if at all, this function can not be integrated.
Intermezzo: Partial fractions
Before we start to explain especial integration methods, we’d like to show you how to dismantle
1/a*b to 1/a + 1/b and 1/a – 1/b
1/a*b = 1/(a + b)(1/a + 1/b)
and
1/a*b =1/(b – a)/(1/a – 1/b)
proof:
1/(a + b)[(b + a)/a*b] = 1/a*b
1/(b – a)[(b – a)/a*b] = 1/a*b
a practical example being used later as an function to be integrated further below:
1/(b – x)(c + x) = 1/(b + c)*[1/(b – x) + 1/(c + x)]
More samples may occure on the chapter of exercises.
Integration methods
Here are some methods presented, which should simplify the integration. However, it must be anticipated that unlike differentiating a success cannot always be expected.
But once you have found a possible solution, you can check whether one is correct by deriving the integrated function.
The following methods will be treated here:
- substitution
- Partial integration
- Integration with the help of partial fraction decomposition
- Summer integration
- Use of integration schemes and tables
In the latter part we come back again to the nonintegratable integrals of type e high-x² etc.
substitution
Substitution as an integration method can be called the inverse rule of the differential calculus chain rule.
It is therefore recommended to repeat this subchapter of the differential calculus.
Here is the short shown formula of the chain rule:
f (g (x)) = function
f (x) = outer function
g (x) = inner function
f ‘(x) = f’ (g (x)) * g ‘(x) = derivative of the function using the chain rule ==> outer function * inner function
In the case of substitution, an inner function must be determined first for the function to be integrated, f (x). We call this inner function u (x).
The right choice of u (x) is the difficult one to substitute. In a function e.g. f (x) = sin (x²) * 2x dx you cannot select 2x as u (x), it will result in an incorrect result.
The substituted function u (x) is to be derived = u ‘(x).
Now one writes after multiplying the result of u ‘(x) by dx, u’ (x)*dx = du
The integral must finally have the form
∫ f (u) du
and may not contain an x.
Now you integrate
∫f (u) du = F (u) + C
and u is again replaced by x. The result of the integral F (x) + C can be differentiated for control by means of a chain rule and, if u is the correct substitution, one obtains
back the function f (x)
Es gibt zur Substitutionsregel auch eine allgemein Formel, diese lautet:
There is also a general formula for the substitution rule:
∫f(u(x))*u'(x)dx = (∫f(u)du) u = u(x)
However, the application of this general formula has its pitfalls.
It is not always possible to determine f (u) in such a way that on the right side of this formula, ∫f (u), you only have the substituent u but no more x is present.
Important! When you have received a result, check by derivation with chain rule.
Examples:
f(x) = cos(x²)*2x
u(x) = x²
u ‘ (x) = 2x
du = 2xdx
The integral now substituted by u looks as follows:
f(u) =∫cos(u)du = F(u) = sin(u) and with u = x² and du = 2xdx you get the integral (2x may be taken away, do not forget!)
F(u) + C = sin(x²)*2x
Proof: Derivation by chain rule:
Inner function g(x) = x², g'(x) = 2x
Outer function f(g(x)) = 1/2sin(u), f ‘ (g(x)) = cos(u), (u = x²)
Outer function * inner function: cos(x²)*2x
an example more easy;
∫f(x) = dx/(b – x)high 5/2 = 1/(b – x)high5/2*dx= (b-x) high -5/2*dx
u(x) = (b – x)
∫u high -5/2*dx = 2/3(u) high -3/2
(see also basics integral calculus)
Solution: u will be replaced by (b-x)
∫f(x) = 2/3(b – x) high -3/2 + C
Control by derivation applying chain rule:
f'(x) = (-3/2)(2/3)(b-x) high (-3/2 – 1)
u(x) = b – x), u ‘(x) = -1
f ‘(x) =(-1)*(-3/2)*2/3(b – x) high -5/2* = (b – x)high -5/2
other examples:
∫y/√(1 – y²)dy
u = 1 – y²
u’ = -2y
du = -2ydy
Now you can write:
ydy = -1/2du
The substituted integral is called now:
-1/2*∫1/√u*du = -√u + C
again u is replaced by 1 – y²
∫y/√(1 – y²)dy = -√(1 – y²) + C
Here you can also directly apply the left part of the substitution formula mentioned above:
∫f(u(y))*u'(y)dy = ∫y/[√(1 – y²)*-2y]dy = -1/2∫1/√(1- y²) = -√(1 – y²) + C
Attention: u’y = -2y is multiplied below the fraction bar!
the y over the fraction bar is being truncated.
Control by derivation applying chain rule:
-√(1 – y²) + C,
External function = -√u, external derivative = -1/2√u = -1/2√(1 – y²)
Internal function = 1 – y², Internal derivative = -2y
External derivative * internal derivative = -1/2√(1 – y²)*(-2y) = y/√(1 – y²)
2 is being truncated.
Other examples:
Exercise: In the following examples, check yourself whether the integrals are correct. Use derivation by chain rule.
∫(5w + 7)²ººdw
u = 5w + 7, u'(w) = 5 and you can write
dw = 1/5du
∫1/5*u²ºº*du = u²º¹/(201*5) + C = u²º¹/1005 + C = (5w + 7)²º¹/1005 + C
∫e hoch (2v + 3)*dv
u = u(v) = 2v + 3, u'(v) = 2 and there you are 2dv = du and dv = 1/2*du
∫e high u *1/2*du = 1/2*e high u + C = 1/2*e high (2v + 3) + C
For integrals with given integration limits or in other words,
for definite integrals, the general formula for the substitution rule is as follows:
Have a look at another example:
∫(3)(-1)√(2x + 3)dx
The integration limits are b = 3, a = -1
There are now 2 methods how to calculate this definite integral:
- directly with the integration limits b and a (top left picture)
- with the pre-calculated integration limits u (b) and u (a)
1st method:
∫√(2x + 3)dx, u(x) = 2x + 3, dx = 1/2du
As shown above:
1/2∫(3)(-1)√(2x + 3)dx = 1/2∫(3)(-1)√udu
1/2∫√udu = 1/2*2/3√u³ = 1/3√u³and the integral of this function is thus:
1/3√(2x + 3)³
Now, instead of x, we set the boundaries of integration and get:
für x = 3, 1/3√(2*3 + 3)³ = 27/3
für x = -1, 1/3√(2*(-1) + 3)³ = 1/3
Value of the integral is thus (27 – 1)/3 = 26/3
2nd method:
∫[u(b)][u(a)]f(u)du according to the picture above right side
We now return to our example with the substituent u
1/2∫[u(3)][u(-1)]√udu
First, we calculate the integration limits u (3) and u (-1)
u(x) = 2x + 3
u(3) = 9, u(-1) = 1
For the 1 method, we have obtained the following integral for this function:
1/3√u³ By adding u(3) = 9 you get 1/3√9³ = 27/3 and analogously for u(-1) =1 = 1/3√1³ = 1/3
27/3 – 1/3 = 26/3
(Side note: sometimes it can help to write a high1/2 instead of √a)
Partial integration
The partial integration is used for functions whose variables: x, y, z or w etc. occur twice for products or quotients or if other integration methods do not succeed.
f(x)dx = ∫xe×dx
f(x)dx = ∫xsinxdx
f(x)dx = ∫x/(a + x)dx
but also
f(x)dx = ∫lnxdx
f(x)dx = ∫sin²dx
Often, partial integration requires certain tricks.
a/2∫1/[(1 + √(1 + ax)]dx
This integral is to be solved by means of both substitution and partial integration, with two integration methods.
The partial integration is based on a derivation rule: the sum rule:
(f(x)*g(x))’ = f’ (x)*g(x) + f(x)*g’ (x)
However the integral of f(x)*g(x)’ is
∫f(x)*g(x)’ = f(x)*g(x)
and thus
f(x)*g(x) = ∫f’ (x)*g(x)dx + ∫f(x)*g’ (x)dx
To simplify the integration, by integrating only f (x) and deriving g (x), the formula is transformed as follows:
∫f ‘ (x)*g(x)dx = f(x)*g(x) – ∫f(x)*g‘ (x)dx
Instead of f and g, u and v are often used.
∫u ‘ (x)*v(x)dx = u(x)*v(x) – ∫u(x)*v ‘ (x)dx
Application of practical examples:
The difficulty is to assign the appropriate subformula to the function for u and v.
It is also highly recommended to deduce the result in the partial integration, to check whether the solution is right.
An example more easy:
∫f(x)dx =∫2x*e×dx (e× = e high x, see Euler)
Choice of u‘(x) und v(x)
u ‘(x) = e ×, which means that u’ (x) = e × has to be integrated! Which should’nt cause any difficulties
∫e × dx = e ×
u (x) = e × (actually one would have to write e × + C, for simplification C = 0 is set)
v(x) = 2x, which means v(x) = 2x has to be derivated!
v ‘ (x) = 2
Now we put these intermediate results in the right part of the formula.
∫u ‘ (x)*v(x)dx = u(x)*v(x) – ∫u(x)*v ‘ (x)dx
e×*2x – ∫e×*2dx
∫2 * e × dx now is integrated = 2 * e × dx and we get for
∫f (x) = ∫2x * e × dx = e × * 2x – 2e × + C = 2e × (x – 1) + C
Control: derivative with sum rule (fg) ‘= f’ g + fg ‘
f = 2e ×, g = (x – 1)
2e × (x – 1) + 2e × = 2x * e × – 2e × + 2e × = 2x * e ×
please recalculate! and there you are.
The next example needs a trick:
∫f(x)dx = ∫2*lnxdx
Choice of u ‘(x) and v (x)
u ‘(x) = 2 and thus
u (x) = 2x
v (x) = lnx
v ‘(x) = 1 / x
Now according to the general formula: ∫u ’(x) * v (x) dx = u (x) * v (x) – ∫u (x) * v‘ (x) dx
2xlnx – ∫2x * 1 / xdx = 2xlnx – 2x + C = 2x (lnx -1) + C
The control by derivation using the sum rule is simple:
f = 2x, g = lnx – 1
2 (lnx – 1) + 2x * 1 / x – 0 = 2 * lnx – 2 + 2 = 2lnx
The next example either can’t be solved without a trick:
∫f(x) = ∫cos²xdx = 1/2 [(x + sin (x) * cos (x)] + C
Choice of u ‘(x) and v (x)
u ‘(x) = cos (x)
u (x) = sin (x)
v (x) = cos (x)
v ‘(x) = -sin (x)
So far everything in the usual manner.
However, if you now want to use the formula ∫u ‘(x) * v (x) dx = u (x) * v (x) – ∫u (x) * v’ (x) dx,
you get into an infinity loop :
sin (x) * cos (x) + ∫sin² (x) dx
If ∫sin² (x) dx is now integrated, the following expression is obtained:
-sin (x) * cos (x) + ∫cos² (x) dx
So far you are as far as at the beginning:
But you know from trigonometry that cos² (x) + sin² (x) = 1.
And thus: sin² (x) = 1 – cos² (x)
Now we replace the function ∫cos²xdx to be integrated with E:
And we now write down what we have achieved so far:
E = ∫cos² (x) dx = sin (x) * cos (x) + ∫sin² (x) dx or = sin (x) * cos (x) + ∫1 – cos² (x) dx
and write the following relationship:
∫ [1 – cos² (x)] dx = x – E
Now we solve the following equation according to E.
E = sin (x) * cos (x) + x – E
2E = sin (x) * cos (x) + x
E = ∫cos²xdx = 1/2 [(x + sin (x) * cos (x)] + C
Control by derivation (sum rule)
1/2 [(1 – sin² (x) + cos² (x)] = 1/2 [cos² (x) + cos² (x)] = 1/2 [2cos² (x)] = cos² (x)
NB: 1 – sin² (x) = cos² (x)
A more complicated example:
∫f(x) = ∫adx/2[(1 + √(1 + ax)] = a/2∫dx/[(1 + √(1 + ax)]
To integrate this function, both of the methods discussed so far are required: substitution and partial integration.
First we substitute 1 + √ (1 + ax) and derive this denominator using the chain rule:
u(x) = 1 + √(1 + ax)
u’ (x) = a/2√(1 + ax)
Now we proceed exactly as in the 2nd example of the subchapter Substitution:
e.g. with the integral like below:
And apply the formula of the substitution:
see formula just below:
Now the Integral looks as follows:
Attention:
u ‘(x) = a / 2√ (1 + ax) must be placed under the fraction line and be multiplied!
∫f(x) = a/2∫dx/[(1 + √(1 + ax)]*a/2√(1 + ax) = a/2∫2/a*√(1 + ax)/[(1 + √(1 + ax)]dx
after fractional division and simplified after being shortened by a / 2
∫f(x) = ∫√(1 + ax)/[(1 + √(1 + ax)]dx
Now we continue to simplify and replace (1 + √(1 + ax) by w
and receive
∫f(w) = ∫(w – 1)/wdw
This integral is now solved by means of partial integration:
u’ (w) = w – 1
u(w) = w²/2 – w
v(w) = 1/w
v’ (w) = -1/w²
And we proceed analogously as usual: With the formula for partial integration already known:
∫u‘ (x)*v(x)dx = u(x)*v(x) – ∫u(x)*v‘ (x)dx
∫f(w) = ∫(w – 1)/wdw
∫f(w) = 1/w(w²/2 – w) + ∫(1/w²(w²/2 – w)dw = 1/w(w²/2 – w) + ∫(1/2 – 1/w)dw
= w/2 – 1 + w/2 – lnw = w – 1 – lnw
Wie place back w = (1 + √(1 + ax) and receive
∫f(x) = ∫adx/2[(1 + √(1 + ax)] = √(1 + ax) – ln(1 + √(1 + ax) + C
Integration using partial fraction decomposition
The partial fraction breakdown was briefly discussed above. If the functions to be integrated are fractions whose denominator consists of one product, it is often helpful for integration if you break these fractions down into summands.
Let’ show an example quite easy:
f(x) = ∫1/(a – x)*(b – x)dx
Let’s do the partial fraction decomposition first:
The 2 summands can now be integrated using the substitution method, which has just been discussed above.
The result of the integrated function looks as follows:
As a check, it can be reshaped as follows.
1 / (b – a) [ln (b – x) – ln (a – x)]
and with the chain rule you get back the original function.
Next example would be some more difficult,
f(x) = 1/(d – 2x)*(e – 3x)dx
because the remaining variables on the fraction left.
We must now perform this function, in order to get rid of the variables on the fraction left:
This integral can be calculated as per each summation term using the substitution method. The result looks like this:
∫1/(3d – 6x) = 1/6*ln(3d – 6x)
and
∫1/(2e – 6x) = 1/6*ln(2e – 6x)
And the result of this integral by substitution, see below:
Attention: ln (3d – 6x) -ln (2e – 6x) = ln [(3d – 6x) / (2e – 6x)]
It is recommended to repeat chapter about Euler, see logarithms
Numerical integration
The Simpson formula
They have a function and should result in a definite integral from a to b (or the area from x = a to x = b).
Now this function cannot be integrated or you lack the necessary knowledge etc.
There is a simplified method to calculate the specific integral without integrating the function.
The Simpson formula calculates 3 function values (f (a), f (b) and f (a + b) / 2) and the integral can be calculated as follows.
Although the value differs somewhat from the usual integration, the results are surprisingly good.
Examples:
The definite integral of
∫(3)(-1)√(2x + 3)dx
that was calculated above using two methods of substitution results in 26/3 = 8,666
We now apply the Simpson formula:
a = -1, f(a) = y0 = 1
b = 3, f(b) = y2 = 3
(a + b)/2 = f(1) = y1 = √5
(b – a)/6 = 4/6 = 2/3
Now we put these results directly into the Simpson formula:
2/3*(1 + 4*√5 + 3) = 8,6295
You see the result is not exactly but quite good.
In the next example we come back to the eFunction, which was integrated at the very beginning using area calculation.
We tried to determine this integral manually using area calculation. Here again the table of values:
For this function we are now trying to integrate from x = 0 to x = 0 2.2 using the Simpson formula. We have adopted this function in the Simpson formula and in addition created an Excel formula.
Consequently:
b = 2,2,
a = 0
y0 = f(b) = f(2,2) = 2.2*ehigh(-2.2*2.2)
y1 = f(a – b)/2 = 2,2/2*ehigh(-2,2/2*2,2/2)
y2 = 0
The Excel formula is: for b = B1 and a = A1 and C1 = result of the integral:
=(B1*EXP(-B1*B1))*(B1-A1)/6+4*(B1+A1)/2*EXP(-(B1+A1)/2*(B1+A1)/2)*(B1-A1)/6+(A1*EXP(-A1*A1))*(B1-A1)/6
If you put in the field B1 = 2.2 you get 0.487469 …. ∼ 0.5 = 1.2
With the Simpson formula, the integration limits can also be determined as desired within the definition range x = 0 to x = 2.2
for a = 1 and b = 2 one gets 0.1728 …
Use of integration schemes and tables
Below is a collection of definite and indefinite integrals:
F (x) is the integrated function of f (x). In textbooks, F (x) is also called the root function.
Indefinite integrals: rational functions (the constant C is omitted)
Indefinite integrals: square root functions
1) arcsin, arccos, arctan and arccot (also known as “arc functions” or “cyclometric functions”) are inverse functions of sin, cos, tan and cot,
of the trigonometric functions that are known.
Example: sin30 = 1/2.
If you use arcsinx for x = 1/2, you get the angle you are looking for in radians = ¶ / 6 = 180/6 = 30.
Example: tan30 = 0.57735
We use this value as x = 0.57735 in arctanx. arctan 0.57735 = 0.52359, this is the desired angle in radians!
The conversion to degrees is 180 * 0.52359 / ¶ ≅ 30 (29.987) and we get back the angle of 30.
Arc functions can be calculated with Excel.
Now some indefinite integrals of trigonometric functions:
Indefinite integrals: some trigonometric functions
Now we come to some definite integrals: Most of the time, the integration limits go from 0 to ∞, from -∞ to ∞ or from 0 to 2¶.
Definite integrals: some trigonometric functions
For the first two functions (see picture above) the definite integral = 0, if m and n differ,
it must be taken into account that m and n must be whole numbers, Z, the same also applies to the third function: ∫sin mx * cos nx dx (= 0), m, n ∈ Z
Definite integrals: Some Euler functions of the Gauss type
In the last function above, only natural numbers should be used for n: n ∈ N
Bibliography
For middle and vocational students: formulas and tables, mathematics – physics, Orell Füssli Verlag Zurich, ISBN 3 280 01496 4 or the yellow book
For students at technical universities: M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions