What is multiple integrals

Integral calculus

Along with differential calculus, integral calculus is the most important branch of the mathematical discipline of analysis. It arose from the problem of area and volume calculation.

The integral is a generic term for the indefinite and the definite integral. The calculation of integrals is called integration.

First we go through the basics of integral calculus again. Afterwards, areas are determined and difficult integration rules such as partial integration are presented.

Table of Contents


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Form an antiderivative

A function F is an antiderivative of a function f if for all $ x \ in \ mathbb {D} $: F '(x) = f (x).

The reverse of deriving is the formation of antiderivatives and is therefore also colloquial Lead up called.

As with deriving, there is also one here Sum rule (= A total is shown "by total") and oneFactor rule (= A constant factor is retained when adding).


Overview of typical antiderivatives in integral calculus

If $ F $ is an antiderivative of $ f $ and $ C $ is any real number (constant), then $ F (x) + C $ is also an antiderivative of $ f $. For example are

\ begin {align *}
F (x) & = \ left (\ frac {x ^ 2} {2} \ right) +5 \
F (x) & = \ left (\ frac {x ^ 2} {2} \ right) +10 \
F (x) & = \ left (\ frac {x ^ 2} {2} \ right) -200
\ end {align *}

all antiderivatives of $ f (x) = x $. Basically the antiderivative for $ f (x) = x $ is thus $ F (x) = \ left (\ frac {x ^ 2} {2} \ right) + C $. If we are only looking for an antiderivative, we can choose $ C = 0 $ for simplicity.

The antiderivative to the power function
\ begin {align *}
f (x) = x ^ n, \ quad n \ in \ mathbb {N}
\ end {align *}
is generally determined by
\ begin {align *}
F (x) = \ frac {1} {n + 1} x ^ {n + 1}.
\ end {align *}

When adding up, the exponent must be increased by 1 and written in the denominator of the fraction!

\ begin {array} {c | c}
f (x) & F (x) \ \
\ hline
1 & x \ \
10 & 10x \ \
x & \ frac {1} {2} x ^ 2 \ \
10x & \ frac {10} {2} x ^ 2 \ \
x ^ 2 & \ frac {1} {3} x ^ 3 \ \
5x ^ 7 & \ frac {5} {8} x ^ 8 \ \
3x ^ 4-2x ^ 3 + 4 & \ frac {3} {5} x ^ 5- \ frac {2} {4} x ^ 4 + 4x
\ end {array}

As already mentioned, there is also a summation rule in integral calculus, which says that each summand is integrated individually. For example, $ F (x) = x ^ 2 + 3x $ is an antiderivative of $ f (x) = 2x + 3 $.

Examples of typical antiderivatives in integral calculus
Form an antiderivative of the given functions:

  1. $ f (x) = 1 $
  2. $ f (x) = 3x ^ 2 + x $
  3. $ f (x) = 3x ^ 5-2x ^ 2 + 1 $
  4. $ f (x) = 3e ^ x $
  5. $ f (x) = 5e ^ {5x + 2} $
  6. $ f (x) = 2e ^ {2x} + 2x $

Solutions to antiderivatives

  1. $ F (x) = x $
  2. $ F (x) = x ^ 3 + \ frac {1} {2} x ^ 2 $
  3. $ F (x) = \ frac {1} {2} x ^ 6- \ frac {2} {3} x ^ 3 + x $
  4. $ F (x) = 3e ^ x $
  5. $ F (x) = e ^ {5x + 2} $
  6. $ F (x) = e ^ {2x} + x ^ 2 $

For more in-depth information, take a look at Daniel's playlist on the topic of antiderivatives and derivation

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Indefinite integral

As already indicated above, the totality of all antiderivatives F (x) + C of a function f (x) is called an indefinite integral. The notation for indefinite integrals is

\ begin {align *}
\ int f (x) \ \ textrm {d} x = F (x) + C
\ end {align *}

$ \ Int $ is the integration sign and $ f (x) $ is the integrand. The variable $ x $ is called the integration variable and $ C $ is the constant of integration. Here two Examples for indefinite integrals:

\ begin {align *}
\ int 2x ~ \ textrm {d} x & = x ^ 2 + C \
\ int x ^ 3 ~ \ textrm {d} x & = \ frac {1} {4} x ^ 4 + C
\ end {align *}

For deepening: tutorial video on the indefinite integral

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Definite integral

If integration limits are given, it is no longer an indefinite integral. One then speaks of a certain integral, since the integration limits are given - consequently determined.
In contrast to the indefinite integral, a definite integral can be combined with the Law of integral calculus to solve!

The result is a concrete numerical value.

\ begin {align *}
\ int_a ^ b f (x) \ \ textrm {d} x = \ left [F (x) \ right] _a ^ b = (F (b) -F (a))
\ end {align *}


example 1
The net inflow speed of a water tank, ie inflow speed minus outflow speed, can be described in the period [0.3] by the function $ f (t) = t ^ 2-2t $, where f (t) is the unit [m $ ^ 3 $ / min] and t is given in minutes.

a) Calculate the filling amount of the water container after one minute if it was filled with 3m $ ^ 3 $ at time t = 0.
b) Calculate the smallest amount of water in the period $ t \ in [1,2] $.

Solution to example 1:
a) The filling quantity is the antiderivative of f, where the initial level is given as c = 3. So the antiderivative of f and the filling quantity are
\ begin {align *}
F (t) = \ frac {1} {3} t ^ 3-t ^ 2 + 3 \ quad \ longrightarrow \ quad F (1) = \ frac {7} {3} \
\ end {align *}
after t = 1 minute $ \ longrightarrow \ frac {7} {3} m ^ 3 $.

b) The smallest amount of water is the lowest point of the antiderivative F. So we investigate the derivative of the antiderivative, the given function $ f_s $,
\ begin {align *}
F '(t) = t ^ 2-2t \ stackrel {!} {=} 0 \ Rightarrow t_1 = 0 \ not \ in [1; 2], \ quad t_2 = 2 \
\ end {align *}

The lowest water level is at t = 2 minutes and the amount of water is approximately $ 1.67 m ^ 3 $.


Example 2

Find the following integrals:
\ begin {align *}
& a) \ int_0 ^ 2 x ^ 2 + 2x-3 \
& b) \ int _ {- 1} ^ 1 x ^ 3 \
& c) \ int_0 ^ 1 e ^ x \
& d) \ int _ {- 1} ^ 2 e ^ {2x} + x \
\ end {align *}

Solution to example 2
\ begin {align *}
& a) \ int_0 ^ 2 x ^ 2 + 2x-3 \ \ textrm {d} x = \ left [\ frac {1} {3} x ^ 3 + x ^ 2-3x \ right] _0 ^ 2 = \ frac {2} {3} \
& b) \ int _ {- 1} ^ 1 x ^ 3 \ \ textrm {d} x = \ left [\ frac {1} {4} x ^ 4 \ right] _ {- 1} ^ 1 = 0 \
& c) \ int_0 ^ 1 e ^ x \ \ textrm {d} x = \ left [e ^ x \ right] _0 ^ 1 = e-1 \
& d) \ int _ {- 1} ^ 2 e ^ {2x} + x \ \ textrm {d} x = \ left [\ frac {1} {2} e ^ {2x} + \ frac {1} {2} x ^ 2 \ right] _ {- 1} ^ 2 \ approx 28.73 \
\ end {align *}

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Integral calculus - determination of areas

The integral calculus can be used to calculate areas. If limit values ​​are given, there is a certain integral. In the following we will show you examples of various problems.

Be sure to take a look at the introduction to determining area!

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Calculation of the area between the graph and the x-axis using the integral calculus

Action:

  • Determine the zeros to get the limits.
  • If the area is always above the x-axis, you can calculate the integral as normal.

Note: If the function has a sign change in the interval to be calculated, part of the area is below the x-axis and an area above the $ x $ axis. The area below the x-axis must then be taken in amount.


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example 1
The function $ f (x) = - x ^ 2 + 7x-10 $ is given (see figure). We want to calculate the area enclosed by the graph and the x-axis. First the zeros are calculated: $ x_1 = 2 $ and $ x_2 = 5 $. At the same time, these are our integration limits.

Solution to example 1
It follows for the area:

\ begin {align *}
\ int \ limits_2 ^ 5 -x ^ 2 + 7x-10 \ \ textrm {d} x & = \ left [- \ frac {x ^ 3} {3} + \ frac {7x ^ 2} {2} -10x \ right] _2 ^ 5 \
& = \ left (- \ frac {5 ^ 3} {3} + \ frac {7 \ cdot 5 ^ 2} {2} -10 \ cdot 5 \ right) - \ left (- \ frac {2 ^ 3} {3} + \ frac {7 \ times 2 ^ 2} {2} - 10 \ times 2 \ right) \
& = 4.5 \ [\ textrm {FE}]
\ end {align *}


Example 2
Find the area enclosed by the graph of the function $ f (x) = - 0.5x ^ 2 + 3x-2.5 $ and the x-axis.

Solution to example 2
First we set the function “equal to 0” and get the zeros $ x_1 = 1 $ and $ x_2 = 5 $ through a skilful calculation. The calculation of the areas then follows:

\ begin {align *}
A = \ int_1 ^ 5-0.5x ^ 2 + 3x-2.5 \ textrm {d} x = \ Bigg [- \ frac {1} {6} x ^ 3 + \ frac {3} {2} x ^ 2-2,5x \ bigg] _1 ^ 5 = \ frac {16} {3} [FE]
\ end {align *}


Example 3
Find the area enclosed by the graph of the function $ f (x) = 0.5x ^ 3-x ^ 2-4x $ and the x-axis.

Solution to example 3
We proceed exactly as in the previous example and first determine the zeros of the function.
Zeroing: $ x_1 = -2, x_2 = 0 $ and $ x_3 = 4 $. It follows for the area:


Example 4
In the following figure the area of ​​a function f (x) is to be determined in the interval [2,4].

Example 4
The expression $ \ int_2 ^ 4 f (x) \ \ textrm {d} x = -6 $ does not indicate the area you are looking for, but the integral value!
For this reason, the calculation of the zeros is important. Since there is a zero at x = 2.5, i.e. within our integration limits, there is a change in sign and part of the graph must lie below the x-axis.

Tip: Calculate the partial area from the lower limit to the zero point $ A_1 $ and from the zero point to the upper limit $ A_2 $. It follows with

\ begin {align *}
A_1 = \ int_2 ^ {2.5} f (x) \ \ textrm {d} x = 1 \ [\ textrm {FE}] \ quad \ textrm {and} \ quad
A_2 = \ int_ {2,5} ^ 4 f (x) \ \ textrm {d} x = | -7 | = 7 \ [\ textrm {FE}]
\ end {align *}

the area you are looking for $ A_ {total} = A_1 + A_2 = 8 $ [FE].


Example 5
Determine for the function $ f (x) = - x ^ 3 + 6x ^ 2-8x $ the area between the graph of the function f (x) and the x-axis in the given interval I = [1,3].

Solution to example 5
First we need to find the zeros of the function. These are $ x_1 $ = 0, $ x_2 $ = 2 and $ x_3 $ = 4.
IMPORTANT! As can be seen on closer inspection, the zero $ x_3 $ is not in the interval I. This results in the following calculation:

\ begin {align *}
& A = \ left \ lvert \ int_1 ^ 2 \ -x ^ 3 + 6x ^ 2-8x \ \ textrm {d} x \ right \ rvert + \ left \ lvert \ int_2 ^ 3 \ -x ^ 3 + 6x ^ 2 -8x \ \ textrm {d} x \ right \ rvert \
& = \ left \ lvert \ Bigg [- \ frac {1} {4} x ^ 4 + 2x ^ 3-4x ^ 2 \ bigg] _1 ^ 2 \ right \ rvert + \ left \ lvert \ Bigg [- \ frac {1} {4} x ^ 4 + 2x ^ 3-4x ^ 2 \ bigg] _2 ^ 3 \ right \ rvert \
& = \ left \ lvert - \ frac {7} {4} \ right \ rvert + \ left \ lvert \ frac {7} {4} \ right \ rvert = \ frac {7} {2} \ textrm {[FE ]}
\ end {align *}


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Calculate the area between two intersecting graphs

If f and g are two functions on the interval [a; b] are continuous and $ f (x) \ ge g (x) $ for all x in [a; b],
then is the area enclosed by both functions

\ begin {align *}
A = \ int_a ^ b (f (x) -g (x)) \ \ textrm {d} x = [F (x) -G (x)] | _a ^ b = (F (b) -G (b )) - (F (a) -G (a)).
\ end {align *}


example 1
Determine the area used by the functions

\ begin {align *}
f (x) = - \ frac {x ^ 2} {12} +5 \ quad \ textrm {and} \ quad
g (x) = \ frac {x ^ 2} {6} +1
\ end {align *}

is included.

Solution to example 1
To do this, we first need the intersections of the two functions.

To do this, we equate both functions and obtain

\ begin {align *}
f (x) & = g (x) \
- \ frac {x ^ 2} {12} +5 & = \ frac {x ^ 2} {6} +1 \
x_1 = -4 \ & \ wedge \ x_2 = 4
\ end {align *}

Now we have all the information to calculate the area between the two graphs using the following integral:

\ begin {align *}
\ int _ {- 4} ^ 4 (f (x) -g (x)) \ \ textrm {d} x = \ int _ {- 4} ^ 4 - \ frac {x ^ 2} {12} +5 - ( \ frac {x ^ 2} {6} +1) \ \ textrm {d} x = \ int _ {- 4} ^ 4 - \ frac {x ^ 2} {4} +4 \ \ textrm {d} x
\ end {align *}

Please note: If two graphs intersect, the upper function becomes the lower from the point of intersection. You would now get a negative surface area, so you have to put absolute dashes.

Action:

  1. Find interfaces
  2. Set up partial integrals and set amount lines.

Then proceed as in the previous example.


Example 2
Find the area between the graphs of f (x) and g (x).

a)
$ f (x) = x ^ 3-6x ^ 2 + 9x \ text {and} g (x) = - 0.5x ^ 2 + 2x $

b)
$ f (x) = - x ^ 2 + 4 \ text {and} g (x) = - x ^ 4 + 4x ^ 2 $

Solution to 2a)

Intersections of f and g: $ x_1 $ = 0, $ x_2 $ = 2 and $ x_3 $ = 3.5. It follows for the area:

\ begin {align *}
A & = \ left | \ int_0 ^ 2 f (x) -g (x) \ \ textrm {d} x \ right | + \ left | \ int_ {2} ^ {3,5} f (x) -g (x) \ \ textrm {d} x \ right | \
& = \ left | \ int_0 ^ 2 x ^ 3-5,5x ^ 2 + 7x \ \ textrm {d} x \ right | + \ left | \ int_ {2} ^ {3,5} x ^ 3-5,5x ^ 2 + 7x \ \ textrm {d} x \ right | \
& = \ left | \ left [\ frac {1} {4} x ^ 4- \ frac {11} {6} x ^ 3 + \ frac {7} {2} x ^ 2 \ right] _0 ^ 2 \ right | + \ left | \ left [\ frac {1} {4} x ^ 4- \ frac {11} {6} x ^ 3 + \ frac {7} {2} x ^ 2 \ right] _ {2} ^ {3.5 } \ right | \
& = \ left | \ frac {10} {3} \ right | + \ left | - \ frac {99} {64} \ right | \ approx 4.88 \ [\ textrm {FE}]
\ end {align *}

Solution to 2b):

Intersections of f and g: $ x_1 = -2 $, $ x_2 = -1 $, $ x_3 = 1 $ and $ x_4 = 2 $. It follows for the area:

\ begin {align *}
A & = \ left | \ int _ {- 2} ^ {- 1} f (x) -g (x) \ \ textrm {d} x \ right | + \ left | \ int _ {- 1} ^ {1} f (x) -g (x) \ \ textrm {d} x \ right | + \ left | \ int_ {1} ^ {2} f (x) -g (x) \ \ textrm {d} x \ right | \
& = \ left | \ int _ {- 2} ^ {- 1} x ^ 4-5x ^ 2 + 4 \ \ textrm {d} x \ right | + \ left | \ int _ {- 1} ^ {1} x ^ 4-5x ^ 2 + 4 \ \ textrm {d} x \ right | + \ left | \ int_ {1} ^ {2} x ^ 4-5x ^ 2 + 4 \ \ textrm {d} x \ right | \
& = \ left | \ left [\ frac {1} {5} x ^ 5- \ frac {5} {3} x ^ 3 + 4x \ right] _ {- 2} ^ {- 1} \ right | + \ left | \ left [\ frac {1} {5} x ^ 5- \ frac {5} {3} x ^ 3 + 4x \ right] _ {- 1} ^ {1} \ right | + \ left | \ left [\ frac {1} {5} x ^ 5- \ frac {5} {3} x ^ 3 + 4x \ right] _ {1} ^ {2} \ right | \
& = \ left | - \ frac {22} {15} \ right | + \ left | \ frac {76} {15} \ right | + \ left | - \ frac {22} {15} \ right | = 8 \ [\ textrm {FE}]
\ end {align *}


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Partial integration

The partial integration, also called product integration, is a way of calculating certain integrals and determining antiderivatives in integral calculus.
It is, so to speak, the counterpart to the product rule when deriving.

\ begin {align *}
& \ int_a ^ bu (x) \ cdot v '(x) \ \ textrm {d} x = \ left [u (x) \ cdot v (x) \ right] _a ^ b - \ int_a ^ b u' ( x) \ cdot v (x) \ \ textrm {d} x
\ end {align *}

Partial integration is always used for a product of two functions, where the antiderivative of one factor is known (v '(x)) and one has the hope that the derivative of the other factor (u (x)) will simplify the integral .

Why is it actually called partial integration? Because part of the integral $ \ left [u (x) \ cdot v (x) \ right] _a ^ b $ is solved and the other part is still an integral $ \ int_a ^ b u '(x) \ cdot v (x ) \ \ textrm {d} x $ includes. The difficulty is to decide which part is u (x) and which part is v '(x).Under certain circumstances it can happen that the integral cannot be solved if the wrong choice is made. The question we have to ask ourselves: Which factor derive simplifies the integral?

General procedure:

  1. Consideration: The derivation of which factor simplifies the integral? Then define u (x) and v '(x).
  2. Determine derivative u '(x).
  3. Determine antiderivative v (x).
  4. Insert results in formula.

example 1
Find the integral of the function $ f (x) = x \ cdot e ^ x $ in the limits [0; 2].

Solution to example 1
First we write down what we should do. The integral should finally be formed.

\ begin {align *}
\ int_0 ^ 2 \ left (x \ cdot e ^ x \ right) \ \ textrm {d} x =?
\ end {align *}

But at this point we get no further with our simple methods of forming the antiderivative. The function f (x) is namely a product of the two functions x and $ e ^ x $. So we apply partial integration to solve the problem. To do this, we go through the above steps from the procedure. 1. We consider: The derivation of which factor simplifies the integral? The derivative of x is 1. The derivative of $ e ^ x $ is $ e ^ x $.
Since $ e ^ x $ can also be easily integrated, it follows:

\ begin {align *}
& u (x) = x \ longrightarrow u '(x) = 1 \ quad \ textrm {and} \ quad v' (x) = e ^ x \ longrightarrow v (x) = e ^ x \
& \ Rightarrow \ int_0 ^ 2 \ left (x \ cdot e ^ x \ right) \ \ textrm {d} x = [x \ cdot e ^ x] _0 ^ 2 - \ int_0 ^ 2 (1 \ cdot e ^ x ) \ \ textrm {d} x = [x \ cdot e ^ x] _0 ^ 2 - [e ^ x] _0 ^ 2 = e ^ 2 +1
\ end {align *}

tip: If the task cannot be solved with the choice of u and v ‘, you should exchange them for each other and try again. Sometimes it helps to partially integrate and rearrange twice. In general, powers $ x ^ n $ or inverse functions like $ \ ln (x) $ or $ \ arcsin (x) $ become simpler by deriving them, and functions like $ e ^ x $ or $ \ sin (x) $ by integrating them do not become more complicated.


Example 2
Find an antiderivative of the function f using partial integration.

\ begin {align *}
& a) f (x) = 2xe ^ x \ \
& b) f (x) = (x-2) e ^ {2x} \ \
& c) f (x) = 5xe ^ {3x + 2} \ \
& d) f (x) = 1 \ cdot \ ln (x) \ \
\ end {align *}

Solutions to example 2
\ begin {align *}
a)
& \ int \ underbrace {2x} _u \ cdot \ underbrace {e ^ x} _ {v ‘} \ \ textrm {d} x \ \
& = 2xe ^ x- \ int 2e ^ x \ \ textrm {d} x \ \
& = 2xe ^ x- 2e ^ x \ \
& = e ^ x (2x-2)
\ end {align *}

\ begin {align *}
b)
& \ int \ underbrace {(x-2)} _ u \ cdot \ underbrace {e ^ {2x}} _ {v ‘} \ \ textrm {d} x \ \
& = \ frac {1} {2} (x-2) e ^ {2x} - \ int \ frac {1} {2} e ^ {2x} \ \ textrm {d} x \ \
& = \ frac {1} {2} (x-2) e ^ {2x} - \ frac {1} {4} e ^ {2x} \ \
& = \ frac {(2x-5) e ^ {2x}} {4} \ \
& = \ left (\ frac {1} {2} x- \ frac {5} {4} \ right) e ^ {2x}
\ end {align *}

\ begin {align *}
c)
& \ int \ underbrace {5x} _u \ cdot \ underbrace {e ^ {3x + 2}} _ {v ‘} \ \ textrm {d} x \ \
& = \ frac {1} {3} 5xe ^ {3x + 2} - \ int \ frac {1} {3} 5e ^ {3x + 2} \ \ textrm {d} x \ \
& = \ frac {5} {3} xe ^ {3x + 2} - \ frac {5} {9} e ^ {3x + 2} \ \ textrm {d} x \ \
& = \ left (\ frac {5} {3} x - \ frac {5} {9} \ right) e ^ {3x + 2}
\ end {align *}

\ begin {align *}
d)
& \ int \ underbrace {1} _ {v ‘} \ cdot \ underbrace {\ ln (x)} _ {u} \ \ textrm {d} x \ \
& = x \ ln (x) - \ int x \ cdot \ frac {1} {x} \ \ textrm {d} x \ \
& = x \ ln (x) - \ int 1 \ \ textrm {d} x \ \
& = x \ ln (x) -x
\ end {align *}

Deepen your knowledge with the learning video on partial integration

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Integration through substitution

Let's come to Integration through substitution. Substitution is generally understood to mean the replacement of one term by another. And that is exactly what we are doing here to carry out an integration.

By introducing a new integration variable, part of the integrand is replaced in order to simplify the integral and thus ultimately lead it back to a known or simpler integral.

The chain rule from differential calculus is the basis of the substitution rule.

\ begin {align *}
\ int_a ^ b \ f (u (x)) \ cdot u '(x) \ \ textrm {d} x = \ int_ {u (a)} ^ {u (b)} f (u) \ \ textrm { you
\ end {align *}

Based on the chain rule, integration by substitution can be said to be used wherever a factor in the integrand is the derivative of another part of the integrand; in principle always where the chain rule would be applied. If the derivative is a constant factor, it can be factored from the integral.

General procedure:

  1. Determine the term to be substituted, derive it and convert it to $ \ textrm {d} x $.
  2. Perform substitution.
  3. Solve integral.
  4. Perform back substitution.

example 1
Find the integral of the function $ f (x) = (x ^ 2-4) ^ 3 \ cdot 2x $ in the interval 4 and 5 and give the set of all antiderivatives.

Solution to example 1
We first write down the integral that is to be determined:

\ begin {align *}
\ int_4 ^ 5 \ underbrace {(x ^ 2-4) ^ 3} _ {f (u (x))} \ cdot \ underbrace {2x} _ {u '(x)} \ textrm {d} x
\ end {align *}

We recognize a concatenation $ (x ^ 2-4) ^ 3 $ and realize that we cannot integrate this part with the methods known so far. Additionally we see that 2x is the derivative of the inner function $ u (x) = x ^ 2-4 $ and that's what we want! So let's replace (substitute) this part with the parameter u:

\ begin {align *}
\ textrm {with} \ u = x ^ 2-4 \ \ textrm {follows}: \ quad \ int_4 ^ 5 u ^ 3 \ cdot 2x \ \ textrm {d} x
\ end {align *}

Since it should be integrated after u, $ \ textrm {d} x $ must be replaced next. We can do this by deriving $ u $ with respect to x, converting it to dx and inserting it into the integral:

\ begin {align *}
u '= \ frac {\ textrm {d} u} {\ textrm {d} x} = 2x \ \ Leftrightarrow \ \ textrm {d} x = \ frac {\ textrm {d} u} {2x} \ \ Rightarrow \ int_4 ^ 5 u ^ 3 \ cdot 2x \ \ frac {\ textrm {d} u} {2x}
\ end {align *}

The 2x is canceled out at this point and the integrand only depends on $ u $. At this point we still have to replace the integral limits with u (4) = 12 and u (5) = 21 and we can determine the integral:

\ begin {align *}
\ int_ {12} ^ {21} u ^ 3 \ \ textrm {d} u = \ left [\ frac {1} {4} u ^ 4 \ right] _ {12} ^ {21} = 43,436.25 \ [\ textrm {FE}]
\ end {align *}

For the antiderivative we have to back-substitute $ u $: $ F (x) = \ frac {1} {4} {\ underbrace {(x ^ 2-4)} _ {= u}} ^ 4 + C $.


Example 2


Example 3
Integrate through substitution.
\ begin {align *}
& a) \ int 2xe ^ {x ^ 2} \ \ text {d} x \
& b) \ int \ frac {4x} {\ sqrt {x ^ 2 + 2}} \ \ text {d} x \
& c) \ int \ frac {1} {x} \ ln (x) \ \ text {d} x \
& d) \ int 3x ^ 2e ^ {3x ^ 3} \ \ text {d} x
\ end {align *}

Solutions to example 3:

a) \ begin {align *} & \ text {With} z = x ^ 2 \ text {and} \ textrm {d} x = \ frac {\ textrm {d} z} {2x}: \
& F (x) = \ int 2xe ^ {x ^ 2} \ \ textrm {d} x = \ int 2xe ^ {x ^ 2} \ \ frac {\ textrm {d} z} {2x} = \ int e ^ z \ \ textrm {d} z = e ^ z = e ^ {x ^ 2} \ \
\ end {align *}

b) \ begin {align *} & \ text {With} z = x ^ 2 + 2 \ text {and} \ textrm {d} x = \ frac {\ textrm {d} z} {2x}: \
& F (x) = \ int \ frac {4x} {\ sqrt {x ^ 2 + 2}} \ \ textrm {d} x = \ int \ frac {4x} {\ sqrt {z}} \ \ frac {\ textrm {d} z} {2x} = \ int \ frac {2} {\ sqrt {z}} \ \ textrm {d} z = 4z ^ {1/2} = 4 \ sqrt {x ^ 2 + 2} \
\ end {align *}

c) \ begin {align *} & \ text {With} z = \ ln (x) \ text {and} \ textrm {d} x = x \ textrm {d} z: \
& F (x) = \ int \ frac {1} {x} \ ln (x) \ \ textrm {d} x = \ int \ frac {1} {x} \ cdot z \ x \ textrm {d} z = \ int z \ \ textrm {d} z = \ frac {1} {2} z ^ 2 = \ frac {1} {2} \ ln ^ 2 (x) \ \
\ end {align *}

d) \ begin {align *} & \ text {With} z = 3x ^ 3 \ text {and} \ textrm {d} x = \ frac {\ textrm {d} z} {9x ^ 2}: \
& F (x) = \ int 3x ^ 2e ^ {3x ^ 3} \ \ textrm {d} x = \ int 3x ^ 2e ^ {z} \ \ frac {\ textrm {d} z} {9x ^ 2} = \ int \ frac {1} {3} e ^ z \ \ textrm {d} z = \ frac {1} {3} e ^ z = \ frac {1} {3} e ^ {3x ^ 3} \ \
\ end {align *}


Special cases of substitution:

  • Linear substitution: \ begin {align *} \ int_a ^ bf (mx + n) \ \ textrm {d} x = \ frac {1} {m} \ left [F (mx + n) \ right] _a ^ b \ end {align *}
  • Logarithmic integration: \ begin {align *} \ int_a ^ b \ frac {g '(x)} {g (x)} \ \ textrm {d} x = \ left [ln | g (x) | \ right] _a ^ b \ end {align *}

Simple explanation in Daniel's tutorial video

Integration through substitution 1, formula, explanation, notation | Math by Daniel Jung

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Interpretation in context

Pupils often have difficulties with the interpretation when the graph shows velocities etc. instead of a quantity. So first look at the axes to see which units are given or read the text.

  • The water flows up to the zero point, because the graph is positive there.
  • from the zero point the water flows off because the graph is in the negative.

Learning video on interpretation in a factual context.

Integrals, integral value, area value in a factual context | Math by Daniel Jung

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Average theorem of integral calculus

Often there is a function that indicates the water level or the speed of the water inflow! If, for example, the average height of the water level in a certain period of time is asked, the is often used Mean theorem the integral calculus:

\ begin {align *}
\ frac {1} {ba} \ int_a ^ bf (x) ~ dx = \ frac {1} {ba} [F (x)] _ a ^ b = \ frac {1} {ba} (F (b) - Fa))
\ end {align *}

The mean theorem is generally the average of all $ y $ values ​​(pay attention to what the function says in context).

Examples 1

$ \ frac {1} {24-0} \ int_0 ^ {24} f (x) ~ dx $ = average height of the water level in 24 hours.

$ \ frac {1} {24-0} \ int_0 ^ {24} f (x) ~ dx $ = average rate of increase of the water in 24 hours.

At this point we have 2 videos on the subject of mean value theorem for you!

Average theorem of integral calculus, average of the Y values ​​| Math by Daiel Jung
Average theorem of integral calculus, average of the Y values ​​| Math by Daniel Jung

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Solid of revolution in integral calculus

A body of revolution is called a body of revolution in geometry, e.g. a sphere, circular cone or cylinder, which is created by rotating a curve around an axis. The curve and the axis of rotation must lie in the same plane.

To calculate the surface and volume of a solid of revolution, you only need the function specification of the curve. A distinction is made between the solid of revolution around the x-axis and the y-axis.

Volume formula with integral for solid of revolution:

\ begin {align *}
\ begin {array} {rl}
\ textrm {around the} x \ textrm {axis:} & V = \ pi \ cdot \ int_a ^ b \ left (f (x) \ right) ^ 2 \ \ text {d} x \ & \
\ textrm {around the} y \ textrm {axis:} & V = \ pi \ cdot \ int_ {f (a)} ^ {f (b)} \ left (f ^ {- 1} (x) \ right ) ^ 2 \ \ text {d} x
\ end {array}
\ end {align *}

As a typical example we would like to introduce you to the function $ f (x) = \ sqrt {x} $ in a predefined interval $ x \ in [0; 10] $. The graph of the function rotates around the x-axis and a volume is created - in a factual context it could represent a champagne glass.

Possible question: How big is the volume of the champagne glass when it is full or half full? Since there is a rotation around the $ x $ axis, we use the corresponding formula from above. The full champagne glass has a volume of

\ begin {align *}
V _ {\ textrm {full}} = \ pi \ cdot \ int_0 ^ {10} \ left (\ sqrt {x} \ right) ^ 2 \ x = \ pi \ cdot \ int_0 ^ {10} x \ x = \ pi \ left [\ frac {x ^ 2} {2} \ right] _0 ^ {10} = 50 \ pi
\ end {align *}

and the half full a volume of

\ begin {align *}
V _ {\ textrm {half}} = \ pi \ cdot \ int_0 ^ {5} \ left (\ sqrt {x} \ right) ^ 2 \ x = \ pi \ cdot \ int_0 ^ {5} x \ x = \ pi \ left [\ frac {x ^ 2} {2} \ right] _0 ^ {5} = 12.5 \ pi.
\ end {align *}


Integral function

The integral of a fixed lower limit a and a variable upper limit x is called the integral function

\ begin {align *}
\ int_a ^ x f (t) ~ dt = F (x) -F (a),
\ end {align *}

where $ F $ is an antiderivative of $ f $.

Integral function learning video

Integral function, only one limit known, math help online, tutorial video | Math by Daniel Jung

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Improper integral

It can happen that a limit is determined (i.e. given, here a) and a limit is infinite $ \ infty $. We then speak of an improper integral.

\ begin {align *}
\ int_a ^ {\ infty} f (x) ~ x
\ end {align *}

To determine the area, one works again with the limit theorem $ \ lim $. At infinity the function converges to a value and we can determine the area.


example 1

\ begin {align *}
\ int_0 ^ {\ infty} e ^ {- x} x & = \ lim_ {b \ to + \ infty} [-e ^ {- x}] _ 0 ^ b \ \
& = \ lim_ {b \ to + \ infty} [-e ^ {- b} - (- e ^ 0)] \ \
& = \ lim_ {b \ to + \ infty} [\ underbrace {-e ^ {- b}} _ {\ rightarrow 0} +1] = 1
\ end {align *}

If b takes on an infinitely high number, then $ e ^ {- b} $ tends towards zero. Always keep the graph of the exponential function in mind. e high what positive goes towards plus infinity and e high what negative goes towards zero.

Improper integral, unknown limit, infinite | Math by Daniel Jung

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Publication of the integral calculus page: March 5th, 2018
Last update: 06.11.2019