# How is isobaric interval calculator

## Simplexy

Integral calculator

The integral calculator from Simplexy can integrate and perform any functions for you. Simply calculate the antiderivative and the areas under a graph.

Integral calculus

In addition to differential calculus, integral calculus is one of the main topics in analysis. Integration is the reverse of the derivative and is mostly used in schools to calculate areas under a graph. The word integral is a generic term for the "definite integral" and "indefinite integral". In the following you will learn the basics of integral calculus and the important integral rules, and you will also see how to calculate the area under a graph.

Basics of integral calculus

As already mentioned, integrating a function is the reverse of the derivative, which is why one also says "derive" to integrate.

\ (\ underbrace {F (x)} _ {\ text {antiderivative}} \ overbrace {\ leftarrow} ^ {\ text {integrate}} f (x) \ overbrace {\ rightarrow} ^ {\ text {derive}} \ underbrace {f '(x)} _ {\ text {1st derivation}} \)

If you integrate (expand) a function \ (f (x) \), you get the so-called antiderivative \ (F (x) \). If you derive the antiderivative then you get the output function

\ (F '(x) = f (x) \).

The following table shows some typical power functions and their antiderivatives.

f (x) | F (x) |

\(1\) | \ (x \) |

\(4\) | \ (4 \ cdot x \) |

\ (x \) | \ (\ frac {1} {2} \) \ (x ^ 2 \) |

\ (x ^ 2 \) | \ (\ frac {1} {3} \) \ (x ^ 3 \) |

\ (2 \ cdot x ^ 5 \) | \ (\ frac {2} {6} \) \ (x ^ 6 \) |

\ (x ^ 2-4 \) | \ (\ frac {1} {3} \) \ (x ^ 3-4 \ cdot x \) |

As you may have noticed, one integrates a power function by increasing the exponent by \ (1 \) and writing it in the denominator.

Rule:

Integration of power functions

The antiderivative to the pontence function

\ (f (x) = x ^ n \) \ (\, \, \, \, \, \, \, \, n \ in \ natnums \)

is calculated using:

\ (F (x) = \) \ (\ frac {1} {n + 1} \) \ (x ^ {n + 1} \)

example

Calculate the antiderivative of \ (f (x) = x ^ 2 + 2x-1 \)

We apply the rule for integrating power functions and get:

\ (F (x) = \) \ (\ frac {1} {3} \) \ (x ^ 3 + \) \ (\ frac {2} {2} \) \ (x ^ 2-x \)

\ (\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \) \ (\ frac {1} {3} \) \ (x ^ 3 + x ^ 2-x \)

To check whether we have integrated correctly, we have to derive the antiderivative and should get the original function. It follows that deriving is the reverse of integration.

\ (F '(x) = \ Big (\) \ (\ frac {1} {3} \) \ (x ^ 3 + \) \ (x ^ 2-x \ Big)' \)

\ (\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = \) \ (\ frac {3} {3} \) \ (x ^ 2 + 2x-1 \)

\ (\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = x ^ 2 + 2x-1 \)

As you can see, we have correctly determined the stem function.

Comment:

We could have added any constant to our antiderivative without getting a wrong antiderivative.

\ (F (x) = \) \ (\ frac {1} {3} \) \ (x ^ 3 + \) \ (x ^ 2-x + \ underbrace {10} _ \ text {constant} \)

Because the derivation of \ (F (x) \) with the constant would still lead to our output function. This is because the constant is dropped when deriving.

Because \ (10 ββ'= 0 \)

In general, the constant in the antiderivative is written as follows:

\ (F (x) = \) \ (\ frac {1} {3} \) \ (x ^ 3 + \) \ (x ^ 2-x + C \)

For the sake of simplicity, the constant is often left out or \ (C = 0 \) is used.

Other typical antiderivatives:

f (x) | F (x) |

\ (e ^ x \) | \ (e ^ x \) |

\ (3e ^ {3x} \) | \ (e ^ {3x} \) |

\ (e ^ {2x} \) | \ (\ frac {1} {2} \) \ (e ^ {2x} \) |

\ (e ^ {- 2x} \) | \ (- \ frac {1} {2} \) \ (e ^ {- 2x} \) |

\ (\ frac {1} {x} \) | \ (ln (x) \) |

\ (\ frac {1} {x ^ 2} \) | \ (- \ frac {1} {x} \) |

\ (sin (x) \) | \ (- cos (x) \) |

\ (cos (x) \) | \ (sin (x) \) |

Integrating functions is more time-consuming than deriving them. To derive a function one only has to stupidly apply the derivation rules, that integrating is more complex because there is not always a rule that can be applied. When looking for the antiderivative of a function, it helps to ask the question:

Which function do I have to derive so that my output function comes out?

You should always do a test and derive your antiderivative to see if you get the output function.

Indefinite integral

The indefinite integral is a mathematical notation with which one expresses that one is looking for the antiderivative of an output function \ (f (x) \). One writes:

\ (\ displaystyle \ int f (x) \, \, \, dx = F (x) + C \)

The character \ (\ displaystyle \ int \) is just a mathematical symbol to make it clear that you want to integrate. \ (f (x) \) is the function of which one would like to calculate the antiderivative and the character \ (dx \) indicates which variable one would like to integrate with. The character \ (dx \) tells you which variable the function \ (f \) depends on.

A variable is just a placeholder. A function can, for example, also depend on \ (y \), then one would write:

\ (\ displaystyle \ int f (y) \, \, \, dy = F (y) + C \)

In this case the integration variable is \ (y \), which is why we wrote \ (dy \).

Examples:

- \ (\ displaystyle \ int \, 1 \, \, \, dx = x + C \)
- \ (\ displaystyle \ int \, x \, \, \, dx = \) \ (\ frac {1} {2} \) \ (x ^ 2 + C \)
- \ (\ displaystyle \ int \, 3 \ cdot y ^ 2 \, \, \, dy = \) \ (y ^ 3 + C \)
- \ (\ displaystyle \ int \, e ^ {2x} \, \, \, dx = \) \ (\ frac {1} {2} \) \ (e ^ {2x} + C \)

Definite integral

If integration limits are given, then it is a specific integral. With a certain integral one can calculate the area under the function \ (f (x) \), which is spanned by the integration limits. A certain integral is solved with the main theorem of integral and differential calculus.

Mathematically, a certain integral is written as follows:

\ (\ displaystyle \ int ^ {b} _ {a} f (x) \, \, \, dx = [F (x)] ^ {b} _ {a} = F (b) -F (a) \)

Area calculation with integrals

example 1

What is the area of ββthe graph of the

Function \ (f (x) = - (x-3) ^ 2 + 4 \) and the \ (x \) axis in the interval \ ([1,5] \) is included?

solution

The area under the graph is calculated using:

\ (\ displaystyle \ int ^ {5} _ {1} - (x-3) ^ 2 + 4 \, \, \, dx = \) \ (\ Big [- \) \ (\ frac {1} { 3} \) \ (x ^ 3 + 3x ^ 2-5x \ Big] ^ {5} _ {1} \)

\ (\, \, = \ Big (\) \ (- \ frac {1} {3} \) \ (5 ^ 3 + 3 \ times 5 ^ 2-5 \ times 5 \ Big) - \) \ ( \ Big (\) \ (- \ frac {1} {3} \) \ (1 ^ 3 + 3 \ cdot 1 ^ 2-5 \ cdot 1 \ Big) \)

\ (\, \, = 8.33 + 2.33 = 10.66 \, \, FE \)

The area under the graph in the interval \ ([1,5] \) is

\ (10.66 \) area units in size

You can also calculate such a certain integral with the online integral calculator with limits from Simplexy.

Here you come to the integral calculator.

Example 2

Calculate the area given by the graph of the function

\ (f (x) = - x ^ 2 + 1 \) and the \ (x \) axis is included.

solution

First of all, we have to determine the limits of integration. Since we are supposed to calculate the area between the graph and the \ (x \) axis, the integration limits are precisely the zeros of the function.

With the pq formula we calculate the zeros of

\ (f (x) = - x ^ 2 + 1 \) and get:

\ (x_1 = -1 \) and \ (x_2 = 1 \)

The area under the graph is obtained from the definite integral

\ (\ displaystyle \ int ^ {1} _ {- 1} -x ^ 2 + 1 \, \, \, dx = [- \ frac {1} {3} x ^ 3 + x] ^ {1} _ {-1} = \ frac {2} {3} + \ frac {2} {3} = \ frac {4} {3} \)

The area between the graph and the \ (x \) axis is \ (\ frac {4} {3} \) area units.

Example 3

Find the area between the graph of the function

\ (f (x) = x ^ 3 \) and the \ (x \) axis in the interval \ ([- 1,1] \).

solution

To get the area under the graph we have to solve the following integral:

\ (\ displaystyle \ int ^ {1} _ {- 1} x ^ 3 \, \, \, dx \)

Danger !

If the function graph lies below the \ (x \) axis in a specific interval, the specific integral in this interval delivers a negative number. If the area is asked for, then one takes the value of the integral in this interval, because an area is always positive.

Since the graph of \ (f (x) = x ^ 3 \) lies in the interval \ ([- 1.0] \) below the \ (x \) axis, we have to divide our integral into two integrals.

\ (\ displaystyle \ int ^ {1} _ {- 1} x ^ 3 \, \, \, dx = \ displaystyle \ int ^ {0} _ {- 1} x ^ 3 \, \, \, dx + \ displaystyle \ int ^ {1} _ {0} x ^ 3 \, \, \, dx \)

Since the function lies in the interval \ ([- 1,0] \) below the \ (x \) axis, the integral over this interval will return a negative number. However, an area must have a positive value, so you have to take the amount of the number.

\ (\ displaystyle \ int ^ {0} _ {- 1} x ^ 3 \, \, \, dx = \ big [\ frac {1} {4} x ^ 4 \ big] _ {- 1} ^ { 0} = - \ frac {1} {4} \)

The area between the \ (x \) axis and the function in the interval \ ([- 1.0] \) is

\ (| - \ frac {1} {4} | = \ frac {1} {4} \)

The integral in the interval \ ([0,1] \) yields:

\ (\ displaystyle \ int ^ {1} _ {0} x ^ 3 \, \, \, dx = \ big [\ frac {1} {4} x ^ 4 \ big] _ {0} ^ {1} = \ frac {1} {4} \)

In total we get for the area:

\ (\ displaystyle \ int ^ {1} _ {- 1} x ^ 3 \, \, \, dx = \ frac {1} {4} + \ frac {1} {4} = \ frac {1} { 2} \)

The area between the function and the \ (x \) axis is \ (\ frac {1} {2} \) area units.

Danger !

If one did not divide one's integral and note that the graph runs in the interval \ ([- 1,0] \) below the \ (x \) axis, one would get the following:

\ (\ displaystyle \ int ^ {1} _ {- 1} x ^ 3 \, \, \, dx = 0 \)

The area between the graph of the function \ (f (x) = x ^ 3 \) and the \ (x \) axis in the interval \ ([- 1,1] \) is \ (\ frac {1} { 2} \) area units large.

Area between two functions

Example 4

What is the area between the following two functions.

\ (f (x) = - \) \ (\ frac {x ^ 2} {12} \) \ (+ 5 \)

\ (g (x) = \) \ (\ frac {x ^ 2} {6} \) \ (+ 4 \)

method

- First, the intersection points \ (x_1 \) and \ (x_2 \) of the two functions have to be calculated.
- Then the area between the \ (x \) axis and the function \ (f (x) \) in the interval \ ([x_1, x_2] \) is calculated. Let's call this area \ (F_1 \).
- Then you need the area between the function \ (g (x) \) and the \ (x \) axis in the interval \ ([x_1, x_2] \). We call this area \ (F_2 \).
- To get the area between the two functions, we have to subtract the areas \ (F_1 \) and \ (F_2 \) from each other, i.e. calculate \ (F_1-F_2 \).

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