2.1 Einführung zur Integralrechnung

Aus Online Mathematik Brückenkurs 2

Version vom 13:55, 17. Sep. 2008

Area below the curve of a function

We previously have found that the slope of a curve of a function is interesting. It gives us information about how the function changes and has great significance in many applications. In a similar way the area between the curve of a function and the x-axis is of importance. It of course is dependent on the curves appearance and thus closely related to the function in question. It is easy to see that this area has practical significance in many different contexts.

If an object is moving, we can illustrate its speed v plotted against time t in a v,t-diagram. We can see in the figure below three different hypothetical examples:

The distance travelled is in each case

 s(6) = 4\cdot 3 + 6\cdot 3 = 30\,\mbox{m},\quad
 s(6) = \frac{6\cdot 6}{2} = 18\,\mbox{m}\,\mbox{.}

In each cases, you see that the distance travelled by the object is matched by the area below the curve.

More examples of what the area below a curve can symbolise are shown below.

The notation for an integral.

In order to describe the area below the curve of a function in symbolic form one introduces the integral sign \displaystyle \,\smallint\, :

 = \int_{a}^{\,c} f(x)\, dx\,\mbox{.}

The primitive function

The function \displaystyle F is a primitive function for \displaystyle f if \displaystyle F'(x) = f(x) in any interval. If \displaystyle F(x) is a primitive function for \displaystyle f(x), it is clear that \displaystyle F(x) + C is as well, for any constant \displaystyle C. In addition, it can be shown that \displaystyle F(x) + C gives all possible primitive functions of \displaystyle f(x).

The relationship between an integral and its primitive function

We have previously found that the area below the curve of a function, i.e. integral of a function, is dependent on the form of the curve. It turns out that this dependence makes use of the primitive function, which also gives rise to the ability to calculate such an area exactly.

Suppose that \displaystyle f is a continuous function in an interval. The value of the integral \displaystyle \ \int_{a}^{b} f(x) \, dx\ then is dependent on the limits of integration \displaystyle a and \displaystyle b, but if one lets \displaystyle a have a fixed value and lets \displaystyle x be the upper limit, the integral will depend only on the upper limit. To clarify this, we prefer to use \displaystyle t as the variable of integration:

We shall now show that \displaystyle A is in fact a primitive function of \displaystyle f.

The total area below the curve from \displaystyle t=a to \displaystyle t=x+h can be written as \displaystyle A(x+h) and is approximately equal to the area up to \displaystyle t=x plus the area of the column between \displaystyle t=x and \displaystyle t=x+h, i.e. .

where \displaystyle c is a number between \displaystyle x and \displaystyle x+h. This expression can be rewriten as

If we let \displaystyle h \rightarrow 0 the the left-hand side tends towards \displaystyle A'(x) and the right-hand side tends towards \displaystyle f(x) , i.e. .

Thus the function \displaystyle A(x) is a primitive function of \displaystyle f(x).

Evaluating integrals

In order to use primitive functions in the calculation of a definite integral, we note first that if \displaystyle F is a primitive function of \displaystyle f then

where the constant \displaystyle C must be chosen so that the right-hand side is zero when \displaystyle b=a, i.e.

which gives that \displaystyle C=-F(a). If we summarise, we have that

 = F(b) - F(a)\,\mbox{.}

We can, of course, just as easily, choose \displaystyle x as the variable of integration and write

 = F(b) - F(a)\,\mbox{.}

Evaluating an integral is performed in two steps. First one determines a primitive function, and then inserts the limits of integration. The usual way of writing this is as follows,

 = \Bigl[\,F(x)\,\Bigr]_{a}^{b} = F(b) - F(a)\,\mbox{.}

Antidifferentiation

To differentiate the common functions is not an insurmountable problem, there are general methods for doing this. To perform the reverse operation, that is, find a primitive function for a given function, however, is much more difficult and in some cases impossible! There is no systematic method that works everywhere, but by exploiting the usual rules of differentiation "in the opposite direction" and also by learning a number of special techniques and tricks one can tackle a large number of the functions that turn up.

The symbol \displaystyle \,\int f(x) \,dx\ is called the indefinite integral of \displaystyle f(x) and is used to denote an arbitrary primitive function for \displaystyle f(x). The usual rules of differentiation give

Compensating for the ”inner derivative”

When differentiating a composite function one makes use of the chain rule, which means that one must multiply by the inner derivative. If the inner function is linear, then the inner derivative is a constant. Thus when integrating such a composite function, one must divide by the inner derivative as a sort of compensation.

Note that this way to compensate for the inner derivative only work if the inner derivative is a constant.

Rules for evaluating integrals

Using the way integration has been defined here, it is easy to show the following properties of integration:

1. \displaystyle \int_{b}^{\,a} f(x) \, dx = - \int_{a}^{\,b} f(x) \, dx\,\mbox{,}\vphantom{\Biggl(}
2. \displaystyle \int_{a}^{\,b} f(x) \, dx + \int_{a}^{\,b} g(x) \, dx = \int_{a}^{\,b} (f(x) + g(x)) \, dx\,\mbox{,}\vphantom{\Biggl(}
3. \displaystyle \int_{a}^{\,b} k \cdot f(x)\, dx = k \int_{a}^{\,b} f(x)\, dx\,\mbox{,}\vphantom{\Biggl(}
4. \displaystyle \int_{a}^{\,b} f(x) \, dx + \int_{b}^{\,c} f(x)\, dx = \int_{a}^{\,c} f(x)\, dx\,\mbox{.}

Moreover, areas below the x-axis are subtracted, that is, if the curve of the function lies below the x-axis in a region, the integral has a negative value in this region:

The total area is \displaystyle \ A_1 + A_2 = \int_{a}^{\,b} f(x)\, dx - \int_{b}^{\,c} f(x)\, dx\,.

Note . The value of an integral can be negative, while an area always has a positive value.

Area between curves

If \displaystyle f(x) \ge g(x) in an interval \displaystyle a\le x\le b then the area between the curves of the function is given by

 - \int_{a}^{b} g(x) \, dx\,\mbox{,}

which can be simplified to

Note that it does not matter whether \displaystyle f(x) < 0 or \displaystyle g(x) < 0 as long as \displaystyle f(x) \ge g(x). The value of the area between the curves is independent of whether the curves are above or below the x-axis, as the following figures illustrate:

 \quad\text{och}\quad A_2 = \int_{b}^{\,c} (2- x) \, dx\,\mbox{.}

 \quad \Leftrightarrow \quad x^2 = \frac{1}{2}
 \quad \Leftrightarrow \quad x = \pm \frac{1}{\sqrt{2}}\,\mbox{.}

 \quad \Leftrightarrow \quad x^3 = 1
 \quad \Leftrightarrow \quad x=1\,\mbox{.}