The properties of integrals

Let be \(f, g\) two functions of class \(\mathcal{C}^1\) on an interval \(I = [a,b]\), and a function \(F\) which is an antiderivative of \(f\).

The integral from \(a\) to \(b\) is the area located between the curve of \(f\) and the abscissa axis, from \(a\) to \(b\) (see link between integrals and antiderivatives).

Function which is defined by an integral

$$ F(x) = \int_{a}^x f(t) \hspace{0.2em}dt \Longrightarrow F(a) = 0$$

Integral from a to a

$$ \int_{a}^a f(t) \hspace{0.2em}dt =0 $$

Integral from b to a

$$ \int_{b}^a f(t) \hspace{0.2em}dt = -\int_{a}^b f(t) \hspace{0.2em}dt $$

Chasles' relation

$$ \int_{a}^{\lambda} f(t) \hspace{0.2em}dt + \int_{\lambda}^b f(t) \hspace{0.2em}dt = \int_{a}^b f(t) \hspace{0.2em}dt $$

Linearity of the integral

$$ \int_{a}^b \biggl(\lambda f(t) + \mu g(t) \hspace{0.2em} \biggr) dt = \lambda \int_{a}^b f(t) \hspace{0.2em}dt + \mu \int_{a}^b g(t) \hspace{0.2em}dt $$

The integral of a linear combination is the linear combination of the integral of each function

Positivity of the integral

$$ \enspace f(x) \geqslant 0 \hspace{0.2em} \Longrightarrow \hspace{0.2em} \int_{a}^b f(t) \hspace{0.2em}dt \geqslant 0 $$

In the same way, if \( f(x) \leqslant 0 \) on \( [a, b] \),

$$ f(x) \textcolor{#8E5B5B}{\leqslant} 0 \hspace{0.2em} \Longrightarrow \hspace{0.2em} \int_{a}^b f(t) \hspace{0.2em}dt\textcolor{#8E5B5B}{\leqslant} 0 $$

Growth of the integral

$$ f(x) \leqslant g(x) \hspace{0.2em} \Longrightarrow \hspace{0.2em} \int_{a}^b f(t) \hspace{0.2em}dt \leqslant \int_{a}^b g(t) \hspace{0.2em}dt $$

Mean value

$$ \mu = \frac{1}{(b-a)} \int_{a}^b f(t) $$

Mean value theorem for definite integrals

$$ \exists c \in \hspace{0.05em} ]a,b[, \enspace f(c) = \frac{1}{(b-a)} \int_{a}^b f(t) \hspace{0.2em}dt$$

The extreme value theorem

$$ m \hspace{0.2em} \leqslant \hspace{0.2em} \frac{1}{(b-a)} \int_{a}^b f(t) \hspace{0.2em}dt \hspace{0.2em} \leqslant \hspace{0.2em} M $$

Recap table of the properties of integrals

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Demonstrations

Function which is defined by an integral

We know from the link between integrals and antiderivatives that:

But, if a function is defined by an integral, we do have:

This necessarily implies that \( F(a) = 0 \).

Thus, \( F(x) \) is the antiderivative of \( f(x) \) which is worth \( 0 \) at \( x = a \).

$$ \Biggl \{ \begin{gather*} F'(x) = f(x) \\ F(a) = 0 \end{gather*}$$

And finally,

$$ F(x) = \int_{a}^x f(t) \hspace{0.2em}dt \Longrightarrow F(a) = 0$$

1. Example

$$ ln(x) = \int_{1}^x \frac{dt}{t} $$

Integral from a to a

We know from the link between integrals and antiderivatives that:

So that,

If \( x = a \), then:

The function \(F \) is the antiderivative of \(f\) which is worth \(0\) at \(x=a\).

And finally,

$$ \int_{a}^a f(t) \hspace{0.2em}dt =0 $$

Integral from b to a

We know from the link between integrals and antiderivatives that:

So that,

Reversing the two bounds, we do have now:

And finally,

$$ \int_{b}^a f(t) \hspace{0.2em}dt = -\int_{a}^b f(t) \hspace{0.2em}dt $$

Chasles' relation

We know from the link between integrals and antiderivatives that:

So that,

Hence,

And finally,

$$ \int_{a}^{\lambda} f(t) \hspace{0.2em}dt + \int_{\lambda}^b f(t) \hspace{0.2em}dt = \int_{a}^b f(t) \hspace{0.2em}dt $$

Linearity of the integral

Let \(f,g\) be two continuous functions on \(I = [a,b]\), and two functions \(F,G\) one of their respective antiderivative.

As well, let be \( (\lambda , \mu) \in \hspace{0.05em} \mathbb{R}^2 \) two real numbers which allows us to build another function \(\Lambda\), which is a linear combination of \(f\) and \(g\):

Calculationg its integral between the two bounds \(a\) and \(b\), we do have:

But,

$$\left\{ \begin{gather*} F(b) - F(a) = \int_{a}^b f(t) \\ G(b) - G(a) = \int_{a}^b g(t) \end{gather*} \right\}$$

And finally,

$$ \int_{a}^b \biggl(\lambda f(t) + \mu g(t) \hspace{0.2em} \biggr) dt = \lambda \int_{a}^b f(t) \hspace{0.2em}dt + \mu \int_{a}^b g(t) \hspace{0.2em}dt $$

The integral of a linear combination is the linear combination of the integral of each function

Positivity of the integral

Let \(f\) be a continuous function on \(I = [a,b]\) and for all \( x \in I, \enspace f(x) \geqslant 0\).

If a function\(F\) is an antiderivative of \(f\) on \(I\), then:

So that,

As our hypothesis is that \( f(x) \geqslant 0\), then \( F'(x) \geqslant 0\).

This implies that \( F(x)\) is growing on \( I\).

$$ \Biggl \{ \begin{gather*} f(x) \geqslant 0 \\ F'(x) = f(x) \end{gather*} \hspace{0.2em} \Longrightarrow \hspace{0.2em} F'(x) \geqslant 0 \hspace{0.2em} \Longrightarrow \hspace{0.2em} F \nearrow $$

And in this case, we do have :

So that,

And finally,

$$ f(x) \geqslant 0 \hspace{0.2em} \Longrightarrow \hspace{0.2em} \int_{a}^b f(t) \hspace{0.2em}dt \geqslant 0 $$

In the same way, if \( f(x) \leqslant 0 \) on \( [a, b] \),

$$ f(x) \textcolor{#8E5B5B}{\leqslant} 0 \hspace{0.2em} \Longrightarrow \hspace{0.2em} \int_{a}^b f(t) \hspace{0.2em}dt \textcolor{#8E5B5B}{\leqslant} 0 $$

1. Consequences

We know that the geometrical interpretation of the integral of a function between two bounds \(a\) and \(b\) is the fact that it is worth the area located between the curve of this function and the abscissa axis, such as the following figure:



But thanks to the property of growth, the sign of an integral will match the sing of integrated function on the studied interval. So that in the case of a negative function, we will have a negative integral.



As a consequence of it, if we want to calculate the area of a function which is both positive and negative on the studied interval, we have to change the sign of the negative part to obtain a positive value.

Indeed, a negative area won't have any physical sense.



Let us call \(S\) the sum of areas \(A\) and \(B\).

$$ S = A + B $$

To obtain \(S\), we have to reverse the sign of the integral where \(f(x) \leqslant 0\). To be precise for \( x \in [a, 0]\).

$$ S = - \hspace{0.2em} \underbrace{\int_{a}^0 f(t) \hspace{0.2em}dt } _\text{negative part} \hspace{0.2em} + \hspace{0.2em} \int_{0}^b f(t) \hspace{0.2em}dt $$

Growth of the integral

Let be \(f,g\) two continuous functions on \(I = [a,b]\), and for all \( x \in I, \enspace f(x) \leqslant g(x) \).

Then,

With the property of positivity of the integral, we do have:

Consequently, with the property of linearity of the integral, we do have now:

And finally,

$$ \forall x \in I, \enspace f(x) \leqslant g(x) \hspace{0.2em} \Longrightarrow \hspace{0.2em} \int_{a}^b g(t) \hspace{0.2em}dt \leqslant \int_{a}^b f(t) \hspace{0.2em}dt $$

Mean value

Here, it is more about a statement-definition.

The mean value of an integral on an interval \( [a,b]\) is worth:

$$ \mu = \frac{1}{(b-a)} \int_{a}^b f(t) $$

Mean value theorem extended to definite integrals

Let \(f\) be a continuous function on \(I = [a,b]\), and \(c \in I\) a real number.

Its antiderivative \(F\) is itself continous, since any function derivable on an interval is continuous on this interval.

In our case,

But,

And as a result,

$$ \exists c \in \hspace{0.05em} ]a,b[, \enspace f(c) = \frac{1}{(b-a)} \int_{a}^b f(t) \hspace{0.2em}dt$$

1. Geometric interpretation

We know that the geometric interpretation of the integral of a function between two bounds \(a\) and \(b\) is the fact that it is equal to the area between the curve of the function and the abscissa axis, such as shown in the following figure:



The previous result allows us to see that there will exist at least one real number \( c \in \hspace{0.05em} ]a,b[\) such as:

$$ f(c)(b-a) = \int_{a}^b f(t) \hspace{0.2em}dt$$

That is to say, the integral from \( a\) to \( b \) will be equal to a rectangle formed by \( f(c) \) and \( (b-a) \).

The extreme value theorem

Let \(f\) be a continuous function on \(I = [a,b]\).

We introduce both numbers \((m, M) \in \hspace{0.05em} \mathbb{R}^2 \), respectively the minimum and maximum values of \(f\) on \(I\), and such as the following figure:

With the growth of an integral, we do have:

And finally,

$$ m \hspace{0.2em} \leqslant \hspace{0.2em} \frac{1}{(b-a)} \int_{a}^b f(t) \hspace{0.2em}dt \hspace{0.2em} \leqslant \hspace{0.2em} M $$

Recap table of the properties of integrals