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
$$ \forall (a, x) \in D_f^2, $$
$$ F(x) = \int_{a}^x f(t) \hspace{0.2em}dt \Longrightarrow F(a) = 0$$
$$ \forall a \in D_f, $$
$$ \int_{a}^a f(t) \hspace{0.2em}dt =0 $$
$$ \forall (a,b) \in D_f^2, $$
$$ \int_{b}^a f(t) \hspace{0.2em}dt = -\int_{a}^b f(t) \hspace{0.2em}dt $$
$$ \forall (a, \lambda ,b) \in D_f^3, \enspace a \leqslant \lambda \leqslant b, $$
$$ \int_{a}^{\lambda} f(t) \hspace{0.2em}dt + \int_{\lambda}^b f(t) \hspace{0.2em}dt = \int_{a}^b f(t) \hspace{0.2em}dt $$
$$ \forall (a,b) \in D_f^2, \enspace \forall (\lambda, \mu) \in \hspace{0.05em} \mathbb{R}^2,$$
$$ \int_{a}^b \biggl(\lambda f(t) + \mu g(t) \hspace{0.2em} \biggr) dt = \lambda \int_{a}^b f(t) \hspace{0.2em}dt + \mu \int_{a}^b g(t) \hspace{0.2em}dt $$
The integral of a linear combination is the linear combination of the integral of each function
$$ \forall (a,x, b) \in D_f^3, \enspace x \in [a, b], $$
$$ \enspace f(x) \geqslant 0 \hspace{0.2em} \Longrightarrow \hspace{0.2em} \int_{a}^b f(t) \hspace{0.2em}dt \geqslant 0 $$
In the same way, if \( f(x) \leqslant 0 \) on \( [a, b] \),
$$ \forall (a,x, b) \in D_f^3, \enspace x \in [a, b], $$
$$ f(x) \textcolor{#A65757}{\leqslant} 0 \hspace{0.2em} \Longrightarrow \hspace{0.2em} \int_{a}^b f(t) \hspace{0.2em}dt\textcolor{#A65757}{\leqslant} 0 $$
$$ \forall (a,x, b) \in D_f^3, \enspace x \in [a, b], $$
$$ f(x) \leqslant g(x) \hspace{0.2em} \Longrightarrow \hspace{0.2em} \int_{a}^b f(t) \hspace{0.2em}dt \leqslant \int_{a}^b g(t) \hspace{0.2em}dt $$
$$ \forall (a,b) \in D_f^2, $$
$$ \mu = \frac{1}{(b-a)} \int_{a}^b f(t) $$
Mean value theorem for definite integrals
$$ \forall (a,b) \in D_f^2, $$
$$ \exists c \in \hspace{0.05em} ]a,b[, \enspace f(c) = \frac{1}{(b-a)} \int_{a}^b f(t) \hspace{0.2em}dt$$
$$ \forall (a,b) \in D_f^2, \ \Bigl[ \forall x \in [a, b], \ m = min\{f\}, \ M = max\{f\} \Bigr], $$
$$ m \hspace{0.2em} \leqslant \hspace{0.2em} \frac{1}{(b-a)} \int_{a}^b f(t) \hspace{0.2em}dt \hspace{0.2em} \leqslant \hspace{0.2em} M $$
Recap table of the properties of integrals
We know from the link between integrals and antiderivatives that:
$$ F(x) = F(a) + \int_{a}^x f(t) \hspace{0.2em}dt $$
But, if a function is defined by an integral, we do have:
$$ F(x) = \int_{a}^x f(t) \hspace{0.2em}dt $$
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,
$$ \forall (a, x) \in D_f^2, $$
$$ F(x) = \int_{a}^x f(t) \hspace{0.2em}dt \Longrightarrow F(a) = 0$$
$$ ln(x) = \int_{1}^x \frac{dt}{t} $$
We know from the link between integrals and antiderivatives that:
$$ F(x) = F(a) + \int_{a}^x f(t) \hspace{0.2em}dt $$
So that,
$$ \int_{a}^x f(t) \hspace{0.2em}dt = F(x) - F(a) $$
If \( x = a \), then:
$$ \int_{a}^a f(t) \hspace{0.2em}dt = F(a) - F(a) = 0 $$
The function \(F \) is the antiderivative of \(f\) which is worth \(0\) at \(x=a\).
And finally,
$$ \forall a \in D_f, $$
$$ \int_{a}^a f(t) \hspace{0.2em}dt =0 $$
We know from the link between integrals and antiderivatives that:
$$ F(x) = F(a) + \int_{a}^x f(t) \hspace{0.2em}dt $$
So that,
$$ \int_{a}^x f(t) \hspace{0.2em}dt = F(x) - F(a) $$
Reversing the two bounds, we do have now:
$$ \int_{a}^x f(t) \hspace{0.2em}dt = F(a) - F(x) = -(F(x) - F(a)) $$
$$ \int_{x}^a f(t) \hspace{0.2em}dt = - \int_{a}^x f(t) \hspace{0.2em}dt $$
And finally,
$$ \forall (a,b) \in D_f^2, $$
$$ \int_{b}^a f(t) \hspace{0.2em}dt = -\int_{a}^b f(t) \hspace{0.2em}dt $$
We know from the link between integrals and antiderivatives that:
$$ F(x) = F(a) + \int_{a}^x f(t) \hspace{0.2em}dt $$
So that,
$$ \int_{a}^x f(t) \hspace{0.2em}dt = F(x) - F(a) $$
Hence,
$$ \int_{a}^b f(t) \hspace{0.2em}dt + \int_{b}^c f(t) \hspace{0.2em}dt = F(b) - F(a) + F(c) - F(b) $$
$$ \int_{a}^b f(t) \hspace{0.2em}dt + \int_{b}^c f(t) \hspace{0.2em}dt = F(c) - F(a) $$
And finally,
$$ \forall (a, \lambda ,b) \in D_f^3, \enspace a \leqslant \lambda \leqslant b, $$
$$ \int_{a}^{\lambda} f(t) \hspace{0.2em}dt + \int_{\lambda}^b f(t) \hspace{0.2em}dt = \int_{a}^b f(t) \hspace{0.2em}dt $$
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 \(g\):
$$ \Lambda = \lambda f(t) + \mu g(t) $$
Calculationg its integral between the two bounds \(a\) and \(b\), we do have:
$$ \int_{a}^b \biggl(\lambda f(t) + \mu g(t) \hspace{0.2em} \biggr) dt = \Biggl[\lambda F(t) + \mu G(t) \Biggr]_a^b $$
$$ \int_{a}^b \biggl(\lambda f(t) + \mu g(t) \hspace{0.2em} \biggr) dt = \lambda F(b) - \lambda F(a) + \mu G(b) - \mu G(a) $$
$$ \int_{a}^b \biggl(\lambda f(t) + \mu g(t) \hspace{0.2em} \biggr) dt = \lambda \bigl(F(b) - F(a)\bigl) \hspace{0.1em} + \hspace{0.1em} \mu \bigl(G(b) - G(a)\bigl) $$
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,
$$ \forall (a,b) \in D_f^2, \enspace \forall (\lambda, \mu) \in \hspace{0.05em} \mathbb{R}^2,$$
$$ \int_{a}^b \biggl(\lambda f(t) + \mu g(t) \hspace{0.2em} \biggr) dt = \lambda \int_{a}^b f(t) \hspace{0.2em}dt + \mu \int_{a}^b g(t) \hspace{0.2em}dt $$
The integral of a linear combination is the linear combination of the integral of each function
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:
$$ \forall x \in I, \enspace F(x) = F(a) + \int^x f(t) \hspace{0.2em}dt $$
$$ F'(x) = \hspace{0.2em} \underbrace {(F(a))'} _\text{= 0} \hspace{0.2em} + \hspace{0.2em} f(x) $$
So that,
$$ F'(x) = f(x) $$
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 :
$$ F(a) \leqslant F(b) \hspace{0.2em} \Longrightarrow \hspace{0.2em} F(b) - F(a) \geqslant 0 $$
So that,
$$ \int_{a}^b f(t) \hspace{0.2em}dt \geqslant 0 $$
And finally,
$$ \forall (a,x, b) \in D_f^3, \enspace x \in [a, b], $$
$$ f(x) \geqslant 0 \hspace{0.2em} \Longrightarrow \hspace{0.2em} \int_{a}^b f(t) \hspace{0.2em}dt \geqslant 0 $$
In the same way, if \( f(x) \leqslant 0 \) on \( [a, b] \),
$$ \forall (a,x, b) \in D_f^3, \enspace x \in [a, b], $$
$$ f(x) \textcolor{#A65757}{\leqslant} 0 \hspace{0.2em} \Longrightarrow \hspace{0.2em} \int_{a}^b f(t) \hspace{0.2em}dt \textcolor{#A65757}{\leqslant} 0 $$
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 \( x \in [a, 0]\).
$$ S = - \hspace{0.2em} \underbrace{\int_{a}^0 f(t) \hspace{0.2em}dt } _\text{negative part} \hspace{0.2em} + \hspace{0.2em} \int_{0}^b f(t) \hspace{0.2em}dt $$
Let be \(f,g\) two continuous functions on \(I = [a,b]\), and for all \( x \in I, \enspace f(x) \leqslant g(x) \).
Then,
$$ f(x) \leqslant g(x) \hspace{0.2em} \Longleftrightarrow \hspace{0.2em} f(x) - g(x) \leqslant 0 $$
With the property of positivity of the integral, we do have:
$$ \int_{a}^b f(t) - g(t) \hspace{0.2em}dt \leqslant 0 $$
Consequently, with the property of linearity of the integral, we do have now:
$$ \int_{a}^b f(t) \hspace{0.2em}dt - \int_{a}^b g(t) \hspace{0.2em}dt \leqslant 0 $$
And finally,
$$ \forall (a,x, b) \in D_f^3, \enspace x \in [a, b],$$
$$ \forall x \in I, \enspace f(x) \leqslant g(x) \hspace{0.2em} \Longrightarrow \hspace{0.2em} \int_{a}^b g(t) \hspace{0.2em}dt \leqslant \int_{a}^b f(t) \hspace{0.2em}dt $$
Here, it is more about a statement-definition.
The mean value of an integral on an interval \( [a,b]\) is worth:
$$ \forall (a,b) \in D_f^2, $$
$$ \mu = \frac{1}{(b-a)} \int_{a}^b f(t) $$
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.
Now, with the mean value theorem, we know that:
For any function \(f\), continuous on \(I = [a,b]\) and derivable on \(]a,b[\),
$$ \exists c \in \hspace{0.05em} ]a,b[, \enspace f'(c) = \frac{f(b) - f(a)}{b-a}$$
In our case,
$$ \exists c \in \hspace{0.05em} ]a,b[, \enspace F'(c) = \frac{F(b) - F(a)}{b-a}$$
But,
$$ F(b) - F(a) = \int_{a}^b f(t) \hspace{0.2em}dt $$
And as a result,
$$ \forall (a,b) \in D_f^2, $$
$$ \exists c \in \hspace{0.05em} ]a,b[, \enspace f(c) = \frac{1}{(b-a)} \int_{a}^b f(t) \hspace{0.2em}dt$$
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 \( (b-a) \).
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:
$$ \forall t \in [a,b], \in m \leqslant f(t) \leqslant M $$
With the growth of an integral, we do have:
$$ \int_{a}^b m \hspace{0.2em}dt \leqslant \int_{a}^b f(t) \hspace{0.2em}dt \leqslant \int_{a}^b M \hspace{0.2em}dt $$
$$ \Bigl[mt\Bigr]_a^b \leqslant \int_{a}^b f(t) \hspace{0.2em}dt \leqslant \Bigl[Mt \Bigr]_a^b $$
$$ m(b-a) \hspace{0.2em} \leqslant \hspace{0.2em} \int_{a}^b f(t) \hspace{0.2em}dt \hspace{0.2em} \leqslant \hspace{0.2em} M(b-a) $$
$$ m \hspace{0.2em} \leqslant \hspace{0.2em} \frac{1}{(b-a)} \int_{a}^b f(t) \hspace{0.2em}dt \hspace{0.2em} \leqslant \hspace{0.2em} M $$
And finally,
$$ \forall (a,b) \in D_f^2, \ \Bigl[ \forall x \in [a, b], \ m = min\{f\}, \ M = max\{f\} \Bigr], $$
$$ m \hspace{0.2em} \leqslant \hspace{0.2em} \frac{1}{(b-a)} \int_{a}^b f(t) \hspace{0.2em}dt \hspace{0.2em} \leqslant \hspace{0.2em} M $$
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