Let be \(f\) a continous function on an interval \(I = [a, b]\).
Fundamental theorem of calculus
From a definite integral, it is possible to determine an antiderivative \(S \) of \(f\).
This antiderivative will be the antiderivative of \(f\) which vanishes at point \(a\):
$$ S(x)= \int_a^x \ f(t) dt $$
We can then define a primitive using this integral:
$$ F(x) = \int_a^x \ f(t) dt + F(a) $$
Conversely, from an antiderivative \(F\) of function \(f\), it is possible to determine the integral between two bounds \(a\) and \(x\).
$$ \int_a^x \ f(t) dt = F(x) - F(a) $$
Or more precisely, for two fixed bounds \(a\) and \(b\) :
$$ S_{a,b}= \int_a^b \ f(t) dt = F(b) - f(a) $$
Let be \(f : x \longmapsto f(x) \) a continuous function, positive and increasing on an interval \(I = [a, b]\).
As well, let be \( n \in \mathbb{N}\) a natural number.
Let \(I = [a,b]\) be an interval, let us subdivide this interval into a series of points \(\bigl \{x_0, \ x_1, \ ...,\ x_i, \ x_{i+1}, \ ..., \ x_{n}, \ x_{n+1} \bigr\} \) theoretically quite small, and such as the following figure:
We will set down \(\Delta_{x} \), the difference between one point and the one after:
$$ \forall i \in [\![0, n ]\!], \ \Delta_{x} = x_{i+1}- x_{i} $$
The mean value theorem tells us that:
$$ f \ continuous \ on \ [a,b] \ and \ derivable \ on \ ]a,b[ \ \Longrightarrow \ \exists c \in \hspace{0.05em} ]a, b[, \ f'(c) = \frac{ f(b) - f(a)}{b-a}$$
Function \(f\) being continuous on \([a,b]\) and therefore derivable, we can apply this theorem to it:
$$\forall i \in [\![0, n ]\!], \ \exists \alpha_i \in \hspace{0.05em} ]x_i, x_{i+1}[, \ f'(\alpha_i) = \frac{ f(x_{i+1}) - f(x_i)}{x_{i+1}-x_{i}}$$
So,
$$f'(\alpha_i) \Delta_{x} = f(x_{i+1}) - f(x_i)$$
By adding all these elements over the interval \([a,b]\), we do have:
$$ \sum_{i=0}^n f'(\alpha_i) \Delta_{x} = \sum_{i=0}^n \Bigl[ f(x_{i+1}) - f(x_i) \Bigr]$$
However, we know that when we are faced with recurring amounts, telescoping will occur:
$$\sum_{k=0}^n \bigl [ a_{k+1} - a_k \bigr] = \underbrace{a_{n+1}} _\text{premier terme} - \underbrace{a_{0}} _\text{dernier terme} $$
So in our specific case:
$$ \sum_{i=0}^n \Bigl[ f(x_{i+1}) - f(x_i) \Bigr] = f(x_{n+1}) - f(x_0) $$
Hence,
$$ \sum_{i=0}^n f'(\alpha_i) \Delta_{x} = f(x_{n+1}) - f(x_0) $$
But by initial assumptions, we have:
$$ \Biggl \{ \begin{align*} f(x_0) = f(a) \\ f(x_{n+1}) = f(b) \end{align*} $$
So,
$$ \sum_{i=0}^n f'(\alpha_i) \Delta_{x} = f(b) - f(a) $$
Now, we can replace each function by its respective primitive and:
$$ \sum_{i=0}^n f(\alpha_i) \Delta_{x} = F(b) - F(a) $$
Performing the limit when \(n \to \infty\), that is to say an infinite number of subdivisions of the interval \([a,b]\), we do have:
$$ lim_{n \to \infty} \ \sum_{i=0}^n f(\alpha_i) \Delta_{x} = lim_{n \to \infty} \ F(b) - F(a) $$
$$ \sum_{i=0}^n f(\alpha_i) dx = F(b) - F(a) \qquad (quand \ n \to \infty, \ \Delta_{x} \to dx) $$
The term on the left corresponds to the integral between \(a\) and \(b\).
Moreover, we do notice that:
$$ \forall i \in [\![0, n ]\!], \ n \to \infty \Longrightarrow \Biggl \{ \begin{align*} \alpha_i \to x_i \Longrightarrow (\alpha_0 \to a, \enspace \alpha_{n+1} \to b)\\ f(\alpha_i) \to f(x_i) \Longrightarrow \Bigl(f(\alpha_0) \to f(a), \enspace f(\alpha_{n+1}) \to f(b)\Bigr) \end{align*}$$
We then obtain the complete area between the abscissa axis and the curve of \(f\) on the interval \([a,b]\).
We will call \( S_{a,b} \) the definite integral of \( f \) on the interval \([a, b]\), and we will note it:
$$ S_{a,b}= \int_a^b \ f(t) dt = F(b) - f(a) $$
The notation \( \int \) historically symbolizes the notion of sum, thus establishing a link between the integrality of a function and its primitive.
This is the reason why we use this notation for primitives, we can also speak of an undefined integral, and we will note this family of antiderivatives, all equal up to a constant:
$$ \int^x \ f(x) dx $$
Now let us consider a variable upper bound \(x\).
To avoid any confusion between \(x\), the function f variable of \(f(x)\), and \(x\) the variable representing the variable upper bound of the integral, it is preferable to introduce a new variable \(t\) inside the integrand, we will then have:
$$ S(x)= \int_a^x \ f(t) dt $$
This function of \(x\) is therefore the antiderivative of \(f\) which vanishes at point \(a\).
The parameter \(t\) being a dummy variable, which will disappear after integration. Furthermore, we can use \(t\) or any other variable, all these writings are equivalent:
$$ S(x)= \int_a^x \ f(t) \ dt = \int_a^x \ f(u) \ du = \int_a^x \ f(\phi) \ d \phi \ ... etc.$$
With what was seen above, we then have for a definite integral:
$$ \int_a^x \ f(t) dt = F(x) - F(a) $$
We can then define an antiderivative with this integral which vanishes at \(a\):
$$ F(x) = \int_a^x \ f(t) dt + F(a) $$
If we know how to calculate the definite integral of a function \(f\) continuous on an interval \([a, x]\), we do know how to find an antiderivative of \(f\).
Conversely, by finding an antiderivative of \(f\), we know how to calculate the definite integral over any interval \([a, x]\).
Although in practice, it is easier to find an antiderivative and then calculate its integral than the opposite.
We will study the following integral of the function \(f : x \longmapsto x^2\).
$$ S(x)= \int_a^x \ t^2 \ dt $$
Function \(f\) is continuous and strictly positive on \([a, x]\).
We have seen that knowing how to calculate the definite integral of a function \(f\) which is continuous on \([a, x]\), we know how to find an antiderivative of \(f\). We will therefore be able to determine an antiderivative of \(f : x \longmapsto x^2\) on the interval \([a, x]\).
And vice versa, by finding an antiderivative of \(f\), we know how to calculate the definite integral over any interval \([a, x]\).
We will then carry out these two processes.
We will calculate the integral \(S(x)\) by the Riemann sum method.
One way to do this is to calculate a sum from the left.
The Riemann sum method by calculating from the left tells us that:
For any function \(f\) and \(n \in \mathbb{N}\) a natural number, which the number of subdivisions of the interval \((x-a)\). We then have:
$$ I_n(x)= \biggl(\frac{x-a}{n} \biggr) \sum_{k=0}^{n-1} \Biggl[ f\biggl(a + k \Bigl(\frac{x-a}{n} \Bigr) \biggr) \Biggr] $$
For the sake of simplicity, we set a new variable down representing the step \(\Delta_{x, n}\):
$$ \Delta_{x, n} = \frac{x-a}{n} $$
Which gives us:
$$ I_n(x)= \Delta_{x, n} \sum_{k=0}^{n-1} \Biggl[ f\bigl(a + k \Delta_{x, n} \bigr) \Biggr] $$
From there, we can reduce the step in a infinitesimal way, by making tend \(n \to +\infty\).
$$ \int_a^x \ f(t)\ dt = lim_{n \to +\infty} \ \Bigl[ I_n(x) \Bigr] $$
Thus, let's calculate in our case:
$$ S_n(x)= \Delta_{x, n} \sum_{k=0}^{n-1} \Biggl[ \bigl(a + k \Delta_{x, n} \bigr)^2 \Biggr] $$
$$ S_n(x)= \Delta_{x, n} \Biggl[ a^2 + a^2 + 2a \Delta_{x, n} + \Delta_{x, n}^2 + a^2 + 4a \Delta_{x, n} + 4\Delta_{x, n}^2 \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a^2 + 2(n-1)a \Delta_{x, n} + (n-1)^2 \Delta_{x, n}^2 \Biggr] $$
By putting a little order, we do have:
$$ S_n(x)= \Delta_{x, n} \Biggl[ (n-1)a^2 + 2a \Delta_{x, n} + 4a \Delta_{x, n} \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} 2(n-1)a \Delta_{x, n} + \Delta_{x, n}^2 + 2^2\Delta_{x, n}^2 \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} (n-1)^2 \Delta_{x, n}^2 \Biggr] $$
$$ S_n(x)= \Delta_{x, n} \Biggl[ (n-1)a^2 + 2a \Delta_{x, n} \Bigl(1 + 2 + \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} (n-1)\Bigr ) + \Delta_{x, n}^2 \Bigl(1 + 2^2 \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} + (n-1)^2 \Bigr) \Biggr] $$
We notice the presence of the sum of natural numbers and the sum of natural squares from \(0\) to \((n-1)\).
$$ S_n(x)= \Delta_{x, n} \Biggl[ (n-1)a^2 + 2a \Delta_{x, n} \Biggl[ \sum_{k=0}^{n-1} k \Biggr] + \Delta_{x, n}^2 \Biggl[ \sum_{k=0}^{n-1} k^2 \Biggr] \Biggr] \qquad (4) $$
We will have to adapt these two sums.
The sum of natural numbers from \(0\) until \(n\) is worth:
$$ \sum_{k = 0}^n k = 1 + 2 \hspace{0.1em} + \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} (n-1) + n \hspace{0.1em} \hspace{0.1em} = \frac{n(n+1)}{2} $$
Then, from \(0\) to \((n-1)\) it is worth now,
$$ \sum_{k = 0}^{n-1} k = \frac{(n-1)n}{2} \qquad (5) $$
In the same way, we will adapt the sum of natural squares:
$$ \sum_{k = 0}^n k^2 = \hspace{0.2em} 1 + \hspace{0.2em} 2^2 \hspace{0.1em} + \hspace{0.1em} ... \hspace{0.1em} + \hspace{0.1em} (n-1)^2 + n^2 \hspace{0.1em} \hspace{0.1em} = \frac{n(n+1)(2n+1)}{6} $$
Now, from \(0\) to \((n-1)\),
$$ \sum_{k = 0}^{n-1} k^2 = \frac{(n-1)n(2n)}{6} \qquad (6) $$
Let us inject \( (5) \) and \( (6) \) into \( (4) \):
$$ S_n(x)= \Delta_{x, n} \Biggl[ (n-1)a^2 + 2a \Delta_{x, n} \frac{(n-1)n}{2} + \Delta_{x, n}^2 \frac{(n-1)n(2n)}{6} \Biggr] $$
Which is worth, under a developped form:
$$ S_n(x)= \Delta_{x, n}(n-1)a^2 + 2a \Delta_{x, n}^2 \frac{(n-1)n}{2} + \Delta_{x, n}^3 \frac{(n-1)n(2n)}{6} $$
$$ S_n(x)= \Delta_{x, n}(n-1)a^2 + 2a \Delta_{x, n}^2 \frac{(n^2 -n)}{2} + \Delta_{x, n}^3 \frac{(2n^3 - n^2)}{6} $$
Now, let us replace \( \Delta_{x, n} \) by its value.
$$ S_n(x)= \frac{x-a}{n}(n-1)a^2 + 2a \biggl( \frac{x-a}{n}\biggr)^2 \frac{(n^2 -n)}{2} + \biggl( \frac{x-a}{n} \biggr)^3 \frac{(2n^3 - n^2)}{6} $$
$$ S_n(x)= a^2(x-a) \biggl[ \frac{n-1}{n} \biggr] + a(x-a)^2\biggl[ \frac{n^2 -n}{n^2} \biggr] + \frac{1}{6}(x-a)^3 \biggl[ \frac{2n^3 - n^3}{n^3} \biggr] $$
By stydying the limit when \(n \to +\infty\):
$$ S(x)= \int_a^x \ t^2 \ dt = lim_{n \to +\infty} \ S_n(x) $$
$$ S(x)= a^2(x-a) + a(x-a)^2+ \frac{1}{3}(x-a)^3 $$
$$ S(x)= a^2x - a^3 + a(x^2 - 2ax + a^2)+ \frac{1}{3}(x^3 - 3x^2 a + 3 xa^2 -a^3) $$
$$ S(x)= a^2x - a^3 + ax^2 - 2a^2x + a^3+ \frac{x^3}{3} -x^2 a + xa^2 - \frac{a^3}{3} $$
$$ S(x)= \frac{x^3}{3} - \frac{a^3}{3} + \hspace{0.2em} \underbrace { a^2x + xa^2 - 2a^2x } _\text{ \( = 0\)} \hspace{0.2em} + \hspace{0.2em} \underbrace { a^3 - a^3 } _\text{ \( = 0\)} \hspace{0.2em} + \hspace{0.2em} \underbrace { ax^2 + -x^2 a } _\text{ \( = 0\)} $$
$$ S(x)= \int_a^x \ t^2 \ dt = \frac{x^3}{3} - \frac{a^3}{3} \qquad (7) $$
$$ S(x)= \int_a^x \ t^2 \ dt = F(x) - F(a)$$
We have determined the general antiderivative \(F\) of the function \(f\), and the latter is worth:
$$ F(x) = \int^x \ t^2 \ dt = \frac{x^3}{3} $$
The general antiderivative of a function of the type \( x^n \) can be calculated with the formula:
$$ \int^x t^n \ dt = \frac{1}{n+1} x^{n+1} $$
So, for our study function \(f : x \longmapsto x^2\), we have as general antiderivative:
$$ \int^x t^2 \ dt = \frac{1}{3} x^{3} $$
From this expression, we can calculate the definite integral:
$$ S(x) = \int_a^x t^2 \ dt =\frac{x^3}{3} - \frac{a^3}{3} \qquad (7) $$
Thanks to the expression \((7) \) previously found, we can calculate the area under the curve of the function \(f : x \longmapsto x^2\) between \(0\) and \(1\).
$$ S_{0, 1} = \int_0^1 t^2 \ dt = \Biggl[ \frac{1}{3} x^{3} \Biggr]_0^1 $$
$$ S_{0, 1} = \int_0^1 t^2 \ dt =\frac{1^3}{3} - \frac{0^3}{3} $$
$$ S_{0, 1} =\frac{1}{3} $$
Go to the top of the page
