The link between integrals and antiderivatives

Demonstration

Let be \(f : x \longmapsto f(x) \) a continuous function, positive and increasing on an interval \(I = \bigl[a, b \bigr]\).

As well, let be \( n \in \mathbb{N}\) a natural number.

Fundamental theorem of calculus

1. Analytic demonstration

Over this interval \(I\), let us subdivide this interval into a series of points \(\bigl \{x_0, \ x_1, \ ...,\ x_i, \ x_{i+1}, \ ..., \ x_{n - 1}, \ x_{n} \bigr \} \) 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:

$$ f \text{ is continuous on } \bigl[a,b \bigr] \text{ and derivable on } \bigl ]a,b \bigr[ \ \Longrightarrow \ \exists c \in \bigl ]a,b \bigr[, \ f'(c) = \frac{ f(b) - f(a)}{b-a} $$

However, since by hypothesis our study function is strictly increasing, this number \(c\) is unique. So in our case :

$$ !\exists c \in \bigl ]a,b \bigr[, \ f'(c) = \frac{ f(b) - f(a)}{b-a}$$

Function \(f\) being continuous on \(\bigl[a,b \bigr]\) and therefore derivable, we can apply this theorem to it:

$$\forall i \in [\![0, n ]\!], \ !\exists \alpha_i \in \hspace{0.04em} \bigl]x_i, x_{i+1}\bigr[, \ f'(\alpha_i) = \frac{ f(x_{i+1}) - f(x_i)}{x_{i+1}-x_{i}}$$


So,

$$f'(\alpha_i) \Delta_{x} = f(x_{i+1}) - f(x_i)$$

Knowing that there is only one element per interval, adding all these elements over the interval \(\bigl[a,b \bigr]\), we do have:

$$ \sum_{i=0}^{n - 1} f'(\alpha_i) \Delta_{x} = \sum_{i=0}^{n - 1} \Bigl[ f(x_{i+1}) - f(x_i) \Bigr] $$

However, we know that when we are faced with recurring amounts, telescoping will occur:

$$\sum_{k=0}^n \bigl [ a_{k+1} - a_k \bigr] = a_{n+1} - a_0 $$

So in our specific case:

$$ \sum_{i=0}^{n - 1} \Bigl[ f(x_{i+1}) - f(x_i) \Bigr] = f(x_{n}) - f(x_0) $$

Hence,

$$ \sum_{i=0}^{n - 1} f'(\alpha_i) \Delta_{x} = f(x_{n+1}) - f(x_0) $$

But by initial assumptions, we have:

$$ \Biggl \{ \begin{gather*} f(x_0) = f(a) \\ f(x_{n}) = f(b) \end{gather*} $$

So,

$$ \sum_{i=0}^{n - 1} f'(\alpha_i) \Delta_{x} = f(b) - f(a) $$

Now, we can replace each function by its respective primitive and:

$$ \sum_{i=0}^{n - 1} f(\alpha_i) \Delta_{x} = F(b) - F(a) $$


Performing the limit when \(n \to \infty\), that is to say an infinite number of subdivisions of the interval \(\bigl[a,b \bigr]\), we do have:

$$ \lim_{n \to \infty} \ \sum_{i=0}^{n - 1} f(\alpha_i) \Delta_{x} = \lim_{n \to \infty} \ \bigl[ F(b) - F(a) \bigr] $$
$$ \sum_{i=0}^{n - 1} f(\alpha_i) dx = F(b) - F(a) \qquad (quand \ n \to \infty, \ \Delta_{x} \to dx) $$

The term on the left corresponds to the integral between \(a\) and \(b\).



Moreover, we do notice that:

$$ \forall i \in [\![0, n ]\!], \ n \to \infty \Longrightarrow \Biggl \{ \begin{gather*} \alpha_i \to x_i \Longrightarrow (\alpha_0 \to a, \enspace \alpha_{n} \to b)\\ f(\alpha_i) \to f(x_i) \Longrightarrow \Bigl(f(\alpha_0) \to f(a), \enspace f(\alpha_{n}) \to f(b)\Bigr) \end{gather*}$$

We then obtain the complete area between the abscissa axis and the curve of \(f\) on the interval \(\bigl[a,b \bigr]\).



We will call \( S_{a,b} \) the definite integral of \( f \) on the interval \(\bigl[a, b \bigr]\), 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:

$$ \int^x \ f(x) dx $$

2. Variable upper bound

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 \(a\):

$$ F(x) = \int_a^x \ f(t) dt + F(a) $$

3. Conclusion

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.

Example

1. Determining a definite integral from an antiderivative

In this sense, it is enough to determine an antiderivative of the function under study, and its definite integral is:

$$ \int_a^x \ f(t) \ dt = F(x) - F(a) $$

2. Determining an antiderivative by calculating the definite integral

We will calculate integrals 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 \(n \to +\infty\).

$$ \int_a^x \ f(t)\ dt = \lim_{n \to +\infty} I_n(x) $$

1. Computation of the primitive of \(f(x) = x^2\)

Thus, let's calculate in our case:

$$ \int_a^x \ t^2 \ dt = \lim_{n \to +\infty} S_n(x) $$

Where \(S_n(x)\) is worth:

$$ S_n(x) = \Delta_{x, n} \sum_{k=0}^{n-1} \Biggl[ \bigl(a + k \Delta_{x, n} \bigr)^2 \Biggr] $$
$$ S_n(x) = \Delta_{x, n} \Biggl[ a^2 + a^2 + 2a \Delta_{x, n} + \Delta_{x, n}^2 + a^2 + 4a \Delta_{x, n} + 4\Delta_{x, n}^2 \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a^2 + 2(n-1)a \Delta_{x, n} + (n-1)^2 \Delta_{x, n}^2 \Biggr] $$

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:

$$ \sum_{k = 0}^n k = \frac{n(n+1)}{2} $$

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:

$$ \sum_{k = 0}^n k^2 = \frac{n(n+1)(2n+1)}{6} $$

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} $$

As we were searching for a general antiderivative \(F\) starting from this expression:

$$ S(x)= \int_a^x \ t^2 \ dt = F(x) - F(a)$$

Therefore, we have determined this primitive \(F\) of the function \(f\), and the latter is worth:

$$ F(x) = \int^x \ t^2 \ dt = \frac{x^3}{3} $$

2. Computation of the primitive of \(g(x) = cos(x)\)

Let's now find a primitive:

$$ \int_a^x \ cos(t) \ dt = \lim_{n \to +\infty} T_n(x) $$

Where \(S_n(x)\) is worth:

$$ T_n(x) = \Delta_{x, n} \sum_{k=0}^{n-1} \Biggl[ cos(a + k \Delta_{x, n} \bigr) \Biggr] $$

We know from addition trigonometric formulas that:

$$ \forall (\alpha, \beta) \in \hspace{0.04em} \mathbb{R}^2, $$

$$ cos(\alpha + \beta) = cos(\alpha) cos(\beta) - sin(\alpha) sin(\beta) $$
$$ T_n(x) = \Delta_{x, n} \Biggl[ \Bigl(cos(a)cos(0) - sin(a)sin(0)\Bigr) + \Bigl(cos(a)cos(\Delta_{x, n}) - sin(a)sin(\Delta_{x, n})\Bigr) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} \Bigl(cos(n)cos(n\Delta_{x, n}) - sin(n)sin(n\Delta_{x, n})\Bigr) \Biggr] $$

Then, by factorizing terms together,

$$ T_n(x) = \Delta_{x, n} \Biggl[ cos(a)\Bigl( cos(0) + cos(\Delta_{x, n}) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} cos(n\Delta_{x, n}) \Bigr) - sin(a)\Bigl( cos(0) + sin(\Delta_{x, n}) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} sin(n\Delta_{x, n}) \Bigr) \Biggr] $$
$$ T_n(x) = \Delta_{x, n} \Biggl[ cos(a) \sum_{k = 0}^{n - 1} cos(k \Delta_{x, n}) - sin(a)\sum_{k = 0}^{n - 1} sin(k \Delta_{x, n}) \Biggr] $$

Now, using trigonometric sums, we do have both relations:

$$\forall n \in \mathbb{N}, $$

$$ \sum_{k = 0}^n cos(k \theta) = cos\left(\frac{n\theta}{2}\right) \times \frac{sin\left(\frac{(n + 1) \theta}{2}\right)}{sin\left(\frac{\theta}{2}\right)} $$

$$ \sum_{k = 0}^n sin(kx) = sin\left(\frac{n\theta}{2}\right) \times \frac{sin\left(\frac{(n + 1) \theta}{2}\right)}{sin\left(\frac{\theta}{2}\right)} $$

So, applying it to our case, that is to say to stop at the \((n - 1)\)-th term of the sum :

$$ T_n(x) = \Delta_{x, n} \left[ cos(a) \left( cos\left(\frac{(n - 1)\Delta_{x, n}}{2}\right) \times \frac{sin\left(\frac{n \Delta_{x, n}}{2}\right)}{sin\left(\frac{\Delta_{x, n}}{2}\right)} \right) - sin(a) \left( sin\left(\frac{(n - 1)\Delta_{x, n}}{2}\right) \times \frac{sin\left(\frac{n \Delta_{x, n}}{2}\right)}{sin\left(\frac{\Delta_{x, n}}{2}\right)} \right) \right] $$
$$ T_n(x) = \frac{x - a}{n} \left[ cos(a) \times cos\left(\frac{(n-1)(x - a)}{2n}\right) \times \frac{sin\left(\frac{x - a}{2}\right)}{sin\left(\frac{(x - a)}{2n}\right)} - sin(a) \times sin\left(\frac{(n-1)(x - a)}{2n}\right) \times \frac{sin\left(\frac{x - a}{2}\right)}{sin\left(\frac{(x - a)}{2n}\right)} \right] $$

Let us use again the addition trigonometric formulas:

$$ T_n(x) = \frac{x - a}{n} \left[ \frac{sin\left(\frac{x - a}{2}\right)}{sin\left(\frac{(x - a)}{2n}\right)} \times cos\left(a + \frac{(n-1)(x - a)}{2n}\right) \right] $$

Now, we are ready to make\(n\) tend to infinity.

.
$$ T(x)= \int_a^x \ cos(t) \ dt = \lim_{n \to +\infty} \ T_n(x) $$

As the quotient \( \frac{(x - a)}{2n} \) tends towards \(0\), then the quotient \(sin\left(\frac{(x - a)}{2n}\right)\) tends towards \(\frac{(x - a)}{2n}\). For the right part, both terms of the same order \(n\) will vanish.

$$ T(x) = \frac{x - a}{n} \times \left( \sim 2n\frac{sin\left(\frac{x - a}{2}\right)}{x - a} \right) \times \left( \sim cos\left(a + \frac{x - a}{2}\right) \right) $$
$$ T(x) = \frac{x - a}{n} \times 2n\frac{sin\left(\frac{x - a}{2}\right)}{x - a} \times cos\left(\frac{2a + x - a}{2}\right) $$
$$ T(x) = \frac{\cancel{x - a}}{\cancel{n}} \times 2\cancel{n}\frac{sin\left(\frac{x - a}{2}\right)}{\cancel{x - a}} \times cos\left(\frac{a + x}{2}\right) $$
$$ T(x) = 2sin\left(\frac{x - a}{2}\right) \times cos\left(\frac{a + x}{2}\right) $$

Finally, still thanks to the addition trigonometric formulas:

$$ \forall (p, q) \in \hspace{0.04em} \mathbb{R}^2, $$

$$ sin(p ) - sin(q) = 2 cos\left(\frac{p+q}{2}\right) sin\left(\frac{p-q}{2}\right) $$

Si, in our case, this leads to it:

$$ T(x) = sin(x) - sin(a) $$

As we were searching for a general antiderivative \(G\) starting from this expression:

$$ T(x)= \int_a^x \ cos(t) \ dt = G(x) - G(a)$$

Therefore, we have determined this primitive \(G\) of the function \(g\), and the latter is worth:

$$ G(x) = \int^x \ cos(t)\ dt = sin(t) $$