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:

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