The definite integral

The definite integral is used to measure the size of the area of shapes that are described by a function.

Motivation

Given a function $f(x) = \sin(x) + \frac{x}{4}$. Now we could ask - how big is the area under the graph curve from the point a = 0 to the point b = 10? So we are interested in the size of the area bounded by the axis x, the graph curve of the function f, and the vertical lines at the points x₁ and x₂. All this is summarized in the figure:

We are interested in the size of the area highlighted in blue. How would we determine it? We can do this by dividing the whole blue area into several smaller areas, and then counting the area of these smaller areas and adding all these partial contents together. In order to calculate the area of the smaller areas, the smaller area must have some reasonable shape so that we can calculate the content using some formula. A rectangle comes to mind.

We first choose an increasing sequence of points x₀ to x_n such that x₀ = a, x_n = b. The larger n, the more accurate the result we get. We choose a simple sequence x₀ = 0, x₁ = 1, x₂ = 2, …, x₁₀ = 10. From adjacent points, we form the intervals <x_i, x_{i + 1}>, i.e. we get the set of intervals

$$ D = \left\{\left<0, 1\right>, \left<1, 2\right>, \left<2, 3\right>, \ldots, \left<9, 10\right>\right\} $$

We denote the first interval by D₁, the second by D₂, etc. Next, we find the minimum values of the function f in each interval D_i. For example, in D₁, the function f has the minimum value for x = 0 and f(0) = 0. In D₂, the minimum value is at x = 1 and $f(1)=\sin(1)+\frac14 \approx 1,0914$. Etc. We mark these function values. We denote the lowest function value in the interval D_i by m_i. So $m_1 = 0, m_2 \approx 1,0914$ etc.

Now we draw rectangles in the graph. i-this rectangle will have width x_i − x_{i − 1} and height m_i:

We can see that each interval has one rectangle that is all under the curve, but is also the tallest. The height of each rectangle in the i-th interval is just m_i. If we now sum the area of these rectangles, we would get an approximate estimate of the size of the area under the curve. The more rectangles we draw under the curve, the more accurate the estimate we get. The area of the i-th rectangle would be calculated as s_i = m_i · x_i − x_{i − 1}, so the sum of all the rectangles would then be:

$$ s = \sum_{i=1}^{10} m_i \cdot x_i - x_{i-1} $$

We can do the reverse: instead of always taking the minimum functional value on the interval D_i, we take the largest functional value. We would label these function values M_i. The rectangles would then look like this:

Again, there is one rectangle for each interval, and it always has height M_i. The top of the rectangle always touches the curve only at the local maxima of the interval. Summing the area of each rectangle would give another estimate of the size of the area under the curve. We would calculate this content as

$$ S = \sum_{i=1}^{10} M_i \cdot x_i - x_{i-1} $$

Again, if we increase the number of rectangles, we would get a more accurate estimate.

We introduce the concepts of lower and upper integral summation. We say that s(D, f) is the lower integral sum and its value is equal to

$$ s(D, f) = \sum_{i=1}^{10} m_i \cdot x_i - x_{i-1}, $$

where f is a continuous function on the interval <a, b>, where x₀ = a, x_n = b, D is the division of the function into intervals, and n is the number of these intervals. We denote the upper integral sum by S(D, f) and it is equal to :

$$ S(D, f) = \sum_{i=1}^{10} M_i \cdot x_i - x_{i-1}, $$

where f is the function and D is the division of the function into n intervals.

However, in simplified terms, the larger D is, the closer the values of s(D, f) and S(D, f) are to each other until they are equal somewhere at infinity. Thus, it is true that there is a single real number A such that

$$ s(D, f) \le A \le S(D, f) $$

for all possible divisions of D. What is A a number? The number A is just the area of the area under the curve, because it is the only value that is greater than or equal to the sum of the area of the rectangles below the curve and less than or equal to the sum of the area of the rectangles "above the curve".

Definition of the definite integral

From the previous section, we know that for a continuous function f on the interval <a, b>, s(D, f) ≤ A ≤ S(D, f) holds for all divisions of D. This number A is called the definite integral of the function f from a to b. We then use the notation

$$ A = \int_a^b f(x) \mathrm{d}x. $$

The Newton-Leibniz formula further tells us that the following relation holds:

$$ \int_a^b f(x) \mathrm{d}x = F(b)-F(a), $$

where F is a primitive function to the function f. The right-hand side of the equation is sometimes written like this:

$$ \int_a^b f(x) \mathrm{d}x = \left[F(x)\right]_a^b $$

The geometric meaning of the definite integral is then the aforementioned area content under the curve of the nonnegative function f. We started with the function $f(x) = \sin(x) + \frac{x}{4}$, so let's try to calculate the area content on the interval <0, 10>. We thus calculate the definite integral

$$ \int_0^{10} \sin(x) + \frac{x}{4} \mathrm{d}x $$

First we integrate the function f:

$$\begin{eqnarray} \int \sin(x) + \frac{x}{4} \mathrm{d}x &=& \int \sin(x) \mathrm{d}x+ \int \frac{x}{4} \mathrm{d}x\\ &=& -\cos(x) + c_1 + \frac{x^2}{8} + c_2 \\ &=& \frac{x^2}{8} - \cos(x) + c_1+c_2 \end{eqnarray}$$

The primitive function F thus takes the form $F(x) = \frac{x^2}{8} - \cos(x) + c$. We will decompose the definite integral:

$$\begin{eqnarray} \int_0^{10} \sin(x) + \frac{x}{4} \mathrm{d}x &=& F(10) - F(0)\\ &=& \frac{10^2}{8} - \cos(10) + c - \left(\frac{0^2}{8} - \cos(0) + c\right)\\ &=& \frac{100}{8} - \cos(10) + \cos(0) + c - c\\ &\approx& 12{,}5 + 0{,}839 + 1\\ &\approx& 14{,}339 \end{eqnarray}$$

The approximate area of the area under the curve on the interval <0, 10> is 14,339.