Integration is needed whenever you want to calculate the total number of something that is constantly divided (not discrete). For example, this is the volume of any piece, or even the accumulated distance, when the function of the place is known.

In high school mathematics, this generally means more complicated area and volume integrals. Let's look at a few of them in this chapter. Further examples with solutions can be found, for example, from old YO tasks.

Example 1: Calculate the volume of a sphere of radius r by integration.

Let's start by looking at the circle

The equation of the circle holds for the points of the circumference of the circle [this is also easy to see from the figure]

Then construct the sphere by taking the upper semicircle, y > 0, and rotating it around the x-axis

[In the figure, only one point has rotated, but you can imagine the entire curve rotating.]

Now the volume is obtained as the volume of the solid of revolution between [-r, r]

Example 1: The work done by the force F(x) is calculated [in one dimension] as the integral of the force with respect to the distance

The “harmonic” force against stretching a spring is defined

Where k is the spring constant of the spring and x is the deviation from the equilibrium position of the spring. Let us fit the equilibrium position to the zero value of the coordinate system and calculate the work to be done when the spring is stretched from the equilibrium position by the distance s:

This work is stored as the energy of the spring, so the energy stored in such a “harmonic oscillator” at elongation x is

There are many such examples in physics. Otherwise, it is good to know that in general, the expression of force is a derivative of the expression of energy with respect to place, energy is the integral of force with respect to place.

Example 2: [YO long mathematics s2002, abbreviated assignment] The bottom of a building has a circle with a diameter of 19,7 meters. The cross section of a building is always a rectangle with a height of half the base. Determine the volume of the building.

Position the building so that the origin is in the center of the building and the “tips” shown in the figure are on the x-axis. In this case, the ends of the roof arch are set at a distance of 19,7 / 2 m = 9,85 m from the origin. Then the integration range is [-9,85 m; 9,85 m]. Because the volume is calculated as an integral

we still construct the slice area function A(x). The bottom of the building is a circle with a radius of 19,7 m, whereby the distance of the edge from the x-axis (denoted by y) at x is obtained from the equation of the circle

The width of the rectangle at x is 2y and the height is half of this, i.e. y. The area function of a rectangle is obtained

The volume of a building can be calculated as an integral [Note that positive and negative x-values also give the same result based on symmetry]

The integral of the example is not in itself very interesting. However, forming an area function is tricky, so you'll find that it's a good idea to count many of these to get the job done.