# Integration of functions

Task: calculate the area of a figure that is bounded by an arbitrary function.

Let some function f be given : D ∈ R^{2} → R on a closed interval D=[a,b].
We divide the definition area into intervals and represent this interval as a series:
D = {x_{0}=a,x_{1},...,x_{n-1},x_{n}=b}.

## Riemann integral

Let be given a function f bounded on the interval [a,b]. The function is integrable on the interval [a,b] and the value of the integral is s if

∫^{b}_{a}f = ∫^{b}_{a}f = s ∴ ∫^{b}_{a}f = s

### Riemann integrability criterion

A function f bounded on the interval [a,b] is integrable if for any ε > 0 there exists such a division of the domain of definition that U(f,D) - L(f,D) < ε

### The highest Riemann sum

U(f,D) = Σ^{n}_{i=1}M_{i}(x_{i}-x_{i-1}), where M_{i} = sup _{x∈[xi-1,xi]}{f(x)}

### Lowest Riemann sum

L(f,D) = Σ^{n}_{i=1}M_{i}(x_{i}-x_{i-1}), where M_{i} = inf _{x∈[xi-1,xi]}{f(x)}

## Theorems of integral calculus

### The mean value theorem

Let's give a continuous function f : [a,b] ⊂ R → R, then there exists c∈(a,b) such that
f(c) =∫^{b}_{a}f(t)dt /(b-a) and this value will be
make sense of the arithmetic mean.

### The main theorem of the analysis

Let a continuous function f(x) be given, then there exists some differentiable function F(x)
such that F(x) =∫^{x}_{a}f(t)dt. At the same time
F'(x) = f(x). The function F is called the primitive function f. If F and G are two primitive
functions of f, then they differ by a constant: G(x) = F(x) + c

∫^{b}_{a}f(x)dx = F(b) - F(a) = F(x) ]^{x=b}_{x=a}

### Integration in parts

d(u⋅v) = du⋅v + dv⋅u

∫udv = u⋅v - ∫v du

∫^{b}_{a}u⋅dv = u⋅v]^{b}_{a} -
∫^{b}_{a}v du

### Variable replacement

Let two functions f and g be given, G is the primitive g, then by the chain rule:

(G○f(x))' = G'(f(x))⋅f'(x) = g(f(x))⋅f'(x)

Replace t = f(x), dt = f'(x)dx and get the following integral:

∫g(f(x))f'(x)dx = ∫g(t)dt = G(t) + C = G(f(x)) + C

## Finding the area using the integral

Task: find the area of a figure bounded by an ellipse with radii a and b.

The ellipse equation looks like this: x^{2}/a^{2}+ y^{2}/b^{2}= 1.

To calculate the area, we need to get the expression of the function y=f(x), we express y:

y = √[b^{2}(1-x^{2}/a^{2})]

Figure area:

A = 2∫^{a}_{-a}f(x)dx = 2 ∫^{a}_{-a}√[b^{2}(1-x^{2}/a^{2})dx] = 2 (b/a)∫√[a^{2}-x^{2}]dx

Let's use the replacement of the variable a⋅sin(t) = x, a⋅cos(t)dt = dx:

= 2(b/a)a^{2}∫^{π/2}_{-π/2}cos^{2}tdt = 2ba∫^{π/2}_{-π/2}[(1+cos2t)/2]dt = ab(t + ½sin2t)^{π/2}_{-π/2}= πab

### The area between the graphs of two functions

The area between two functions on the closed interval [a,b] is defined as ∫^{b}_{a}|f(x)-g(x)|dx.
In practice, it is easier to split the integral into intervals in which the sign does not change and integrate the found sections separately.

### Figure volume *disk method*

Let some function f be given : [a,b] → R. The volume of the figure formed by rotation
functions around the X axis can be found using the integral: V = ∫^{b}_{a}y^{2}dx

### Curve length

The length of the curve formed by some function f between points a and b is equal to the integral: L =∫^{b}_{a}
√[1+f'(x)]dx.

### Surface area of the body of rotation

The surface area of the body of rotation formed as a result of the rotation of the function f(x) around the x axis
is equal to the integral: A = 2π∫^{b}_{a}f(x)√[1+f'(x)^{2}]dx