Integration of functions

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

Let some function f be given : D ∈ R² → R on a closed interval D=[a,b]. We divide the definition area into intervals and represent this interval as a series: D = {x₀=a,x₁,...,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) = Σⁿ_i=1M_i(x_i-x_i-1), where M_i = sup _{x∈[xi-1,xi]}{f(x)}

Lowest Riemann sum

L(f,D) = Σⁿ_i=1M_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_af(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_af(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_af(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_au⋅dv = u⋅v]^b_a - ∫^b_av 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: x²/a² + y²/b² = 1.
To calculate the area, we need to get the expression of the function y=f(x), we express y:
y = √[b²(1-x²/a²)]
Figure area:
A = 2∫^a_-af(x)dx = 2 ∫^a_-a √[b²(1-x²/a²)dx] = 2 (b/a)∫√[a²-x²]dx
Let's use the replacement of the variable a⋅sin(t) = x, a⋅cos(t)dt = dx:
= 2(b/a)a² ∫^π/2_-π/2cos²tdt = 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_ay²dx

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_af(x)√[1+f'(x)²]dx