Integral Calculus – Definition, Properties, and Calculation
Integral calculus is the branch of calculus that deals with integrals and their characteristics. The integral is the reverse function of the derivative. Therefore, the integral is also called anti-derivative. It helps us calculate the area under a curve, the volume, and the solution of a differential equation. This may be challenging to determine without it.
In this article, we will examine the definition of integral with its types. We will learn how to calculate the integration of the given function.
Definition of integral calculus
The integral, denoted by ∫, represents the area under a curve. It consists of two components: Integrand, which is the function being integrated, and the differential, which represents the variable with respect to which the integration is performed.
The integral / anti-derivative is referred to as the inverse process of differentiation. Differentiation calculates a function’s rate of change/slope, while integration allows us to find the original function from its derivative. If g is the differentiable function and g’ express the derivative of g, then the integrating g’ gives the original function g.
Types of Integral
Integral is classified into two types:
- Definite integral
- Indefinite integral
Definite Integral
Definite integral calculates the area under the curves of the function between the specified limits. The definite integral is represented as ∫ab f(x) dx, where ‘a’ is lower and ‘b’ is the upper limit of integration (a < b). A definite integral gives a numeric value, representing the net area between the curve and the x-axis within the specified interval.
Indefinite Integral
The indefinite integral is also known as an antiderivative. This type of integral does not contain specific limits of integration. Integration of f(x) is given by F(x), and it can be written as ∫ f(x) dx = F(x) + C, where C is the constant of integration.
Properties of Integral
The following are some basics properties of integrals:
- Constant Property: Pull out the constant factors from the integrand and evaluate the integral.
⇒ ∫ (k * f(x)) dx = k * ∫ f(x) dx (k is any constant)
- Linearity of integral Property: Sum or difference of two or more functions is written as:
⇒ ∫ (f(x) ± g(x)) dx = ∫ f(x) dx ± ∫ g(x) dx
- Reversal Property: Reversing the limits of the integration changes the sign of the integral.
⇒ ∫ab f(x) dx = -∫ba f(x) dx
- Interval Additivity: Divide the integration interval and evaluate integrals over subintervals.
⇒ ∫ab f(x) dx = ∫ac f(x) dx +∫c b f(x) dx
Fundamental Theorems (First and Second) of Integral Calculus
If F(x) is an anti-derivative of the continuous function f(x) on the interval [a, b]
The First Fundamental Theorem: Consider F(x) = ∫a b f(x) dx
d/dx F(x) = d/dx (∫a b f(x) dx) Or F’(x) = f(x)
Second Fundamental Theorem: ∫a b f(x) dx = F (b) – F (a)
Integral formulas
It is necessary to remember the following integral formulas to master integral calculus. Some basic formulas of the integral are given below:
Different methods for calculating the Integral
Different techniques are used to determine the integral. Some methods to evaluate integration are given below:
- Integration by using a Substitution method
- The Integration by Part
- The Integration by Partial Fraction method
Integration by using the substituting method
The substitution method involves the replacement of variables or expressions with new variables or expressions. By selecting a suitable substitution, the integral is converted into a simple form that is easier to evaluate. The mathematical representation of the substitution method is as follows:
∫ f (g(x))g’(x) dx = ∫ (u)du (where u = g(x), and du = g’(x)dx)
Integration by Part
The integration-by-part technique is used to determine the integration of the product of two functions. Let f(x) is first and g(x) is the second function, then the formula of integration by part is as:
∫ (f(x) * g(x)) dx = f(x) ∫ g(x) dx – ∫ (f’(x) *∫ g(x) dx) dx
The order of selecting the functions follows the ILATE rule (which stands for inverse function, logarithm, algebraic expression, trigonometric function, and exponential function).
Integration by using the Partial Fraction method
This method is helpful when integrating rational functions that cannot be integrated directly. By decomposing a rational function into partial fractions, we can split it into simpler fractions that can be integrated separately.
You can get help from online tools like an integral calculator by MeraCalculator to find the results of integral problems according to the above methods.
Solution of Integral Examples
Let’s learn how to calculate the integral with the help of examples.
Example 1:
Solve the following integral
∫ (x3 + 4x2 + x + 3) dx
Solution
∫ (x3 + 4x2 + x + 3) dx
= ∫ x3 dx + ∫ 4x2 dx + ∫ x dx + ∫ 3 dx (By using linearity property)
= ∫ x3 dx + 4 ∫ x2 dx + ∫ x dx + 3 ∫1dx (Pull out the constant by using constant property)
∴∫ xn dx = ( xn+1 / (n + 1)) + C
⇒ (x3 +1 / 3 + 1) + 4(x2+1 / 2 + 1) + (x1+1 / 1 + 1) + 3x + C
⇒ (x4 / 4) + 4(x3 / 3) + (x2 / 2) + 3x + C
Hence, ∫ (x3 + 4x2 + x + 3) dx = (x4 / 4) + (4x3 / 3) + (x2 / 2) + 3x + C
Example 2:
Solve ∫x. cos (x) dx
By using the integration-by-a-part technique.
Solution
Here, the first function is ‘x’ and the second function is ‘cos(x) ’.
∴ ∫ (f(x) * g(x)) dx = f(x) ∫ g(x) dx – ∫ (f’(x) *∫ g(x) dx) dx
∫x. cos (x) dx = x ∫ cos (x) dx – ∫ (d/dx (x) *∫ cos (x) dx) dx
∴ ∫ cos (x) dx = sin (x)
∴ d/dx (x) = 1
= x sin (x) – ∫ sin (x) dx
∴ ∫ sin (x) dx = – cos (x)
= x sin (x) + cos (x) + C
Hence, ∫x. cos (x) dx = x sin (x) + cos (x) + C
Example 3:
Evaluate ∫0𝜋 sin (3x) dx
Solution
∫0𝜋 sin (3x) dx
∴ ∫ sin (kx) = – cos (kx / k)
= [- cos (3x) / 3]0𝜋 = (- 1/3) [cos (3x)] 0𝜋
∴ ∫ab f(x) dx = F (b) – F (a) (Second Fundamental theorem of integral)
= (- 1/3) [cos (3𝜋) – cos (3(0))]
= (-1/3) [-1 – 1] = 2 / 3
Hence, ∫0𝜋 sin (3x) dx = 2 /3
Conclusion
In this article, we have discussed the definition of integral in depth. We talked about its types. Then we discussed the important properties of the integral. The first and second fundamental theorems have been covered in this article. We learned different methods to evaluate integral problems.
MBA,BBA.
I am Smirti Bam, an enthusiastic edu blogger with a passion for sharing insights into the dynamic world of business and management through this website.
I hold a MBA degree from Presidential Business School, Kathmandu, and a BBA degree with a specialization in Finance from Apex College,
