Integral calculus is a fundamental subtype of mathematics that is used to study definite and indefinite integrals. The area under the curve between two boundary points will be evaluated with the help of a definite integral.
While the indefinite integral is used to evaluate the antiderivative of a function. The definite and indefinite integrals are frequently used in many fields such as physics, engineering, economics, etc.
In this article, we’ll cover the basics of definite and indefinite integrals with the help of examples.
What is integral calculus?
In calculus, a mathematical concept that is used to evaluate the antiderivative of a function and the area under the curve among two points is said to be integral. The way of finding and evaluating the area under the curve and antiderivatives of function is known as integration.
The integral can be imagined as the sum of infinitely small rectangular areas under a curve. The area of each rectangle is equal to the value of the function at that point multiplied by the width of the rectangle.
By summing up these infinitely small rectangles, we can find the area under the curve. It is extensively used in different fields of science. There are two types of integration that are helpful in finding the area under the curve and the antiderivative of a function.
 Definite integral
 Indefinite integral
What is the definite integral?
The definite integral is a kind of integration that is used to evaluate the area under the curve with the help of interval points of the function. The limits of integration are specified in this subtype of integral calculus.
The expression for the definite integral can be written as:
∫^{y}_{x} f(u) du
Where, x & y are the specified limits of integration, f(u) is the function and du is the integrating variable of the function.
The formula expression of the definite integral can be evaluated with the help of the fundamental theorem of calculus. It states that “the definite integral of a function f(u) between two interval points x & y is equivalent to the difference between the antiderivative of f(u) evaluated at “y” and the antiderivative of f(u) evaluated at “x””. Such as
∫^{y}_{x} f(u) du = [F(u)]^{y}_{x}

According to the fundamental theorem 
What is the indefinite integral?
In calculus, the indefinite integral is a way of determining a new function (original function) whose integrand is equal to the derivative. In this subtype of integral, the limits are not specified. The process of evaluating a new function (indefinite integral) is said to be the antiderivative of a function.
The expression of the indefinite integral will be written without upper and lower limits such as:
∫ f(u) du
Where f(u) is a differential function that is to be integrated to find the antiderivative. The constant of integration must be written along with the antiderivative of a function. such as:
∫ f(u) du = F(u) + C 

Relation between Area under the curve and Antiderivative
The calculation of areas under the curve and antiderivative can be done with the help of definite and indefinite integrals respectively. Both types of integrals involve integrating a function that’s why they are closely related.
The fundamental theorem of calculus is a way of showing the relation between the definite and indefinite integrals as first we have to find the antiderivative of a function and then by placing the interval point, we get the area under the curve.
How to calculate the definite and indefinite integrals?
The calculations of the definite and indefinite integral will be explained with the help of examples. Here are a few solved examples to understand the concept of evaluating the area under the curve and finding the antiderivative of a function.
Example 1
Find the antiderivative of the given function with respect to “u”
f(u) = 5sin(u)– 12u^{3} + 4u + 3u^{2} – 50
Solution
Step 1:Write the given function according to the general expression of the indefinite integral.
ʃ f(u) du = ʃ [5sin(u)– 12u^{3} + 4u + 3u^{2} – 50] du
Step 2:Now use the properties of the integral calculus to write the notation of integration with each term separately.
ʃ [5sin(u)– 12u^{3} + 4u + 3u^{2} – 50] du = ʃ [5sin(u)] du– ʃ [12u^{3}] du + ʃ [4u] du + ʃ [3u^{2}] du – ʃ [50] du
Step 3: Now take out the constant coefficients outside the integral notation to make each term standard.
ʃ [5sin(u)– 12u^{3} + 4u + 3u^{2} – 50] du = 5ʃ [sin(u)] du– 12ʃ [u^{3}] du + 4ʃ [u] du + 3ʃ [u^{2}] du – ʃ [50] du
Step 4: Find the antiderivative with the help of power and trigonometry rules.
ʃ [5sin(u)– 12u^{3} + 4u + 3u^{2} – 50] du = 5 [cos(u)]– 12 [u^{3+1} / 3 + 1] + 4 [u^{1+1} / 1 + 1] + 3 [u^{2+1} / 2 + 1] – [50u] + C
ʃ [5sin(u)– 12u^{3} + 4u + 3u^{2} – 50] du = 5 [cos(u)]– 12 [u^{4} / 4] + 4 [u^{2} / 2] + 3 [u^{3} / 3] – [50u] + C
ʃ [5sin(u)– 12u^{3} + 4u + 3u^{2} – 50] du = 5 [cos(u)]– 12/4 [u^{4}] + 4/2 [u^{2}] + 3/3 [u^{3}] – [50u] + C
ʃ [5sin(u)– 12u^{3} + 4u + 3u^{2} – 50] du = 5 [cos(u)]– 3 [u^{4}] + 2/1 [u^{2}] + 1 [u^{3}] – [50u] + C
ʃ [5sin(u)– 12u^{3} + 4u + 3u^{2} – 50] du = 5cos(u)– 3u^{4} + 2u^{2} + u^{3} – 50u + C
An online integral calculator is an alternate way to find the result of the above example without involving in lengthy calculations.
Example 2
Calculate the area under the curve between the interval of [2, 3] with respect to “v”
f(v) = 2v^{5 }+ 3v^{2} + 5v^{3} – v + 20
Solution
Step 1:Write the given function according to the general expression of the definite integral.
ʃ_{x}^{y} f(v) dv = ʃ_{2}^{3} [2v^{5 }+ 3v^{2} + 5v^{3} – v + 20] dv
Step 2:Now use the properties of the integral calculus to write the notation of integration with each term separately.
ʃ_{2}^{3} [2v^{5 }+ 3v^{2} + 5v^{3} – v + 20] dv = ʃ_{2}^{3} [2v^{5}] dv+ ʃ_{2}^{3} [3v^{2}] dv + ʃ_{2}^{3} [5v^{3}] dv – ʃ_{2}^{3} [v] dv + ʃ_{2}^{3} [20] dv
Step 3: Now take out the constant coefficients outside the integral notation to make each term standard.
ʃ_{2}^{3} [2v^{5 }+ 3v^{2} + 5v^{3} – v + 20] dv = 2ʃ_{2}^{3} [v^{5}] dv+ 3ʃ_{2}^{3} [v^{2}] dv + 5ʃ_{2}^{3} [v^{3}] dv – ʃ_{2}^{3} [v] dv + ʃ_{2}^{3} [20] dv
Step 4: Find the antiderivative with the help of power and trigonometry rules.
ʃ_{2}^{3} [2v^{5 }+ 3v^{2} + 5v^{3} – v + 20] dv = 2 [v^{5+1} / 5 + 1]_{2}^{3 }+ 3 [v^{2+1} / 2 + 1]_{2}^{3} + 5 [v^{3+1} / 3 + 1]_{2}^{3} – [v^{1+1} / 1 + 1]_{2}^{3} + [20v]_{2}^{3}
ʃ_{2}^{3} [2v^{5 }+ 3v^{2} + 5v^{3} – v + 20] dv = 2 [v^{6} / 6]_{2}^{3 }+ 3 [v^{3} / 3]_{2}^{3} + 5 [v^{4} / 4]_{2}^{3} – [v^{2} / 2]_{2}^{3} + 20 [v]_{2}^{3}
ʃ_{2}^{3} [2v^{5 }+ 3v^{2} + 5v^{3} – v + 20] dv = 2/6 [v^{6}]_{2}^{3 }+ 3/3 [v^{3}]_{2}^{3} + 5/4 [v^{4}]_{2}^{3} – 1/2 [v^{2}]_{2}^{3} + 20 [v]_{2}^{3}
ʃ_{2}^{3} [2v^{5 }+ 3v^{2} + 5v^{3} – v + 20] dv = 1/3 [v^{6}]_{2}^{3 }+ [v^{3}]_{2}^{3} + 5/4 [v^{4}]_{2}^{3} – 1/2 [v^{2}]_{2}^{3} + 20 [v]_{2}^{3}
Step 5:Now find the area under the curve by putting the interval points to the antiderivative of the function.
ʃ_{2}^{3} [2v^{5 }+ 3v^{2} + 5v^{3} – v + 20] dv = 1/3 [3^{6} – 2^{6}]+ [3^{3} – 2^{3}] + 5/4 [3^{4} – 2^{4}] – 1/2 [3^{2} – 2^{2}] + 20 [3  2]
ʃ_{2}^{3} [2v^{5 }+ 3v^{2} + 5v^{3} – v + 20] dv = 1/3 [729 – 64]+ [27 – 8] + 5/4 [81 – 16] – 1/2 [9 – 4] + 20 [3  2]
ʃ_{2}^{3} [2v^{5 }+ 3v^{2} + 5v^{3} – v + 20] dv = 1/3 [665]+ [19] + 5/4 [65] – 1/2 [5] + 20 [1]
ʃ_{2}^{3} [2v^{5 }+ 3v^{2} + 5v^{3} – v + 20] dv = 665/3+ 19 + 325/4 – 5/2 + 20
ʃ_{2}^{3} [2v^{5 }+ 3v^{2} + 5v^{3} – v + 20] dv = 221.67+ 19 + 325/4 – 5/2 + 20
ʃ_{2}^{3} [2v^{5 }+ 3v^{2} + 5v^{3} – v + 20] dv = 221.67+ 19 + 81.25 – 2.5 + 20
ʃ_{2}^{3} [2v^{5 }+ 3v^{2} + 5v^{3} – v + 20] dv = 339.42
Final Words
The area under the curve and antiderivative is to key terms in calculus that are evaluated with the help of definite and indefinite integrals. The examples of these kinds of integrals will able you to solve any problem of finding antiderivative and calculating the area under the curve.