Integral is one of the frequently used branches of calculus that is used to evaluate the integrals of the functions with respect to independent variables. It is the inverse process of finding the other branch of calculus such as differential.
The family of the curves is represented with the help of integral calculus and it is frequently used to evaluate the area and volumes under the curves. In this article, we are going to explain this representation of the family of curves with the help of types and examples.
What is the Integral Calculus?
The term integral is used in calculus for finding the antiderivative and the area under the curve. As calculus is fundamental in dealing with the properties of integral calculus and differential calculus.
The process of integration is the way to find the integral of the function. This branch of calculus is used in various fields of real life and educational sectors such as engineering and physics. Where in engineering it is used to find the geometry of buildings by the engineers.
While in physics, it is used to explain the middle of gravity along with the other terms. 3 dimensional models are displayed in graphical representation with the help of the integral calculus.
There are two types of integral calculus:
- Definite integral
- Indefinite Integral
Definite integral
It is a type of integration that is used to find the area under the curve, arc length, pressure, the center of mass, and the volume of various terms. It is a branch of integral that involves the upper and lower limit values of the function.
The definite integral limits are evaluated to the integral of the function with the help of fundamental theorem of calculus. According to this theorem of calculus the upper limit will be applied first to the integral function and then the lower limit with a negative sign among them.
Indefinite Integral
The indefinite integral is the other type of integration that is used to find the antiderivative of the function whose original function is the differential of a function as it is the inverse to find the original function.
The term antiderivative is also known as the primitive. There are no boundary values involved in this type of integral. The new function will be obtained with the help of this type of integration and the constant of integration will be applied with it.
Formulas of Integral
Here are the formulas for calculating the area under the curve and the antiderivative of the function.
Type | Formula | Components |
|---|---|---|
Definite Integral | p∫q f(u) du = [F(u)]qp = F(q) – F(q) | Where,
|
Indefinite Integral | ∫ f(u) du = F(u) + C | Where,
|
Properties of Integral Calculus
Below are a few well-known properties of integral calculus.
Properties | Expressions |
|---|---|
Power Property | ʃ f(u) m du = f(u) m + 1/ (m + 1) + C |
Constant Property | ʃ A du = A * u + C |
Constant Function Property | ʃ A * f(u) du = A * ʃ f(u) du + C |
Sum Property | ʃ [f(u) + g(u)] du = ʃ [f(u)] du + ʃ [g(u)] du + C |
Difference Property | ʃ [f(u) – g(u)] du = ʃ [f(u)] du - ʃ [g(u)] du + C |
Trigonometry Property | ʃ sin(u) du = -cos(u) + C ʃ cos(u) du = sin(u) + C ʃ tan(u) du = ln |sec(u)| + C |
Exponential Property | ʃ eu du = eu + C |
Product Property | ʃ f(u) * g’(u) du = f(u) g(u) - ʃ g(u) * f’(u) du |
How to calculate integral calculus problems?
Follow the below examples to learn how to evaluate integral calculus.
Example 1
Calculate the indefinite integral of h(u) = 2u3 + 2u4 – 6u + 2tan(u) – 2t with respect to “u”
Solution
Step 1: Write the given function h(u) and apply the notation of indefinite integral to it.
ʃ h(u) du = ʃ [2u3 + 2u4 – 6u + 2tan(u) – 2t] du
Step 2: Now take the above expression and use the integral properties to write them separately.
ʃ [2u3 + 2u4 – 6u + 2tan(u) – 2t] du = ʃ [2u3] du + ʃ [2u4] du – ʃ [6u] du + ʃ [2tan(u)] du – ʃ [2t] du
Step 3: Now use the constant function property of integral calculus.
ʃ [2u3 + 2u4 – 6u + 2tan(u) – 2t] du = 2ʃ [u3] du + 2ʃ [u4] du – 6ʃ [u] du + 2ʃ [tan(u)] du – 2tʃ [] du
Step 4: Now integrate the above expression with the help of power and trigonometry properties.
ʃ [2u3 + 2u4 – 6u + 2tan(u) – 2t] du = 2 [u3 + 1 / 3 + 1] + 2 [u4 + 1 / 4 + 1] – 6 [u1 + 1 / 1 + 1] + 2 [-log |cos(u)|] – 2t [u] + C
ʃ [2u3 + 2u4 – 6u + 2tan(u) – 2t] du = 2 [u4 / 4] + 2 [u5 / 5] – 6 [u2 / 2] + 2 [-log |cos(u)|] – 2t [u] + C
ʃ [2u3 + 2u4 – 6u + 2tan(u) – 2t] du = 2/4 [u4] + 2/5 [u5] – 6/2 [u2] + 2 [-log |cos(u)|] – 2t [u] + C
ʃ [2u3 + 2u4 – 6u + 2tan(u) – 2t] du = 1/2 [u4] + 2/5 [u5] – 3 [u2] + 2 [-log |cos(u)|] – 2t [u] + C
ʃ [2u3 + 2u4 – 6u + 2tan(u) – 2t] du = u4/2 + 2 u5/5 – 3u2 – 2log |cos(u)| – 2tu + C
An integral calculator could be a handy tool to find the integral of the function in a fraction of a second. Below is a screenshot of the above example solved by this calculator.
Example 2
Calculate the definite integral of g(v) = 3v2 – 4v3 + 2v5 + 6v – 12v4 with respect to “v” in the interval of [1, 3]
Solution
Step 1: Write the given differential function and apply the notation of integral to it with the interval.
pʃq g(v) dv = 1ʃ3 [3v2 – 4v3 + 2v5 + 6v – 12v4] dv
Step 2: Now apply the sum and difference properties of integral calculus to the above expression.
1ʃ3 [3v2 – 4v3 + 2v5 + 6v – 12v4] dv = 1ʃ3 [3v2] dv – 1ʃ3 [4v3] dv + 1ʃ3 [2v5] dv + 1ʃ3 [6v] dv – 1ʃ3 [12v4] dv
Step 3: Now use the constant function property of integral calculus.
1ʃ3 [3v2 – 4v3 + 2v5 + 6v – 12v4] dv = 31ʃ3 [v2] dv – 41ʃ3 [v3] dv + 21ʃ3 [v5] dv + 61ʃ3 [v] dv – 121ʃ3 [v4] dv
Step 4: Now integrate the above expression with the help of power and trigonometry properties.
1ʃ3 [3v2 – 4v3 + 2v5 + 6v – 12v4] dv = 3 [v2 + 1 / 2 + 1]31 – 4 [v3 + 1 / 3 + 1]31 + 2 [v5 + 1 / 5 + 1]31 + 6 [v1 + 1 / 1 + 1]31 – 12 [v4 + 1 / 4 + 1]31
1ʃ3 [3v2 – 4v3 + 2v5 + 6v – 12v4] dv = 3 [v3 / 3]31 – 4 [v4 / 4]31 + 2 [v6 / 6]31 + 6 [v2 / 2]31 – 12 [v5 / 5]31
1ʃ3 [3v2 – 4v3 + 2v5 + 6v – 12v4] dv = 3/3 [v3]31 – 4/4 [v4]31 + 2/6 [v6]31 + 6/2 [v2]31 – 12/5 [v5]31
1ʃ3 [3v2 – 4v3 + 2v5 + 6v – 12v4] dv = [v3]31 – [v4]31 + 1/3 [v6]31 + 3 [v2]31 – 12/5 [v5]31
Step 5: Now use the fundamental theorem of calculus to apply the upper and lower limits of the integral.
1ʃ3 [3v2 – 4v3 + 2v5 + 6v – 12v4] dv = [33 – 13] + [34 – 14] – 1/3 [36 – 16] + 3 [32 – 12] – 12/5 [35 – 15]
1ʃ3 [3v2 – 4v3 + 2v5 + 6v – 12v4] dv = [27 – 1] + [81 – 1] – 1/3 [729 – 1] + 3 [9 – 1] – 12/5 [243 – 1]
1ʃ3 [3v2 – 4v3 + 2v5 + 6v – 12v4] dv = [26] + [80] – 1/3 [728] + 3 [8] – 12/5 [242]
1ʃ3 [3v2 – 4v3 + 2v5 + 6v – 12v4] dv = 26 + 80 – 728/3 + 24 – 2904/5
1ʃ3 [3v2 – 4v3 + 2v5 + 6v – 12v4] dv = 26 + 80 – 9.33 + 24 – 2904/5
1ʃ3 [3v2 – 4v3 + 2v5 + 6v – 12v4] dv = 26 + 80 – 9.33 + 24 – 580.8
1ʃ3 [3v2 – 4v3 + 2v5 + 6v – 12v4] dv = 106 – 9.33 + 24 – 580.8
1ʃ3 [3v2 – 4v3 + 2v5 + 6v – 12v4] dv = 96.67 + 24 – 580.8
1ʃ3 [3v2 – 4v3 + 2v5 + 6v – 12v4] dv = -46.13
Conclusion
Calculus is the fundamental branch of mathematics that deals with the properties of integral and differential. Integral is one of the two main branches of calculus that is used to find the antiderivative and the area under the curve.