What is integral calculus? Its types and methods to evaluate it.
Integral Calculus deals with evaluating integrals and basic components of calculus learning once students can easily simply solve multiple problems and assist with its use in real-life examples fundamental theorem are used to calculate the speed of a car at a specific time not only the average but overall going.
As the car can change its speed the graph would include a curved line here fundamental theorem will also be used.
However, it is easy to calculate the area when all the lines are straight, this is not possible in all cases.
Calculus and the fundamental theorem of calculus help a student to calculate the area in graphs that has a function in place of a straight line.
Moreover, in this article basic definition of integral its formula, and methods to calculate it with the help of examples will be discussed.
What is integral Calculus?
Integral calculus deals with finding integrals. Finding derivatives process uses dividing while finding integrals involves multiplying. These are the major part of calculus and so long as a person understands the algebra and trig formulas used to solve the above things, it will help to solve many different calculus problems.
The basic formula for integration is
∫ f(y) dy = g(y) + c
Where c is an arbitrary constant.
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The First fundamental theorem of the integral calculus:
This theorem tells us that the function is on the graph and that two points are showing starting and ending of the area to be calculated, and the space between them is known as integral and can be split into two further parts. The first part tells us what is required and the second part is how to calculate it.
It tells us the theorem states the integral from a to b of g(y) dy is equal to f(b)-f(a) where f is the integral of g(y). it means that the area is calculated by subtracting the integral of the function calculated for a from the integral of the function calculated for b.
Steps for evaluating integrals:
Calculating the integral depends on the function and logic of how to find it and it is also helpful for solving the fundamental theorem of calculus. For example, if the function has an exponent so here power rule will be helpful to evaluate and similarly for other functions reverse rules will be used to determine the integral.
Types of Integral:
There are two types of integral calculus one is definite integral and the other is indefinite.
Definite Integral
Reiman Integral also known as Definite integral in this limit is involved in the integration of upper limit and lower limit
A definite integral is shown as:
f(y) dy
Indefinite Integral
The indefinite integral is the other type of integral in this upper and lower limit is not included.
The indefinite integral can be shown as:
∫ f(y) dy = g(y) + c
Here c is an arbitrary constant and f(y) is the integrand.
Rules of integration:
These are some basic rules for integration discussed below in table section 1:
Statement |
Formula |
Sum Rule |
∫ [f(y) + g (y)] dy = ∫ [f(y)] dy + ∫ [g(y)] dy |
Difference Rule |
∫ [f(y) – g (y)] dy = ∫ [f(y)] dy – ∫ [g(y)] dy |
Power Rule |
∫ [f(y)]^{ n} dy = [f(y)]^{ n+1} / (n + 1) + C |
Constant function Rule |
∫ [P f(y)] dy = P ∫ [f(y)] dy |
Formulae:
The integrals of the basic trigonometric identities are mentioned in table section 2:
Function | Formula |
Cos(y) | ∫ [cos(y)] dy = sin (y) + C |
Sin (y) | ∫ [sin(y)] dy = -sin (y) + C |
Tan(y) | ∫ [tan(y)] dy = -ln(cos(y)) + C |
Sec(y) | ∫ [sec (y) dy = ln |sec(y) + tan(y)|+ C |
Cosec(y) | ∫ [cosec (y) dy = – ln |cosec(y) + cot(y)|+ C |
Cot(y) | ∫ [cot (y) dy = ln |sin(y)| + C |
Example Section:
In this section, we will discuss the steps to evaluate the integrals.
Example 1:
Find the integral of y (y – 1)
Solution:
Step 1: Supposition
Let, f(y) = y (y – 1)
f(y) = y^{2 }– y
Step 2: Apply integral
f(y) = ∫ (y^{2 }– y) dy
Step 3: Apply appropriate rule
∫ [f(y) – g (y)] dy = ∫ [f(y)] dy – ∫ [g(y)] dy
∫ (y^{2 }– y) dy = ∫ y^{2 }dy – ∫ y dy
Now the help of power rule
∫ [f(y)]^{ n} dy = [f(y)]^{ n+1} / (n + 1) + C
∫ (y^{2 }– y) dy = y^{3 }/ 3 – y^{2}/2 + C
∫ (y^{2 }– y) dy = y^{3 }/ 3 – y^{2}/2 + C
You can evaluate the integrals using Allmath’s integral calculator and can save your precious time.
Example 2:
Find the integral of 7siny + 3cosy
Solution:
Let,
f (y) = 7siny + 3cosy
f(y) = 7siny + 3cosy
With the help of a formula
∫ f(y) dy = F(y) + C, where c is an arbitrary constant.
f(y) = ∫ [7siny + 3cosy] dy
With the help of the sum rule
∫ [f(y) + g (y)] dy = ∫ [f(y)] dy + ∫ [g(y)] dy
∫ [sin(y) + cos(y)] dy = ∫ [7sin(y)] dy + ∫ [3cos(y)] dy
Now using the Constant function rule
∫ [P f(y)] dy = P ∫ [f(y)] dy
∫ [sin(y) + cos(y)] dy = 7∫ [sin(y)] dy + 3∫ [cos(y)] dy
∫ [sin(y) + cos(y)] dy = 7(-cosy) + 3(siny) + C
∫ [sin(y) + cos(y)] dy = -7cosy + 3siny + C
The antiderivative calculator is an amazing tool that can help you in solving the complex and trigonometric integrals. You can solve the problems without any hassle.
Summary
In this article, the basic detail of integral calculus with the help of the first fundamental theorem of integral calculus real-world examples are discussed formula rules and ways to calculate the integral by using the sum rule, difference rule, constant rule, and also for power rule.
Hope you have gone through this basic concept of integration, now you are able to tackle all the basic problems related to the integral calculus.
FAQ's
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