The branch of calculus where we learn about integrals and their properties is called integral calculus. Integration is the basic concept that is the converse of differentiation. The fundamental theorem of calculus connects differentiation and integration.
Integral involves finding the signed area under a curve and helping us understand changes over intervals. The history of calculus is a tale of two giants: Isaac Newton and Gottfried Wilhelm Leibniz. In the late 17th century; both independently developed the fundamentals of calculus; including the integral.
This blog aims to cover the concept of integral calculus; its definition; and its properties and provide some examples to understand the calculation of integration in calculus.
What is integral?
Integral involves determining the region’s area under a curve. This is attained by describing the area with many small rectangles and then adding their areas. The total approaches a limit equivalent to the area under a function’s curve. Integration is the method for uncovering the antiderivative of a function.
The symbol denotes integrals ∫.
Rules of integration:
Linearity: The integral of a sum of functions is the sum of their integrals
Constant Rule: ∫ k f(x) dx = k ∫ f (x) dx + c (k = constant).
Reciprocal rule: ∫ (1 / x) dx = ln (x) + c.
Power rule: ∫ xn dx = (xn+1 / n +1) + c (n ≠ -1).
Addition and subtraction rule: ∫ [f (x) ± g (x)] dx = ∫ f (x) dx ± ∫ g (x) dx.
Trigonometric Integrals Formulas
- ∫ cosxdx = sin(x) + C
- ∫ sinx dx = -cos(x) + C
- ∫ sec2x dx = tan(x) + C
- ∫ cosec2x dx = -cot(x) + C
- ∫ sec2x dx = tan(x) + C
- ∫ secx tanx dx = sec(x) + C
- ∫ cscx cotx dx = -csc(x) + C
Inverse Trigonometric Integration Formulas:
- ∫1/(√(1-x2)) = sin-1 x + C
- ∫-1/(√(1-x2)) = cos-1 x + C
- ∫1/(1+x2) = tan-1 x + C
- ∫-1/(1+x2) = cot-1 x + C
- ∫1/ (x√ (x2 -1)) = sec-1 x + C
- ∫-1/ (x√ (x2 -1)) = cosec-1 x + C
Exponential and logarithm integral Formulas:
- ∫ ex dx=ex + C
- ∫dx/x=ln |x| + C
- ∫ ax dx=ax/ln a + C
Classification of integral:
We classified the integral into two main types which are given below:
- Definite integral
- Indefinite integral
Indefinite Integral:
This type doesn’t have specified limits of integration and it represents a function. Indefinite integrals are also known as antiderivatives. Finding the indefinite integral of a function is the converse of taking the derivative of a function.
Mathematically:
∫ f (x) dx = F (x) + c
Where:
- The symbol ∫ represents the integral operation.
- The function f(x) being integrated is referred to as the integrand.
- The variable x is known as the variable of integration.
- The arbitrary constant of integration is represented by C.
Definite integral:
This type has specified limits of integration and it represents a numerical value. This can be used to calculate quantities such as the area under a curve; the volume of a solid; and the length of the curve.
Mathematically:
a∫b f (x) dx = F (b) – F (a)
Calculation of Integral Calculus:
There are many methods to calculate integral of the function. Some noteworthy are the following.
- Direct integration.
- Trigonometric substitution.
- Integration by parts.
- Trigonometric Substitution:
Trigonometric substitution has three main types which are as follows below:
- Substitution for √ (a² − x²)
- Substitution for √ (a² + x²)
- Substitution for √ (x² – a²)
- Substitution for √ (x² – a²):
We can use x = a sec(θ) when dealing with √ (x² – a²). The result in √ (x² – a²) = atan (θ).
- Substitution for √ (a² + x²):
We can use x = atan(θ) for √ (a² + x²). This substitution leads to √ (a² + x²) = a sec(θ).
- Substitution for √ (a² − x²):
We can use the substitution x = a sin (θ) when we face √ (a² − x²) in the integral. This substitution leads to √ (a² − x²) = a cos(θ)
- Direct integration:
Differential equations are solved using this technique. It involves integrating the equation one or more times to arrive at a solution. This approach is sometimes employed to solve higher-order differential equations in addition to the first-order differential equations it is frequently used to solve.
If we want to solve the differential equation by direct substitution then we need to rewrite the equation in the form:
dy / dx = f (x).
once the equation is in this form; we can integrate both sides of the equation to get:
y = ∫ f (x) dx.
This is the indefinite integral of the equation. We need to specify the limits of integration to find the definite integral.
- Integration by parts:
This method for finding the integral of a product of two functions. It is useful when neither function has an antiderivative that can be easily found. This can be expressed as follows:
∫u(x)v′(x) dx = u(x)v(x) − ∫u′(x)v(x) dx
Where:
- u(x) is the first function
- v(x) is the second function
- u′(x) is the derivative of u(x)
- v′(x) is the derivative of v(x)
Choosing which function to assign u(x) and which function to assign v(x) in this method is necessary. There is no universal guideline for this but a few plans can be beneficial:
- Select the function that has the simpler antiderivative.
- Select the function that has the more complicated derivative.
- Select the function that is more similar to the function that you are trying to integrate.
Integral’s Properties:
The most important properties of indefinite integrals include:
- The derivative of an indefinite integral is equal to the integrand.
- The integral of a sum of functions is equal to the sum of the integrals of the individual functions.
- The integral of a constant multiple of a function is equal to the constant multiple of the integral of the function.
- The integral of a function over a closed interval is equal to the change in the value of the function over the interval.
Solved Problems of Integral Calculus:
Let’s solve some examples to gain more understanding of the integral calculus.
Problem 1:
How can we find the area under the curve of the function f(x) = x4 over the interval [0; 2]?
Solution:
Step 1:
To find the area under the curve; we need to calculate the definite integral of the function f(x) = x4 over the given interval [0; 2].
Step 2:
Start by writing down the definite integral
0∫2 x4 dx
Step 3:
Apply the power rule for integration.
(1 / 5) × x5
Step 4:
Substitute the limit into the antiderivative we obtained in the previous step.
0∫2 x4 dx = (1 / 5) × |x5 0|2
Step 5:
Substitute the upper limit (2) into the antiderivative and subtract the result with the antiderivative substituted with the lower limit (0).
= (1/5) ×35 – (1 / 5) × 05
= (1 / 5) × 243 – 0
= 243 / 5
Therefore; the area under the curve of the function f(x) = x4 over the interval [0; 3] is 243 / 4.
Problem 2:
Evaluate the integral∫1/ √(4−x2) dx using trigonometric substitution.
Solution:
Let x=2sin(θ). Then, dx=2cos(θ) dθ.
Substitute x and dx in terms of θ:
= ∫1/ √(4−x2) dx = ∫ [1/ √(4−(2sin(θ)))2]. 2cos(θ) dθ
Simplify the expression inside the square root:
= ∫ [1/ √ (4−4sin2 θ)]. 2cos(θ) dθ
= ∫ [1/ √4(1−sin2 θ)]. 2cos(θ) dθ
= ∫ [(1/2cos(θ))]. 2cos(θ) dθ.
= ∫ 1 dθ
= θ + C
Substitute back x=2sin(θ):
=sin−1(x / 2) + C
Therefore, ∫1/ √(4−x2) dx =sin−1(2x) + C.
Wrap up:
This article covered integral calculus, delving into its definitions, properties, and calculation methods. We covered rules, trigonometric, exponential, and logarithmic integral formulas, and classified integrals into definite and indefinite types. Methods like direct integration, trigonometric substitution, and integration by parts were discussed. Solved problems and provided practical insights.