Methods to Find Integrals
There are multiple types of integrals which can be solved using different methods. Some integrals can be directly solved by applying formulas. To solve some integrals, we use the following methods:
Integration by Substitution
To understand the method of Integration by Substitution, we can see the following example or we can explore the article mentioned above.
Example: Evaluate: ∫[(2x)/ {5x2 +1}]dx
Solution:
Let 5x2 + 1 = t. Then, d(5x2 + 1) = dt ⇒ 10 x dx = dt
⇒ ∫[(2x)/ {5x2 +1}]dx = ∫[2x / (t×10x)]dt
⇒ ∫[(2x)/ {5x2 +1}]dx = ∫[1 / (5t)]dt
⇒ ∫[(2x)/ {5x2 +1}]dx = (1/5) ∫[1 / t]dt
⇒ ∫[(2x)/ {5x2 +1}]dx = [(log t) / 5] + C
Integration by Parts
Let’s consider an example for better understanding.
Example: Evaluate the integral ∫ex x dx
Solution:
Let I = ∫ex x dx
This integral can be solved by integration by parts (ILATE rule)
According to ILATE rule the first function u = x (algebraic), v = ex (exponent)
The formula of integration by parts
⇒ ∫u.v dx = u∫v dx – ∫[(du/dx)∫v dx] dx
⇒ ∫x ex dx = x∫ex dx – ∫[(dx/dx)∫ex dx] dx
⇒ ∫x ex dx = x ex – ∫[(1)ex] dx + C1
⇒ ∫x ex dx = x ex – ∫ex dx + C1
⇒ ∫x ex dx = x ex – ex + C1 + C2
⇒ ∫x ex dx = ex(x – 1) + C [C = C1 + C2]
Integration by Partial Fraction
For better understanding, let’s consider the following example.
Example: Evaluate the integral ∫[x/ {(x – 1)(x – 2)}]dx
Solution:
I = ∫[x/ {(x – 1)(x – 2)}]dx
These types of integrals can be solved using partial fraction method.
[x/ {(x – 1)(x – 2)}] = A/(x – 1) + B /(x – 2)
⇒ [x/ {(x – 1)(x – 2)}] = [A(x-2) + B(x-1)]/[(x – 1) (x – 2)]
Equating the numerators
x = A(x-2) + B(x – 1)
⇒ x = Ax + Bx -2A – B
⇒ x = (A + B)x – (2A + B)
Comparing coefficients
A + B = 1 . . . (1)
-2A – B = 0 . . . (2)
From (1) and (2)
B = -2A
Thus, A +(-2A) = 1
⇒ -A = 1
⇒ A = -1
Thus, B = 2
Putting values of A and B in (I)
[x/ {(x – 1)(x – 2)}] = (-1)/(x – 1) + 2 /(x – 2)
⇒ ∫[x/ {(x – 1)(x – 2)}] = ∫[(-1)/(x – 1) + 2 /(x – 2)] dx
⇒ ∫[x/ {(x – 1)(x – 2)}] = ∫[(-1)/(x – 1)]dx + ∫[2 /(x – 2)] dx
⇒ ∫[x/ {(x – 1)(x – 2)}] = -∫[1/(x – 1)]dx + 2∫[1 /(x – 2)] dx
⇒ ∫[x/ {(x – 1)(x – 2)}] = -ln(x – 1) + 2 ln(x – 2) + C1 + C2
⇒ ∫[x/ {(x – 1)(x – 2)}] = -ln(x – 1) + 2 ln(x – 2) + C [C=C1 + C2]
Integral Calculus
Integral Calculus is the branch of calculus that deals with topics related to integration. Integrals are major components of calculus and are very useful in solving various problems based on real life. Some of such problems are the Basel problem, the problem of squaring the circle, the Gaussian integral, etc. Integral Calculus is directly related to differential calculus.
This article is a brief introduction to Integral Calculus, including topics such as fundamental theorems of integral calculus, types of integral, and integral calculus formulas, definite and indefinite integrals with their properties, applications of integral calculus, and their examples.
Table of Content
- What is Integral Calculus?
- Fundamental Theorems of Integral Calculus
- Integral Definition
- Types of Integrals
- Definite Integrals
- Definite Integral Formula
- Properties of Definite Integrals
- Indefinite Integrals
- Properties of Indefinite Integrals
- Improper Integrals
- Multiple Integrals
- Integral Calculus Formulas
- Methods to Find Integrals
- Applications of Integral Calculus
- Differential vs Integral Calculus
- Integral Calculus Examples
- Practice Problems on Integral Calculus