Integrals of Particular Functions
Integration is a method to sum functions on a large scale. Here, we discuss integrals of some particular functions commonly used in calculations. These integrals have significant real-life applications, such as finding areas between curves, volumes, average values of functions, kinetic energy, centers of mass, and work done, among others.
1.0Integrals of Particular Functions
2.0Proofs of Integral Functions
- We have
Therefore,
- In view of (1) above, we have
Therefore,
- Put x = a tan θ. Then dx = a sec2 θ dθ.
Therefore,
- Let x = a sec θ. Then dx = a sec θ tan θ dθ.
Therefore,
- Let x = a sinθ. Then dx = a cos θ dθ.
Therefore,
- Let x = a tan θ. Then dx = a sec2 θ dθ.
Applying these standard formulae, we now obtain some more formulae which are useful from applications point of view and can be applied directly to evaluate other integrals.
3.0Some more Integrals
- To find the integral , we write
Now, put so that dx = dt and writing . We find the integral reduced to the form depending upon the sign of and hence can be evaluated.
- To find the integral of the type , proceeding as in (1), we obtain the integral using the standard formulae.
- To find the integral of the type , where given p, q, a, b, c are constants. Now, we have to find real numbers, A, B such that:
To determine A and B, we equate the coefficients of x and the constant term from both sides. By solving these equations, we obtain A and B, thus reducing the integral to a known form.
- For the evaluation of the integral of the type , we proceed as in (3) and reduce the integral into known standard forms.
4.0Difference Between a Particular Solution and a Particular Integral
5.0Solved Examples of Integrals of Particular Functions
Example 1:
Solution: put cos x = t ⇒ –sin x dx = dt
Example 2:
Solution: ⇒
⇒
Example 3:
Solution:
Example 4:
Solution: put x2 = t ⇒ 2x dx = t
t = x2
Example 5: Evaluate
Solution: Put x = a cos2θ + b sin2θ, the given integral becomes
Ans.
Example 6: Evaluate
Solution:
Ans.
Example 7: Evaluate
Solution: Express 3x + 2 = ℓ (d.c. of 4x2 + 4x + 5) + m
or, 3x + 2 = ℓ (8x + 4) + m
Comparing the coefficients, we get
8ℓ = 3 and 4ℓ + m = 2 ⇒ ℓ = 3/8 and m = 2 – 4ℓ = 1/2
⇒
Ans.
6.0Practice Questions of Integrals of Particular Functions
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Answers:
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7.0Sample Questions on Integrals of Particular Functions
Q. What is the integral of ?
Ans:
Q. What is the integral of ?
Ans:
Table of Contents
- 1.0Integrals of Particular Functions
- 2.0Proofs of Integral Functions
- 3.0Some more Integrals
- 4.0Difference Between a Particular Solution and a Particular Integral
- 5.0Solved Examples of Integrals of Particular Functions
- 6.0Practice Questions of Integrals of Particular Functions
- 7.0Sample Questions on Integrals of Particular Functions
Frequently Asked Questions
An integral is a mathematical operation that combines the values of a function over an interval, effectively reversing the process of differentiation. Integrals are used to find areas, volumes, and accumulated quantities.
Particular Solution: In differential equations, a particular solution is a specific solution that meets the given initial or boundary conditions. Particular Integral: In the context of integration, a particular integral refers to the antiderivative of a function, often found without considering any specific initial conditions.
Definite integrals calculate the accumulated value of a function over a given interval [a, b]. The result is a numerical value representing the area under the curve between the limits a and b.
Common techniques include: Substitution: Simplifies the integral by changing variables. Integration by Parts: Derived from the product rule for differentiation. Partial Fractions: Decomposes rational functions into simpler fractions.
Integrals are used in various fields such as physics (to calculate work and energy), engineering (for signal processing and system optimization), and economics (to model growth and compute total surplus).
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