quickconverts.org

Integral Of Ln X 3

Image related to integral-of-ln-x-3

Unveiling the Integral of ln(x³)



This article delves into the intricacies of evaluating the definite and indefinite integral of ln(x³), a seemingly simple yet subtly challenging problem in calculus. We will explore various integration techniques, provide step-by-step solutions, and address common misconceptions surrounding this integral. Our goal is to equip readers with a thorough understanding of the process and the underlying mathematical principles.

1. Understanding the Problem: ∫ln(x³) dx



The integral ∫ln(x³) dx represents the area under the curve of the function y = ln(x³) with respect to x. Directly integrating ln(x³) is not straightforward; we need to employ a technique called integration by parts. Before we do that, let's simplify the problem using logarithmic properties. Recall that ln(aᵇ) = b ln(a). Therefore, ln(x³) simplifies to 3ln(x). Our problem now becomes:

∫3ln(x) dx = 3∫ln(x) dx

This simplification significantly streamlines the integration process.

2. Integration by Parts: The Key Technique



Integration by parts is a powerful technique used to integrate the product of two functions. It's derived from the product rule of differentiation and is stated as follows:

∫u dv = uv - ∫v du

To apply this to ∫ln(x) dx, we strategically choose our 'u' and 'dv':

u = ln(x) => du = (1/x) dx
dv = dx => v = x

Substituting these into the integration by parts formula, we get:

∫ln(x) dx = x ln(x) - ∫x (1/x) dx

This simplifies to:

∫ln(x) dx = x ln(x) - ∫1 dx = x ln(x) - x + C

Where 'C' is the constant of integration. Remember, this is the integral of ln(x), not ln(x³).

3. Solving the Original Integral: ∫ln(x³) dx



Now, we can substitute our result back into the simplified equation from section 1:

3∫ln(x) dx = 3(x ln(x) - x + C) = 3x ln(x) - 3x + 3C

Since 3C is still an arbitrary constant, we can simplify it to just 'C':

∫ln(x³) dx = 3x ln(x) - 3x + C


4. Definite Integrals and Practical Applications



Let's consider a definite integral: Find the area under the curve y = ln(x³) from x = 1 to x = e.

We use the result from the previous section:

∫₁ᵉ ln(x³) dx = [3x ln(x) - 3x]₁ᵉ = (3e ln(e) - 3e) - (3(1) ln(1) - 3(1)) = (3e - 3e) - (0 - 3) = 3

Therefore, the area under the curve y = ln(x³) from x = 1 to x = e is 3 square units. This illustrates a practical application of the integral – calculating areas under curves, which has applications in various fields like physics and engineering.


5. Conclusion



Evaluating the integral of ln(x³) requires a combination of logarithmic properties and integration by parts. By simplifying the integrand and carefully applying the integration by parts formula, we arrived at the solution: 3x ln(x) - 3x + C. Understanding this process allows us to solve a range of similar integrals involving logarithmic functions.


FAQs



1. Why is the constant of integration 'C' important? The constant 'C' accounts for the fact that the derivative of any constant is zero. Therefore, infinitely many functions can have the same derivative, differing only by a constant.

2. Can I use other integration techniques besides integration by parts? For this specific integral, integration by parts is the most efficient and straightforward method. Other techniques are less suitable.

3. What is the domain of ln(x³)? The natural logarithm is only defined for positive arguments, so the domain of ln(x³) is (0, ∞).

4. What happens if I try to integrate ln(x³) directly without simplification? You would still arrive at the same answer, but the process would be significantly more cumbersome and prone to error.

5. How can I check my answer? Differentiate your result (3x ln(x) - 3x + C). If the derivative is ln(x³), your integration is correct. Applying the product rule and chain rule will verify this.

Links:

Converter Tool

Conversion Result:

=

Note: Conversion is based on the latest values and formulas.

Formatted Text:

undisputedly meaning
equation to convert celsius to fahrenheit
another word for understand
1cm in inches
first man to walk on the moon
adverse synonym
centripetal
convert pints to litres
isaac physics
fool me once quote
what does congruent mean
33 kg in pounds
askew meaning
what countries fought in ww1
pnd medical abbreviation

Search Results:

Home - Integral Maths Integral is an award-winning online teaching and learning platform designed to develop deep independent mathematical understanding.

Integral Maths for Higher Education Integral for higher education Review, recap, excel Integral for Higher Education is designed for undergraduates who want to review and recap topics studied in A level maths to help build a …

Our organisation - Integral Maths Our organisation Integral is developed by MEI, an independent charity committed to improving maths education. We aim to develop understanding, confidence and enjoyment in using …

Integral Maths for Universities Integral for Higher Education is designed to help refresh A level and Further Mathematics knowledge before approaching new topics.

For Learners - Integral Maths If you’re preparing for your A levels, Integral offers the best experience for interactive, independent maths learning.

UK and international A-Levels Educators - Integral Maths Integral provides extensive UK and International A level Mathematics and Further Mathematics resources, for use in the classroom and online.

Pricing - Integral Maths Get full access to Integral Maths resources through their flexible subscription model. Learn more about the pricing of Integral Maths.

Integral and the AMSP Integral and the Advanced Mathematics Support Programme To support the Advanced Mathematics Support Programme (AMSP), MEI provides free access to certain sections of …

Get in touch - Integral Maths If you have any queries or questions about Integral Maths, do reach out to the team by using the contact form.

UK & International A levels - Integral Maths InTEGRAL FOR A-Levels STUDENTS Maximise your maths learning wherever you are in life Whether you’re preparing for your A levels, or you are simply a maths enthusiast in search of a …