Integrating Exponentials with Coefficients

  • Thread starter Thread starter JFonseka
  • Start date Start date
  • Tags Tags
    Integrals
Click For Summary
SUMMARY

The discussion focuses on integrating the function x.e2x2. The initial attempt incorrectly applies the product rule of differentiation to integration, leading to confusion about the correct answer. The correct approach involves using substitution, specifically letting u = 2x2, which simplifies the integration process. The final result of the integration is e2x2/4, confirming that the initial method was flawed.

PREREQUISITES
  • Understanding of integration techniques, specifically substitution.
  • Familiarity with exponential functions and their properties.
  • Knowledge of differentiation rules, particularly the product rule.
  • Basic algebra skills for manipulating equations.
NEXT STEPS
  • Study integration by substitution techniques in calculus.
  • Review the properties of exponential functions and their derivatives.
  • Practice problems involving the product rule of differentiation.
  • Explore advanced integration methods, such as integration by parts.
USEFUL FOR

Students studying calculus, particularly those focusing on integration techniques, as well as educators seeking to clarify common misconceptions in integration methods.

JFonseka
Messages
117
Reaction score
0

Homework Statement


Integrate: x.e[tex]^{2x^{2}}[/tex]

Homework Equations



None.

The Attempt at a Solution



I first thought that the coefficient 2 in "2x" would become the denominator:

[tex]\frac{x.e^{2x^{2}}}{2}[/tex]

and then integrating the x would mean x^2 and division by another two making the answer:

[tex]\frac{x^{2}.e^{2x^{2}}}{4}[/tex]

But the answer listed in the book doesn't have an x^2, it's only
[tex]\frac{e^{2x^{2}}}{4}[/tex]

What did I do wrong?
 
Physics news on Phys.org
You didn't do anything right! You know, I hope, that the derivative of product is NOT just the product of the derivatives: (fg)'= f'g+ fg', not f'g'. So you can't expect that the integral of a product will just be the product of the integrals: the integral of fg is not just the integral of f times the integral of g.

Here, you need to make a substitution: if u= x2, what is du/dx? What is du in terms of dx?
 
It would be 2x.dx of course.

So I guess the substitution you want me to make is u = 2x^{2} ? Therefore du/dx= 4x.dx
And then it would be [tex]du/4.e^{u}[/tex]

But then I don't see how that becomes the correct answer.
 

Similar threads

  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
20
Views
2K
Replies
19
Views
2K
  • · Replies 27 ·
Replies
27
Views
3K