MHB 2.6.62 inverse integrals with substitution

Click For Summary
The discussion centers on the use of custom macros in an Overleaf document related to inverse integrals and substitution. The original poster expresses uncertainty about the correct application of variables b, x, and u in their work. A participant points out a specific error in the solution where the differential was incorrectly stated as du=dx instead of the correct form, du/4=dx. The conversation highlights the importance of careful substitution in integral calculus. Overall, the thread emphasizes the need for clarity and accuracy in mathematical notation.
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
View attachment 9266
ok this is from my overleaf doc
so too many custorm macros to just paste in code
but I think its ok,,, not sure about all details.
appreciate comments...

I got ? somewhat on b and x and u being used in the right places
 

Attachments

  • 253a.PNG
    253a.PNG
    5.1 KB · Views: 164
Last edited:
Physics news on Phys.org
karush said:
ok this is from my overleaf doc
so too many custorm macros to just paste in code
but I think its ok,,, not sure about all details.
appreciate comments...

I got ? somewhat on b and x and u being used in the right places
Ummm...

What exactly is your question?

-Dan
 
OK this one was kinda obvious but usually that is where I make the errors especially where there is substitution.
I post another one here soon.:cool:
 
karush said:
View attachment 9267
That all looks correct to me, except that at one point in the solution of 6.6.62 you have written $du=dx$. That should be $\dfrac{du}4 = dx$ (which is what you have correctly used in the following line).
 

Attachments

  • 253a.PNG
    253a.PNG
    5.1 KB · Views: 120

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
View full post »

Similar threads

Undergrad Different Substitution for Solving an Integral
  • · Replies 6 ·
Replies
6
Views
2K
Undergrad Integrating a product of exponential and trigonometric functions
  • · Replies 21 ·
Replies
21
Views
4K
Undergrad Having trouble understanding a seemingly simple u-substitution
  • · Replies 5 ·
Replies
5
Views
2K
MHB T1.14 Integral: trigonometric u-substitution
  • · Replies 1 ·
Replies
1
Views
1K
MHB 7.3.5 Integral with trig substitution
  • · Replies 1 ·
Replies
1
Views
2K
MHB Understanding Integral Substitution: Finding Equivalent Ranges for Functions
  • · Replies 12 ·
Replies
12
Views
3K
High School Some questions regarding the integral of cos(ax), "a" not zero
  • · Replies 17 ·
Replies
17
Views
4K
Undergrad Did Math DF's integral calculator make a glaring mistake?
  • · Replies 8 ·
Replies
8
Views
3K
High School Justification for cancelling dx in an integral
  • · Replies 4 ·
Replies
4
Views
4K
MHB How can substitution make solving integrals easier?
  • · Replies 1 ·
Replies
1
Views
1K