Understanding Integrating Lambda with l1(y) and l0(y)

[Ad VanderVen]

TL;DR

Forms (I/a) (L(y + a t) - L( y)) and - (I/a) (L(y - a t) - L( y)).

I have the following function l₁(y)=c₁. Integrating lambda(t) = l₁(y + a₁t) from 0 to t gives (I/a₁) (L₁(y + a₁t) - L₁(y)), where L'₁(x) = l₁₍x). Now I don't understand why that is.

Similarly, I have the following function l₀(y)=c₀/y. Integrating lambda(t) = l₀(y - a₀t) from 0 to t gives - (I/a₀) (L₀(y - a₀t) - L₁(y)), where L'₀(x) = l₀₍x). Now I don't understand again why that is.

 

[PeroK]

Perhaps you need to use Latex there:

https://www.physicsforums.com/help/latexhelp/

 

[Ad VanderVen]

I think it is readable now.

 

[Delta2]

Hmmmm, do you understand the method of Integration by substitution? If not then try to read the following link:
https://en.wikipedia.org/wiki/Integration_by_substitution

You might find other links in the web, as well as youtube videos. Just search by "Integration by substitution" or "Integration by change of variables".

Integration by substitution is a consequence of the fundamental theorem of calculus and the chain rule of derivatives.

 

[Ad VanderVen]

Hmmmm, do you understand the method of Integration by substitution? If not then try to read the following link:
https://en.wikipedia.org/wiki/Integration_by_substitution

You might find other links in the web, as well as youtube videos. Just search by "Integration by substitution" or "Integration by change of variables".

Integration by substitution is a consequence of the fundamental theorem of calculus and the chain rule of derivatives.

What would I get if I have the following function l(y)=c and I integrate lambda(t) = l(1/(1/y + at)) from 0 to t?

 

[Delta2]

What would I get if I have the following function l(y)=c and I integrate lambda(t) = l(1/(1/y + at)) from 0 to t?

Is $c$ constant with respect to $y$?

 

[Ad VanderVen]

Yes.

 

[Delta2]

Yes.

Then you would get $ct$ which is (if you do the algebraic operations) equal to $$\frac{1}{a}(L(y+at)-L(y))$$ with $$L'(x)=I(x)=c$$ or simply $$L(x)=cx+d$$.

But you haven't answered, have you looked for resources for integration by substitution? Both of the things you mention at the OP can be derived relatively easy with integration by substitution.