Integration problem...?

Question

In my book it is written that:
[-1/r](where upper limit is r and lower limit is ∞]
then write,
[-1/r+1/∞]
But i don't understand how [-1/r+1/∞] comes ?any rule?
please explain me.

Answer 1

-1/r evaluated between (lower limit)r=b and (upper limit)r=a is
-1/a-(-1/b) is
1/b-1/a
Here a=R and b=infty
so this is
lim(b-->infty, a-->R)[1/b-1/a]=-1/R

Answer 2

conider the definite integral
integrate cosx dx between the limits 0 to pi/2
how will you go about it?
integral cosx dx=sinx and now you will applythe limits
cospi/2-cos0=>0-1=>-1
conider integrating x dx fr0m -1 to 1
integral x dx=x^2/2 applying the limits (1^2)/2)-(-1)^2/2=0
apply the same norm if the integral is 1/r and if you have to apply the limits from r to -infinity,substitute r and - infinity as follows
(1/r)-1/-infinity) and it becomes (1/r)+(1/infinity)

Answer 3

I dont undestand the question but this is how integration goes :

Integral ( f(x) ) ; limits between a,b

if F(x) is a primitive of f(x) then

Integral ( f(x) ) ; limits between upper a,lower b
is : F(a) - F(b)

the interesting thing to note is that the integral doesnot depend on some points between a and b, only of a and b self.

Answer 4

Do you know that plagiarism is illegal? You do have a studied approach to procure sympathetic assistance. Ask questions of your teacher in class, pay a tutor, and learn to do your own homework. Or stop aspirations of becoming a "cool" nerd, drop out of school, learn ethics, and go to trade school.

Best of luck with your problems.

Answer 5

F(a)-F(b)
-1/r - -(1/infinity)=-1/r + (1/infinity)