Reverse diferentiation problem

  • Thread starter Thread starter cd246
  • Start date Start date
  • Tags Tags
    Reverse
cd246
Messages
30
Reaction score
0

Homework Statement


(x^2+3x-1)/x^4 the answer is,(-x^-1)-(3/2)(x^-2)+(1/3)(x^-3)+C

Homework Equations


Just reverse diferentiation

The Attempt at a Solution


=x^2+3x-1+x^-4
=(x^3)/(3)+(3x^2)/(2)-x+(1/3)(x^-3+C)
(1/3)(x^3)+(3/2)(x^2)-x+(1/3)(x^-3)+C
I know i got the answer but somehow it is backwards(exponents mostly). What am I doing wrong?
 
Physics news on Phys.org
cd246 said:

Homework Statement


(x^2+3x-1)/x^4 the answer is,(-x^-1)-(3/2)(x^-2)+(1/3)(x^-3)+C

Homework Equations


Just reverse diferentiation

The Attempt at a Solution


=x^2+3x-1+x^-4
=(x^3)/(3)+(3x^2)/(2)-x+(1/3)(x^-3+C)
(1/3)(x^3)+(3/2)(x^2)-x+(1/3)(x^-3)+C
I know i got the answer but somehow it is backwards(exponents mostly). What am I doing wrong?
Look at your signs on the exponents before you intergrate and be careful, that should take care of it.
 
Remember in "backwards differentiation":
i) Exponent +1
ii) divide with the new exponent.

(x^2+3x-1)/x^4 = x^{-2} + 3x^{-3} - x^{-4}

the first term:
x^{-2} \rightarrow (x^{-2+1})/(-2+1) = -x^{-1}
etc.
 
cd246 said:

Homework Statement


(x^2+3x-1)/x^4 the answer is,(-x^-1)-(3/2)(x^-2)+(1/3)(x^-3)+C


Homework Equations


Just reverse diferentiation


The Attempt at a Solution


=x^2+3x-1+x^-4
=(x^3)/(3)+(3x^2)/(2)-x+(1/3)(x^-3+C)
(1/3)(x^3)+(3/2)(x^2)-x+(1/3)(x^-3)+C
I know i got the answer but somehow it is backwards(exponents mostly). What am I doing wrong?

Perhaps it would help if you told us what the question is!

Are you trying to integrate (x^2+ 3x-1)/x^4?

You then write "= x^2+ 3x-1+ x^-4" which is not at all the same thing: you are adding x^-4 rather than dividing by x^4.

(x^2+ 3x- 1)/x^4= (x^2/x^4)+ 3(x/x^4)- 1/x^4= x^-2+ 3x^-3- x^-4.
Now integrate.
 
malawi_glenn said:
Remember in "backwards differentiation":
i) Exponent +1
ii) divide with the new exponent.

(x^2+3x-1)/x^4 = x^{-2} + 3x^{-3} - x^{-4}

the first term:
x^{-2} \rightarrow (x^{-2+1})/(-2+1) = -x^{-1}
etc.

Now I know what I did wrong now, I just put x^-4 on top instead of subtracting the lower from the upper which I should have done. Thanks glenn
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top