The discussion revolves around a reverse differentiation problem involving the expression (x^2 + 3x - 1)/x^4. Participants are exploring the process of integrating this expression and addressing issues related to the manipulation of exponents.
Several participants have provided insights into the integration process, noting the importance of correctly handling exponents and the operations involved. There is an acknowledgment of mistakes made in the initial attempts, and some participants are beginning to clarify their understanding of the integration steps.
There is a mention of potential confusion stemming from the original expression's format and the operations performed, particularly regarding the division by x^4 and the subsequent integration steps. Some participants suggest that clearer communication of the problem statement may aid in resolving misunderstandings.
Look at your signs on the exponents before you intergrate and be careful, that should take care of it.cd246 said:Homework Statement
(x^2+3x-1)/x^4 the answer is,(-x^-1)-(3/2)(x^-2)+(1/3)(x^-3)+CHomework Equations
Just reverse diferentiationThe 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?
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?
malawi_glenn said:Remember in "backwards differentiation":
i) Exponent +1
ii) divide with the new exponent.
[tex](x^2+3x-1)/x^4 = x^{-2} + 3x^{-3} - x^{-4}[/tex]
the first term:
[tex]x^{-2} \rightarrow (x^{-2+1})/(-2+1) = -x^{-1}[/tex]
etc.