Applying integration to math problems

[chwala]

Homework Statement

I am looking at the integration of ##(x+2)^2## with respect to ##x##

Relevant Equations

Integration

Ok i know that,
##\int (x+2)^2 dx= \int [x^2+4x+4] dx= \dfrac{x^3}{3}+2x^2+4x+c##

when i use substitution;

i.e letting ##u=x+2## i end up with;

##\int u^2 du= \dfrac{u^3}{3}+c=\dfrac {(x+2)^3}{3}+c=\dfrac{x^3+6x^2+12x+8}{3} +c##

clearly the two solutions are not the same...

appreciate your insight...which approach is more concrete? note that when we differentiate both solutions we get the same function i.e ##x^2+4x+4##.

 

[Frabjous]

clearly the two solutions are not the same...

They are the same to within a constant which is what one expects. Try integrating with bounds from 0 to a.

 

[Alex Schaller]

Homework Statement:: I am looking at the integration of ##(x+2)^2## with respect to ##x##
Relevant Equations:: Integration

Ok i know that,
##\int (x+2)^2 dx= \int [x^2+4x+4] dx= \dfrac{x^3}{3}+2x^2+4x+c##

when i use substitution;

i.e letting ##u=x+2## i end up with;

##\int u^2 du= \dfrac{u^3}{3}+c=\dfrac {(x+2)^3}{3}+c=\dfrac{x^3+6x^2+12x+8}{3} +c##

clearly the two solutions are not the same...

appreciate your insight...which approach is more concrete? note that when we differentiate both solutions we get the same function i.e ##x^2+4x+4##.

Indefinite integrals can be regarded as a set (family) of curves, and each of the curves can be obtained by shifting in a parallel the curve, upwards or downwards (along the "Y" axis).

 

[scottdave]

The constant "c" in the first example is not necessarily the same "number" as the constant in the second example.

 

[malawi_glenn]

clearly the two solutions are not the same...

Why are they not the same? 8/3 + c is a constant right?

You wrote that upon differentiating, we get the same original function.
Well, that is the definition of indefinite integral, i.e. primitive function my friend.

 

[chwala]

Why are they not the same? 8/3 + c is a constant right?

You wrote that upon differentiating, we get the same original function.
Well, that is the definition of indefinite integral, i.e. primitive function my friend.

True, that ought to have been pretty obvious to me......cheers man!