whats this about???!!!

Calculus Cal said:"When you go with a constant k and a function f[x], you know that
D[f[x],x]=D[f[x]+k,x].
as a result , the fundamental formula tells you that when you go with a function f[x] and a constant k, then
∫from a to b f[x] dx =∫from a to b (f[x]+k)dx
what do you say?? :confused:

Reply 1

AlphaNumeric

You've got that mixed up slightly.

D[f(x),x] = D[f(x)+k,x] means that the derivative of a function is the same as the derivative of a function plus a constant

The reverse of that is that the indefinite integral of a function is the same as the indefinite integral of a function plus a constant, since when you put in limits to both those indefinite integrals, the constants vanish (a result I think I proved in another thread of yours).

You are using specific limits there, which isn't the same, because ∫from a to b (f[x]+k)dx = ∫from a to b f[x] dx + k(b-a)

Reply 2

zee22

Hi Alpha
This question was wrote to me in exactley these words .now after you wrote wat you had to say i began to think.. something is wrong about this question
Now do have to agree or disagree with wats in the information in the question??and i dont understand why you wrote your last sentence why? k (b-a)??

Reply 3

AlphaNumeric

What exactly is the question, because you don't put quotation marks at the end of it, so I'm not sure which is the question and which bit is you putting your thoughts.

\frac{d}{dx}\big[ f(x) + k \big] = \frac{d}{dx}f(x)

That is what your first line says.

Your next line seems to then say that

Unparseable latex formula:

\int_{a}^{b}\big{f(x)+k\big} dx = \int_{a}^{b}f(x) dx

which isn't true.

Unparseable latex formula:

\int_{a}^{b}\big{f(x)+k\big} dx = \int_{a}^{b}f(x)dx + \int_{a}^{b}k dx = \int_{a}^{b}f(x)dx + [kx]_{a}^{b} = \int_{a}^{b}f(x)dx + kb-ka = \int_{a}^{b}f(x)dx + k(b-a)

Reply 4

zee22

hello Alpha:
there is nothing in that question thats my thoughts. this is exactley what i have on my question paper.. Maybe im suppose to explain it??
Im gonna ask to double check...what is there to do

Reply 5

zee22

HEllo:
this statement is incorrect , so the question is why??

Reply 6

AlphaNumeric

Because integrating a function is not the same as differentiating a function.

They are inverses of one another, so that means one does the opposite to the other. Differentiation gives the same answer if you differentiate a function or a function plus a constant. Therefore the inverse of differentiation (integration) should take a single function, and give you multiple answers which it does because integrating indefinitlely gives you a function plus a constant of integration.

It is wrong to say that integration gives the same result if you add on a constant within the integrand (inside the dx). What the fundamental formula tels you is that if you integrate indefinitely you can add on any constant you like to your indefinite integral and it's still correct.