How Can I Understand the Integration Process of this Fraction?

[ampakine]

I came across this problem + solution:
http://imageshack.us/m/715/2962/inteyk.png
but I don't understand the calculus there. How can you integrate the stuff inside the brackets but not integrate that fraction?

 

[tiny-tim]

hi ampakine!

because (1/18) ∫ F(x) dx is the same as ∫ (F(x)/18) dx

(just like (1/18) df(x)/dx is the same as d(f(x)/18)dx)

 

[HallsofIvy]

A number is a number is a number. A constant is a constant is a constant.

If "C" is a constant (number), whether it happens to have be given as a fraction or not is irrelevant:
$$\int C f(x)dx= C\int f(x)dx$$

That should have been one of the very first "rules of integration" you learned.

 

[ampakine]

Ah right, thanks a lot. Any idea how I'd apply that to this one?
http://imageshack.us/m/84/7308/45762731.png
I need to find the value of k so I tried moving the k over like this:
$$z \int x^2 (1 - \frac{x}{3})$$
can I do that?

 

[tiny-tim]

if that z is a k … then yes of course

 

[HallsofIvy]

Ah right, thanks a lot. Any idea how I'd apply that to this one?
http://imageshack.us/m/84/7308/45762731.png
I need to find the value of k so I tried moving the k over like this:
$$z \int x^2 (1 - \frac{x}{3})$$
can I do that?

You should have
$$k \int x^2(1- \frac{x}{3})dx$$
That is, you mean "k" not z and you need that "dx".
And, of course, that is
$$k \int x^2- \frac{x^3}{3} dx= k\left(\int x^2 dx- \frac{1}{3}\int x^3dx\right)$$

 

[TylerH]

You should have some basic rules in your book, if you have one, or on the site you're studying.

$$\int cf(x)dx=c\int f(x)dx$$ where c is constant.
$$\int f(x)+g(x)dx=\int f(x)dx + \int g(x)dx$$

Those are what allow you to do what Halls did above.

 

[ampakine]

You should have
$$k \int x^2(1- \frac{x}{3})dx$$
That is, you mean "k" not z and you need that "dx".
And, of course, that is
$$k \int x^2- \frac{x^3}{3} dx= k\left(\int x^2 dx- \frac{1}{3}\int x^3dx\right)$$

Oh yeah I meant k not z. I don't know how you broke it up into 2 integrals like that, I just integrated it as $$k \int x^2- \frac{x^3}{3} dx$$.

You should have some basic rules in your book, if you have one, or on the site you're studying.

$$\int cf(x)dx=c\int f(x)dx$$ where c is constant.
$$\int f(x)+g(x)dx=\int f(x)dx + \int g(x)dx$$

Those are what allow you to do what Halls did above.

Yeah I obviously haven't practiced integration enough or these things would be 2nd nature to me. When I did calculus in college I focussed mainly on differentiation to get good at it so my integration skills are severely limited.

 

[Mute]

Oh yeah I meant k not z. I don't know how you broke it up into 2 integrals like that, I just integrated it as $$k \int x^2- \frac{x^3}{3} dx$$.
Yeah I obviously haven't practiced integration enough or these things would be 2nd nature to me. When I did calculus in college I focussed mainly on differentiation to get good at it so my integration skills are severely limited.

Integration is a linear operation. That means it satisfies two properties:

If c is a constant,

$$\int dx~(cf(x)) = c\left(\int dx f(x)\right)$$

and if $g(x) = g_1(x) + g_2(x) + \dots g_N(x)$

$$\int dx (g_1(x) + g_2(x) + \dots g_N(x)) = \int dx g_1(x) + \int g_2(x) + \dots + \int dx g_N(x)$$

 

[ampakine]

Heres something confusing me. Say I have this function:
$$P(X) = kx^2 (1 - \frac{x}{3})$$
and I want to solve for k. I can rearrange it and integrate the function like this: $$k \int x^2- \frac{x^3}{3} dx$$
and get
$$k \frac{x^3}{3} - \frac{x^4}{12}$$
but what's on the other side of the equation? I've seen that k will be the reciprocal of the result I get when I solve the definite integral so does that mean that
$$k \frac{x^3}{3} - \frac{x^4}{12} = 1$$?

 

[TylerH]

Couple of things: Avoid capitol letters, they traditionally are used for sets, in exception of when used with integration. Variables, like x, should always be lowercase. Also, you integrated incorrectly; it's $$k \left( \frac{x^3}{3} - \frac{x^4}{12} \right) + C$$ where C is an arbitrary constant. (Oh yeah... those are traditionally uppercase too.)

As for solving for k: There's nothing to solve for. You can solve for k in terms of x and P(x), but until you set P(x) equal to a number, there is no number solution to be found. Were did you get such a problem? I've never heard of using integrals in that way.

 

[magicarpet512]

not really.

result I get when I solve the definite integral so does that mean that
$$k \frac{x^3}{3} - \frac{x^4}{12} = 1$$?

When you are solving a definite integral, you are evaluating the integral at certain endpoints on an interval $$(a,b)$$

So if you have the definite integral
$$k\int^{a}_{b} x^{2} - \frac{x^{3}}{3}dx$$

then for an answer you get

$$k[\frac{x^{3}}{3} - \frac{x^{4}}{12}]^{a}_{b}$$

$$= k[(\frac{a^{3}}{3} - \frac{a^{4}}{12}) - (\frac{b^{3}}{3} - \frac{b^{4}}{12})]$$



It is not necesarily true that

$$k (\frac{x^3}{3} - \frac{x^4}{12} = 1)$$

unless you have a question that says the definite integral

$$k\int^{a}_{b} x^{2} - \frac{x^{3}}{3}dx = 1$$
solve for $$k$$

then you can multiply both sides by the recripocal of

$$[\frac{x^{3}}{3} - \frac{x^{4}}{12}]^{a}_{b}$$
to solve for $$k$$.

Hope this helps.

 

[HallsofIvy]

Heres something confusing me. Say I have this function:
$$P(X) = kx^2 (1 - \frac{x}{3})$$
and I want to solve for k. I can rearrange it and integrate the function like this: $$k \int x^2- \frac{x^3}{3} dx$$
and get
$$k \frac{x^3}{3} - \frac{x^4}{12}$$
but what's on the other side of the equation? I've seen that k will be the reciprocal of the result I get when I solve the definite integral so does that mean that
$$k \frac{x^3}{3} - \frac{x^4}{12} = 1$$?

The other side of what equation? The only equation you give before your question is
$$P(x)= kx^2(1- \frac{x}{3})$$
defining the function.

If you are referring to integrating both sides of that equation, then you would have
$$\int P(x)dx= \frac{k}{3}x^3- \frac{k}{12}x^4+ C$$

But you said "Say I have this function:
$$P(X) = kx^2 (1 - \frac{x}{3})$$
and I want to solve for k." I can see no reason to integrate anything. Solving that equation for k gives just
$$\frac{P(x)}{x^2(1- \frac{x}{3})}$$