Integral of 0 (1 Viewer)

asianese · Jul 24, 2012

The integral of 0 is simply a constant?

Can the idea of the 'area under the curve' be applied here? - Seems not?

Aesytic · Jul 24, 2012

the indefinite integral of 0 is a constant - think when you derive the equations of projectile motion in regards to horizontal motion

the area under the curve idea can be applied with the idea of a definite integral, except you're finding the area of the curve bound by the x-axis and the line y=0 (which is also the x-axis), so therefore your "area" will always be 0

Sy123 · Jul 24, 2012

Well when you find the area underneath the curve when you integrate it and apply the Fundemental Theorem of Calculus:
i.e

$\int_{a}^{b} f(x) \ dx=F(b)-F(a) \\ F'(x)=f(x)$

When you do this, the +C's cancel out when you subtract them. And because the C's cancel out, people dont put them in for simplicity's sake (or maybe a deeper mathematical reason)

For example: Find the area underneath the curve of y=x^2 from x=1 to x=0:

$\int_0^1 x^2 \ dx = \left [ \frac{x^3}{3}+C \right ]_0^1 \\ = (\frac{1}{3}+C)-(0+C) = \frac{1}{3}$

In a similar vein:

$\int_0^a 0 \ dx = [C]_0^a = C-C=0$

Well thats how Ive seen it anyway.

iSplicer · Jul 24, 2012

Sy123 said:
Well when you find the area underneath the curve when you integrate it and apply the Fundemental Theorem of Calculus:
i.e

$\int_{a}^{b} f(x) \ dx=F(b)-F(a) \\ F'(x)=f(x)$

When you do this, the +C's cancel out when you subtract them.

For example: Find the area underneath the curve of y=x^2 from x=1 to x=0:

$\int_0^1 x^2 \ dx = \left [ \frac{x^3}{3}+C \right ]_0^1 \\ = (\frac{1}{3}+C)-(0+C) = \frac{1}{3}$

In a similar vein:

$\int_0^a 0 \ dx = [C]_0^a = C-C=0$

Well thats how Ive seen it anyway.

+1, well articulated.

asianese · Jul 24, 2012

I see, thankyou.

Nooblet94 · Jul 25, 2012

Another way to think of it is to consider a function whos derivative is 0 - clearly, it must be a constant.

Carrotsticks · Jul 25, 2012

Sy123 said:
Well when you find the area underneath the curve when you integrate it and apply the Fundemental Theorem of Calculus:
i.e

$\int_{a}^{b} f(x) \ dx=F(b)-F(a) \\ F'(x)=f(x)$

When you do this, the +C's cancel out when you subtract them. And because the C's cancel out, people dont put them in for simplicity's sake (or maybe a deeper mathematical reason)

For example: Find the area underneath the curve of y=x^2 from x=1 to x=0:

$\int_0^1 x^2 \ dx = \left [ \frac{x^3}{3}+C \right ]_0^1 \\ = (\frac{1}{3}+C)-(0+C) = \frac{1}{3}$

In a similar vein:

$\int_0^a 0 \ dx = [C]_0^a = C-C=0$

Well thats how Ive seen it anyway.

I'm a little confused as to what you mean here, but what guarantee is there that the C's are the same, such that they cancel out?

iBibah · Jul 25, 2012

Carrotsticks said:
I'm a little confused as to what you mean here, but what guarantee is there that the C's are the same, such that they cancel out?

Precisely what I was wondering.

lolcakes52 · Jul 25, 2012

Yes, what sy123 said is what we do when we find the area usually. We just skip the step because its trivial.

Sy123 · Jul 25, 2012

Is this the correct explanation then?

$\int_0^a 0 \ dx = [0x]_0^a=0 \times a - 0 \times 0=0$

Also, Ive read from a number of sources that say that it is that reason that we dont write Cs when doing a definite integral. But if its wrong its wrong I guess...

D94 · Jul 25, 2012

Sy123 said:
Also, Ive read from a number of sources that say that it is that reason that we dont write Cs when doing a definite integral. But if its wrong its wrong I guess...

Well, the C is the constant of integration. The derivative of x²+4 is the same as the derivative of x²+7, but the integral of 2x is not necessarily x²+4 or x²+7. The C is sort of a placeholder for the antiderivative constant - we don't know it until we put in limits. So in the case where we bound the integrand, there is no constant of integration as we know it, but rather a specific integral solution.

I don't see why you added the + C in your original reply.

Trebla · Jul 25, 2012

Carrotsticks said:
I'm a little confused as to what you mean here, but what guarantee is there that the C's are the same, such that they cancel out?

The C's must be the same, otherwise the definite integral will always contain an arbitrary constant. From the fundamental theorem of calculus:

$\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)$

Notice that the first term and the second term of the RHS use the same primitive function F(x), therefore it must be the same constant. For example, suppose we have this situation with differently labelled constants:

$\displaystyle\int_a^b x\,dx=\left(\dfrac{b^2}{2}+C_1\right) - \left(\dfrac{a^2}{2}+C_2\right)$

then

$F(b) = \dfrac{b^2}{2}+C_1$

and

$F(a) = \dfrac{a^2}{2}+C_2$

Since a and b only appear in the first term, and we're referring to the same function F, the constants C₁ and C₂ must clearly be equal.

The constant is there to demonstrate that the primitive function is not unique. However, when we refer to F(x) we are implictly referring to a 'specific' primitive function with C taking some fixed value and not varying. Hence, in the fundamental theorem of calculus, the C's must be fixed otherwise it behaves like a variable.

RealiseNothing · Jul 25, 2012

Carrotsticks said:
I'm a little confused as to what you mean here, but what guarantee is there that the C's are the same, such that they cancel out?

But you're still integrating the same function (ie f(x)), you just use different x values which since it's a constant won't change them, so they have to be equal don't they?

Carrotsticks · Jul 25, 2012

Ohhhh right I understand what he is saying now. I thought he was saying this:

$\\ \frac{d}{dx} F(x) = f(x) \\\\ $Therefore integrating both sides with respect to $ x $ yields:$ \\\\ F(x) + C = g(x) + C \\\\ $Therefore $ F(x) = g(x).$

Which is of course not necessarily the case since as I said before (in response to an incorrect interpretation anyway haha), there's no guarantee that the constants are the same.

This is because if the derivative of F(x) is f(x), it does NOT mean that the integral of f(x) is F(x).

cutemouse · Jul 26, 2012

Guys, I think you are overcomplicating things...

Simply,

$\int^b_a 0 \ dx = 0 \int^b_a k \ dx = 0$

In the case of indefinite integrals,

$\int 0 \ dx = 0 \int k \ dx = 0 + C = C$

Carrotsticks · Jul 26, 2012

cutemouse said:
Guys, I think you are overcomplicating things...

Simply,

$\int^b_a 0 \ dx = 0 \int^b_a k \ dx = 0$

In the case of indefinite integrals,

$\int 0 \ dx = 0 \int k \ dx = 0 + C = C$

I don't think the concept of 'taking out the 0' works. Consider the following scenario then:

$\int_{-1}^{2} 0 dx = 0 \int_{-1}^{2} \frac{1}{x} dx $ but the former integral is undefined.$$

Using a Cauchy Principal Value will of course give us 0 as the final answer (not quite happy about having to use it though), but then you run into problems with divergent indefinite integrals, say:

$\int_{0}^{\infty} 0 dx = 0 \int_{0}^{\infty} x^2 dx$

Also, your last line doesn't quite work in this case because suppose k were some arbitrary constant, then by your working out, we should have:

$\int 0 \ dx = 0 \left ( \int k \ dx \right ) = 0 (kx + C) = 0 $ instead of the required answer $ C$

asianese · Jul 26, 2012

Well the reason why I raised the question was because in an actuaries careers talk the guy asked a question:

$\int \tan x dx = \int \sec x \sin x dx \\ $Integrate the RHS by parts using $ u=\sec x , dv=\sin x dx \\ RHS = \sec x \times -\cos x + \int \cos x \sec x \tan x= \\ -1+\int \tan x dx$ And he was trying to get at +C, which would solve the anomaly of 0 = -1, but he said that $\int 0 dx = C$ but yeah I imagined the curve $y=0$ and it's 'infinite 0' area..

RealiseNothing · Jul 26, 2012

asianese said:
Well the reason why I raised the question was because in an actuaries careers talk the guy asked a question:

$\int \tan x dx = \int \sec x \sin x dx \\ $Integrate the RHS by parts using $ u=\sec x , dv=\sin x dx \\ RHS = \sec x \times -\cos x + \int \cos x \sec x = \\ -1+\int \tan x dx$ And he was trying to get at +C, which would solve the anomaly of 0 = -1, but he said that $\int 0 dx = C$ but yeah I imagined the curve $y=0$ and it's 'infinite 0' area..

The integrals aren't right.

asianese · Jul 26, 2012

Yeah sorry, I fixed it

cutemouse · Jul 27, 2012

Carrotsticks said:
$\int 0 \ dx = 0 \left ( \int k \ dx \right ) = 0 (kx + C) = 0 $ instead of the required answer $ C$

Uhh, usually when I evaluate something like $3 \int x \ dx$ I write that it =3(x^2/2) + C ...

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