Endercreeper01

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Part 10. Arc Length of A Curve The arc length of a curve from a to b is how long the function graph from a to b would be. To approximate the arc length, we can break the function up into n lines, each with length [latex]\Delta s[/latex]. An estimate for the arc length would be to write it as the sum of the lengths of these lines, I.e [latex]s \approx \sum_{i=1}^{n} \Delta s_i[/latex]. As we have more and more lines, the approximation becomes more and more accurate, and so we take the limit as n approaches infinity of the above sum. We find that [latex]s = \lim_{n \to \infty} \sum_{i=1}^{n} \Delta s_i[/latex]. We know that an integral is an infinite sum, so we have to integrate over all of the infinitesimal lengths. We now obtain the equation for arc length of a curve: [latex]s=\int_a^b ds[/latex] Because we know that [latex]ds=\sqrt {dx^2+dy^2}[/latex] and that ds is also [latex]\sqrt {1 + \left( \frac {dy} {dx} \right)^2} dx[/latex] we can rewrite the previous equation as [latex]s=\int_a^b \sqrt {1 + \left( \frac {dy} {dx} \right) ^2} dx[/latex] which gives us the equation for arc length. You can also define x = f(y) and use the equation: [latex]s=\int_a^b \sqrt {1 + \left( \frac {dx} {dy} \right) ^2} dy[/latex] as this would also work. For example, we will work out the arc length from 1 to 3 of [latex]y=\frac{2\sqrt {x^3}}{3}[/latex]. The derivative of this function will be [latex]\sqrt x[/latex]. When we square this derivative, we obtain x. The integral now becomes [latex]\int_1^3 \sqrt {1+x} dx[/latex]. The substitution will be u = 1+x, and so du = dx. This integral now becomes [latex]\int_1^3 \sqrt u du[/latex]. The antiderivative of this function is [latex]\frac{2}{3}u^{3/2}[/latex], which is equal to [latex]\frac{2}{3}(1+x)^{3/2}[/latex]. In order to get the answer, we must work this out at x=3, then subtract by the antiderivative value at x=1. Our final answer ends up as [latex]\int_1^3 \sqrt {1+x} dx = \frac{16}{3} - \frac{4\sqrt 2}{3}[/latex]. The next tutorial will be on the area between two curves. Remember, if you need any help with this, you can just ask our calculus forum.

What do you mean by dry corn?

Do black holes gain mass when matter falls into them?

When black holes radiate, they loose some mass. The reason black holes evaporate is because they radiate, since they eventually loose all of their mass.

It would not be a violation of conservation of mass and energy, since the black hole gains mass. When the black hole evaporates, the mass it absorbed would also be evaporated, and this would not violate the conservation of mass. Energy and mass are interchangeable, since [latex]E=mc^2[/latex]. A black hole evaporating would basically mean that all of the energy of a black hole would be converted into radiation. I don't see how this violates the conservation of energy. If white holes exist, they would have negative entropy. Therefore, the entropy is negative of that of a black hole, i.e [latex]S=\frac{-Akc^3}{4G\hbar}[/latex]. That would be a violation of the 2nd law of thermodynamics in a closed system universe. Meanwhile, the metric would still be the same, as the metric for a spherical, non-rotating body (Schwarzschild metric) is [latex]ds^2=(1-\frac{Rs}{r})dt^2-(1-\frac{Rs}{r})^{-1}dr^2-r^2d \theta ^2 - r^2sin^2(\theta )d \phi ^2[/latex]. But, a white hole needs to have everything away from it. It has to move in a direction opposite of that of black holes, i.e negative gravity. In order for this to work, we need to have the metric in the direction outwards. Since a white hole would have negative gravity, we find that the mass of a white hole must be negative. And because black holes emit hawking radiation, white holes would have to absorb it, making them eventually evaporate, since it would gain mass from hawking radiation. White holes would be a violation of thermodynamics if the universe is a closed system. If black holes actually do make mass disappear, there would have to be white holes to keep the amount of mass and energy in the universe the same. But, I don't see how black holes violate the law of conservation of mass/energy, since the mass and energy would turn into radiation with the black hole.

What are you talking about?

What exactly do you mean by "the lord's science"? Pharmacies are made to help people, and there is plenty of evidence they do. Just because doctors are trying to help people with medicine does not mean pharmacies are drug dealers.

No, medicine's purpose is to help patients be healthier. The pros of science outweigh the cons. Science has benefited mankind for centuries. What the government does with science is unrelated to scientists themselves. Scientists do research, and few work with the government.

Religious answers do not count as scientific. You should not answer science questions with religion, as this is a science forum. Nobody knows how life started, but scientists are trying to figure it out.

Haven't you heard of medicine?

Psychology should not be treated like pseudoscience, since psychology follows the scientific method. Pseudoscience is something that is claimed to be scientific, but it does not use the scientific method. Therefore, psychology is not pseudoscience, but things like psychokinesis are.

Thanks for the support! Now, we will discuss improper integrals. Part 8. Improper integrals An improper integral is when one or two limit(s) of a definite integral is infinity. For example, [latex]\int_{-\infty}^{5} \frac{1}{e^x} dx[/latex] and [latex]\int_0^\infty x^{3x} dx[/latex] are both improper integrals. Of course, you can't work this out by just plugging in infinity. For improper integrals, you have to use a limit. You could write the above integral as [latex]\lim_{n \to \infty} \int_0^n x^{3x} dx.[/latex], and this would mean you would have to take the limit as n approaches infinity or negative infinity for that function. Improper integrals are categorized 2 ways: divergent and convergent. It is convergent if the integral exists and is not infinite, and divergent if it is infinite or it doesn't exist. For instance, let's work out the integral [latex]\int_0^\infty \frac{1}{x}[/latex]. We know the antiderivative of this function is [latex]ln(x)[/latex]. We also know that [latex]ln(x)[/latex] approaches infinity as x approaches infinity. Therefore, this integral is divergent. That concludes the tutorial on calculating integrals. Next, you will learn about the applications of integrals. Part 9. Average Function Value We know that in order to calculate the average of a set of numbers, you must add together all of the parts of the set and divide by the number of numbers in the set. If you had an infinite number of numbers in a set, such as function values from a to b, you would have to use a different formula. The equation for the average function value from a to b is that [latex]f_{avg}=\frac{\int_a^b f(x) dx}{b-a}[/latex]. This is because we know that in a Riemann sum, [latex]\Delta x = \frac{b-a}{n}[/latex], and when n is infinity, we represent it by dx (an infinitesimal). Dividing by b-a in the denominator gets rid of the b-a in the dx, and so, it is the infinite sum of the infinitesimals of [latex]\frac{f(x)}{\infty}[/latex]. For example, let's try [latex]\frac{\int_1^5 x^2 dx}{4}[/latex]. The denominator is 4, since the difference between the top and bottom limit is 4, and because the antiderivative is [latex]\frac{x^3}{3}[/latex], we find the difference between this at x = 5 and x = 1, then divide by 4. Working this out, we find that [latex]f_{avg} = 10 + 1/3[/latex]. Next, you will learn about calculating arc length. Remember, if you need any help with this, you can ask our calculus forum.

What exactly do you mean by a god?

I agree. IQ tests aren't always accurate. The score can vary from day to day, depending on a large amount of factors. Also, people's IQ's can change over time as a result of having better or worse cognitive abilities, and that makes IQ less accurate for the future. IQ scores are simply an estimate of a person's intelligence.

Oh yes, thanks for reminding me. I will make sure to mention that when we get to the applications of integral calculus. I have decided to skip the practice problems. Upon request, I will provide them, but I have decided to skip them for now. Part 7. Integration By Parts Integration by parts is like the product rule for integrals. Let's define f(x) as being [latex]f(x) = uv[/latex], where u and v are both functions of x. We know that the derivative of this function will be [latex]f'(x) = (uv)' = u'v + v'u[/latex]. Because of this, we can integrate u'v + v'u to get the original function, i.e [latex]\int {u'v + v'u} dx = uv[/latex]. Integrals are distributive, so we can write the previous equation as [latex]uv = \int u'v dx + \int v'u dx[/latex]. In order to integrate uv', you have to subtract the integration of vu' from the equation, and this would have to be from both sides of the equation. We find that [latex]\int uv' dx = uv - \int vu' dx[/latex]. Using the fact that [latex]u' dx = du[/latex] and [latex] v' dx = dv[/latex], we obtain the equation for integration by parts: [latex]\int u dv = uv - \int v du[/latex] For example, let's integrate [latex]2x ln(x)[/latex]. We can set the value of dv to 2x dx, and therefore, [latex]v = x^2[/latex]. From here, you can see that [latex]u = ln(x)[/latex], since that is the only other part of the function. We now need to plug the value of u and v into the equation. Because u = ln(x) and v = x2, we find that [latex]\int 2x ln(x) dx = x^2 ln(x) - \frac{x^2}{2}[/latex], since du = 1/x dx and [latex]\frac{x^2}{x}=x[/latex]. Because of that, the integral turns into [latex]\int x dx[/latex], which turns out to be x2/2.

The laws of physics are not made by us. It is just that we discover them using scientific evidence and logic.

We don't exactly know, but we do know that the total amount of energy and mass in the universe is conserved.

Because I am unable to edit the post now, I will add a short part about the constant of integration: Part 4.5:The constant of integration When you take antiderivatives, you need to add something called a constant of integration. During differentiation, you get rid of any constant terms, so when you antidifferentiate, you must add something called a constant of integration to include any constant terms lost when differentiating. This is only for indefinite integrals. If you have a definite integral, you do not need to include the constant of integration, since when you work out [latex]F(b) - F(a)[/latex], F(b) and F(a) have the same constant of integration. Because of this, the two constants of integration would cancel out when you are subtracting, and this means there is no constant of integration in definite integrals.

The post has been edited. For some reason, the negative didn't show up. Also, I explained the constant of integration and why there should be one in the part about antiderivatives.

Ok, so let's continue. Part 3: Integral properties Before we move on, you must know the properties of integrals. [latex]\int_a^b f(x) + g(x) dx = \int_a^b f(x) dx + \int_a^b g(x) dx[/latex] [latex]\int_b^a f(x) dx = -\int_a^b f(x) dx[/latex] [latex]\int_a^b cf(x) dx = c\int_a^b f(x) dx[/latex] where c is a constant [latex]\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx[/latex] where c is a number. Part 4: Antiderivatives Antiderivatives are the opposite of derivatives. It's like being given [latex]f'(x)[/latex] and trying to figure out [latex]f(x)[/latex]. Usually, they are written as [latex]\int f(x) dx[/latex], and this type of integral is called an improper integral. They are also written as [latex]F(x)[/latex]. You find antiderivatives by figuring out which function's derivative would be your original function, and that function would be the antiderivative. For example, if we wanted to find the antiderivative of [latex]sin(x)[/latex], we know that the derivative of [latex]cos(x)[/latex] is [latex]-sin(x)[/latex], so therefore, the derivative of [latex]-cos(x)[/latex] is [latex]sin(x)[/latex]. This means that the antiderivative of [latex]sin(x)[/latex] is cos(x). We can do this with much more functions. The antiderivatives of simple functions will be in the next section, and you will need these for more complex integration. This method of antidifferentiating is how mathematicians found these antiderivatives. Part 5: Simple Antiderivatives Here will be a list of simple antiderivatives. You will need to know these before we learn more complex integration. [latex]\int x^n dx = \frac {x^{n+1}}{n+1}[/latex] (except for n = -1) [latex]\int e^x dx = e^x [/latex] [latex]\int a^x dx = \frac{a^x}{ln(a)}[/latex] [latex]\int ln(x) = xln(x) - x[/latex] [latex]\int \frac{1}{x} dx = ln|x| [/latex] [latex]\int sin(x) dx = -cos(x) [/latex] [latex]\int cos(x) dx = sin(x)[/latex] [latex]\int tan(x) dx = ln|sec(x)|[/latex] [latex]\int sec(x) dx = ln|sec(x) + tan(x)| [/latex] Part 6. U subsitution U substitution is a useful integration technique for complex integrals. For example, say you wanted to find [latex]\int sin(3x + 8) dx[/latex]. You could easily solve his problem with u substitution. You would need to set a value to u, and then integrate with respect to u. In this example, we would set [latex] u = 3x + 8[/latex], making this [latex]\int sin(u) dx[/latex]. What we need to do now is make this integral with respect to u. If you imagine [latex]\frac{du}{dx}[/latex] as being the ratio between 2 infinitesimals, you can multiply this by dx to get du. In this case, since [latex]u = 3x + 8[/latex], this means [latex]\frac{du}{dx} = 3[/latex] , and [latex]du=3dx[/latex]. Because we need to take the integral with respect to u, we change [latex]dx[/latex] to [latex]\frac{1}{3} du[/latex], since [latex]\frac{1}{3} du = dx[/latex]. We bring the 1/3 out front, and this becomes [latex]\frac{1}{3} \int sin(u) du[/latex]. This gives you [latex]-\frac{1}{3} cos(u) + C[/latex]. We now have to plug in the value of u, and because [latex] u = 3x + 8[/latex], the integral works out to be [latex]-\frac{1}{3}cos(3x+8)+C[/latex]. Don't forget the constant of integration, since you always need to add that. The next lesson will be on u substitution practice. Remember, if you need help with any of this, you can just ask our calculus forum.

Your welcome. Do you also want a tutorial for the rest of integral calculus?

Here will be a tutorial on integral calculus. Part 1. Riemann Sums An integral is the area under a curve. You can approximate an integral by adding together the area of a number of rectangles, each with a height of [latex]f(x)[/latex] and a length of [latex]\Delta x[/latex]. They each have an area that is equal to [latex]f(x)\Delta x[/latex]. You take this sum from [latex]x = a[/latex] to [latex]x = b[/latex], and since you are summing n rectangles, [latex]\Delta x[/latex] will be the length from a to b, or a - b divided by the number of rectangles. This is written as [latex]\Delta x = \frac{a-b}{n}[/latex] The sum of the areas of each rectangle will give you something close to the value of the integral. This is called a Riemann Sum. It is useful in approximations of integrals. As the rectangles have smaller and smaller lengths, the sum becomes more and more accurate, and you get a closer and closer approximation. This means that you have to use a limit. Therefore, the value of an integral is [latex]\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\Delta x[/latex] You can imagine a function as being made up of an infinite number of these rectangles, each with an infinitesimal length of dx and height of f(x). An integral is written as [latex]\int_a^b f(x) dx[/latex] which represents the infinite sum from a to b of the areas of each infinitesimal area. Part 2. The Second Part of The Fundamental Theorem of Calculus There is another formula for the integral, which was worked out by Issac Newton. Issac Newton Found out that integrating is the inverse function of differentiating. Because the area under a curve is evaluated from a to b, this means that the exact value of the integral is [latex]\int_a^b f(x) dx = F(b) - F(a)[/latex] Where F(b) and F(a) are the antiderivatives of the function f(x) with respect to a and b. This lets you know the exact area under a curve without working out a Riemann Sum. That is the introduction to integration. In the next tutorial, you will learn about integration techniques.

What is it exactly that you are trying to say?