Abhi_KS 1 Posted December 28, 2013 Share Posted December 28, 2013 Can someone put up a tutorial on Integral Calculus just like there is one for differentiation? I need it. Link to post Share on other sites

EdEarl 677 Posted December 28, 2013 Share Posted December 28, 2013 See https://www.khanacademy.org/ 2 Link to post Share on other sites

Abhi_KS 1 Posted December 28, 2013 Author Share Posted December 28, 2013 Thanks, but I need more text oriented and Khan Academy teach in videos, which take eons to load on my slow internet. Link to post Share on other sites

John 165 Posted December 28, 2013 Share Posted December 28, 2013 (edited) One good resource is Paul's Online Math Notes. The last couple of sections of Calculus I introduce integration, substitution (one of several techniques of integration), and applications of the integral. Calculus II then continues with more integration techniques and applications before getting into other calculus topics. There are also notes for Calculus III (which is multivariable calculus, though some of that material is introduced at the end of Calculus II) and Differential Equations. PDFs of the notes are available if you prefer those over the website, and Calculus I and II (but not III or DE) also come with practice problems and solutions.There's also William Chen's lecture notes, which cover quite a few different mathematical subjects, including calculus. The notes include exercises, but I don't think worked solutions are available. As for tutorials here, there has been some talk (though not really recently, that I know of) of adding more tutorials, but with the huge number of resources already available online for pretty much any math topic, adding more tutorials here probably isn't a priority. Edited December 28, 2013 by John Link to post Share on other sites

Endercreeper01 47 Posted December 29, 2013 Share Posted December 29, 2013 Do you want someone on science forums to explain it to you? Or do you want a website? Link to post Share on other sites

Abhi_KS 1 Posted December 29, 2013 Author Share Posted December 29, 2013 Well, I would prefer it here, and more if it was just like the previous one. Link to post Share on other sites

Endercreeper01 47 Posted December 30, 2013 Share Posted December 30, 2013 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. 3 Link to post Share on other sites

Abhi_KS 1 Posted December 30, 2013 Author Share Posted December 30, 2013 @Endercreeper01 : That was what I wanted. Thanks! Link to post Share on other sites

Endercreeper01 47 Posted December 30, 2013 Share Posted December 30, 2013 (edited) Your welcome. Do you also want a tutorial for the rest of integral calculus? Edited December 30, 2013 by Endercreeper01 Link to post Share on other sites

Abhi_KS 1 Posted December 31, 2013 Author Share Posted December 31, 2013 (edited) Yes, why not? This was exactly in the format and language which I wanted. Edited December 31, 2013 by Abhi_KS Link to post Share on other sites

Endercreeper01 47 Posted December 31, 2013 Share Posted December 31, 2013 (edited) 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. Edited December 31, 2013 by Endercreeper01 Link to post Share on other sites

imatfaal 2481 Posted December 31, 2013 Share Posted December 31, 2013 (edited) Ok, so let's continue....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)[/latex].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. surely the integral works out as -1/3 cos(3x+8) . Also you need to add a constant - or explain why you should add one Edited December 31, 2013 by imatfaal horrid horrid editor 2 Link to post Share on other sites

Endercreeper01 47 Posted January 1, 2014 Share Posted January 1, 2014 surely the integral works out as -1/3 cos(3x+8) . Also you need to add a constant - or explain why you should add one 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. 1 Link to post Share on other sites

John 165 Posted January 1, 2014 Share Posted January 1, 2014 (edited) I don't see it. Also, another slight edit: In the antiderivatives section, you refer to the antiderivative as an "improper" integral. This should be "indefinite" integral. I'm assuming you'll discuss improper integrals later. Edited January 1, 2014 by John 1 Link to post Share on other sites

Endercreeper01 47 Posted January 1, 2014 Share Posted January 1, 2014 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. 1 Link to post Share on other sites

studiot 2240 Posted January 1, 2014 Share Posted January 1, 2014 (edited) 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 , 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. There is more to it than this. Others have already said that you should be comparing the indefinite integral and the definite integral. The anti-derivative is something else again. The definite integral is a pure number it is not a function. The indefinite integral is a function. I suggest accepting these for now until the significance of this distinction really becomes apparent when you proceed to the applications of calculus. And well done (+1) to endercreeper for being prepared to put up the work. Edited January 1, 2014 by studiot 2 Link to post Share on other sites

Endercreeper01 47 Posted January 1, 2014 Share Posted January 1, 2014 (edited) 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 = x^{2}, 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 x^{2}/2. Edited January 1, 2014 by Endercreeper01 2 Link to post Share on other sites

Abhi_KS 1 Posted January 3, 2014 Author Share Posted January 3, 2014 Endercreeper01 rulez! I think now this thread can be counted in Tutorial Section. 1 Link to post Share on other sites

Endercreeper01 47 Posted January 3, 2014 Share Posted January 3, 2014 (edited) 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. Edited January 3, 2014 by Endercreeper01 1 Link to post Share on other sites

Endercreeper01 47 Posted January 5, 2014 Share Posted January 5, 2014 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. 1 Link to post Share on other sites

Endercreeper01 47 Posted January 7, 2014 Share Posted January 7, 2014 Also, I might not post tutorials as often because of school, as school gives me less time to post tutorials. I will still post tutorials, just not as often. Link to post Share on other sites

Endercreeper01 47 Posted January 9, 2014 Share Posted January 9, 2014 I will now be able to continue posting tutorials. I now have more free time then I did before, so now I will continue to make tutorials. Part 11. Area Between Two Curves Today, you will learn about the area between two curves. If you had 2 functions f(x) and g(x), you can determine the area between them by finding the difference between the top function area and the bottom function area, i.e [latex]\int_a^b (f(x) - g(x)) dx[/latex], since the integral distributes to both f(x) and g(x). Because it distributes, it is the same as the integral over the top function minus the integral over the integral over the bottom function. For example, we will work out the area between e^{x} and x^{2} from 3 to 7. First, we have to determine the integral. The integral becomes [latex]\int_3^7 (e^x - x^2) dx[/latex]. We find that this will be [latex]e^x - \frac{1}{3}x^3[/latex] at x = 7 minus this value at x = 3. This is because the antiderivative (not including the constant of integration) of e^{x} is e^{x}, and this minus the antiderivative of x^{2} will be [latex]e^x - \frac{1}{3}x^3[/latex]. Working this out, we find that this is about 971 square units. In the next tutorial, you will learn about the volumes of solids of revolutions. Remember, if you need any help with this, you can just ask our calculus forum. 1 Link to post Share on other sites

Endercreeper01 47 Posted January 10, 2014 Share Posted January 10, 2014 (edited) That concludes the 2 dimensional part of our tutorial. We will move on to 3 dimensions. Part 12. Volumes of Solids of Revolution A solid of revolution is simply a function wrapped around an axis. In order to get the volume, we need to sum all of the infinitesimal volumes from x = a to x = b. Each infinitesimal volume is represented by [latex]dV=A(x) dx = A(y) dy[/latex], where A is area and V is volume, and it can be taken with respect to dx or dy. We will have to do an integral, which can be written as [latex]V=\int_a^b A(x) dx=\int_c^d A(y) dy[/latex], where a and b are the limits for the x axis, and c and d are the limits for the y axis. Because each cross-sectional area is circular, we can define the cross sectional areas as being [latex]A=\pi r^2[/latex]. The radius would be f(x) if you are finding A(x) and f(y) if you are finding A(y), so we can write this as [latex]A(x)=\pi f^2(x)[/latex] and [latex]A(y) = \pi f^2(y)[/latex], depending on which axis the function is wrapped around. The radius is the original function because the function is rotated, and it is rotated among each function value. Therefore, the radius will be the original function. This is only if the solid of revolution is not the result of wrapping the difference in 2 functions around an axis. It would still be considered a solid of revolution if you have f(x) (or f(y)) and g(x) (or g(y)) and you wrap the change between them around an axis. In that case, [latex]A=\pi (f^2(x) - g^2(x)[/latex], since it would be the difference in the cross sectional areas of f(x) and g(x), and this is the same for f(y). This gives us the equation for volume of a solid of revolution: [latex]V = \pi \int_a^b f^2(x) - g^2(x) dx = \pi \int_c^d f^2(y) - g^2(y) dy[/latex] For example, we will work out the volume from 2 to 9 of [latex]f(x)=e^{5x+7}[/latex] and [latex]g(x) = 0[/latex]. In this case, the square of this function will be [latex]e^{10x+14}[/latex]. Using u substitution, we find that the volume is [latex]\frac{\pi}{10}e^{10x+14}[/latex] evaluated from x = 9 to x = 2. Our final answer for the volume is that V = 1.2204033 x 10^{39 }units^{3} Our next tutorial will be on the surface area of solids of revolution. Remember, if you need any help with this, you can just ask our calculus forum. Edited January 10, 2014 by Endercreeper01 1 Link to post Share on other sites

Endercreeper01 47 Posted January 10, 2014 Share Posted January 10, 2014 Part 13. Surface Areas of Solids of Revolution To find the surface area of a solid of revolution, we must sum all of the infinitesimal areas [latex]dA[/latex] among the length of the solid, written as [latex]A=\int_a^b dA[/latex]. The infinitesimal dA is the circumference times ds, i.e [latex]dA=Cds[/latex] where C is the circumference and s is length. The circumference is [latex]2\pi y[/latex], making [latex]A_s=2 \pi \int_a^b y ds[/latex] when we plug in the value of C. Since [latex]ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx[/latex], we plug in the value of ds, giving us the equation for surface area of a solid of revolution: [latex]A_s=2\pi \int_a^b y\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx[/latex] If it is rotated about the y axis, it will be: [latex]A_s=\int_a^b x\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy[/latex] Our next tutorial will be on double integrals. Remember, if you need any help with this, you can just ask our calculus forum. 1 Link to post Share on other sites

Endercreeper01 47 Posted January 10, 2014 Share Posted January 10, 2014 Part 14. Integrals over n dimensions Let's start with a function z = f(x, y). To find the volume under this region, we need to have 2 limits of this integral, an x limit and a y limit. Because of this, we must integrate over both the x and y regions. Since we integrate over 2 regions, we have to integrate with respect to 2 variables, x and y. This is called a double integral. Therefore, the volume under a function f(x, y) is: [latex]V=\int_a^b \int_c^d f(x, y) dx dy[/latex] If you are finding the volume between 2 areas, you would use the equation: [latex]V=\int_a^b \int_c^d f(x, y) - g(x, y) dx dy[/latex] since you would be finding the difference between 2 volumes. In order to calculate definite integrals, we need to first find the antiderivative with respect to all the variables. In this case, there are 2 possible orders it can be done in: You could do it with respect to x first, hen y, or take the antiderivative with respect to y, then x. There are n! possible ways you could find the antiderivative, where n is the number of dimensions you are integrating over. I will take all antiderivatives in multiple dimensions in the order of the variables, just to avoid confusion. After the antiderivative is found, we need to evaluate it at x = d and y = b and then subtract from the antiderivative evaluated at x = c and y = a. For example, we are going to compute [latex]\int_1^3 \int_5^9 e^{xy} + sin(7x+2y) - 4xy dx dy[/latex] The antiderivative of this function (not including C) is [latex]\frac{e^{xy}}{xy} - x^2 y^2 - \frac{cos(7x+2y)}{14xy}[/latex]. Substituting the values of the limits and evaluating gives us our answer: [latex]V = 19705489659[/latex] This can be extended to n dimensions. Generally, the n volume over an n dimensional region is: [latex]V_n = \int_a^b \int_c^d \int_e^f ... f(x, y, z... x_n) dx dy dz... dx_n[/latex] Our next tutorial will be on the surface area of functions in n dimensions. Remember, if you need any help with this, you can just ask our calculus forum. 1 Link to post Share on other sites

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now