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Can someone put up a tutorial on Integral Calculus?


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#21 Endercreeper01

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Posted 7 January 2014 - 10:19 PM

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.


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#22 Endercreeper01

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Posted 9 January 2014 - 01:22 AM

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 \int_a^b (f(x) - g(x)) dx, 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 ex and x2 from 3 to 7. First, we have to determine the integral. The integral becomes \int_3^7 (e^x - x^2) dx. We find that this will be e^x - \frac{1}{3}x^3 at x = 7 minus this value at x = 3. This is because the antiderivative (not including the constant of integration) of ex is ex, and this minus the antiderivative of x2 will be e^x - \frac{1}{3}x^3. 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.


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#23 Endercreeper01

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Posted 10 January 2014 - 12:03 AM

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 dV=A(x) dx = A(y) dy, 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 V=\int_a^b A(x) dx=\int_c^d A(y) dy, 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 A=\pi r^2. 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 A(x)=\pi f^2(x) and A(y) = \pi f^2(y), 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, A=\pi (f^2(x) - g^2(x), 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:

V = \pi \int_a^b f^2(x) - g^2(x) dx = \pi \int_c^d f^2(y) - g^2(y) dy

For example, we will work out the volume from 2 to 9 of f(x)=e^{5x+7} and g(x) = 0. In this case, the square of this function will be e^{10x+14}. Using u substitution, we find that the volume is \frac{\pi}{10}e^{10x+14} evaluated from x = 9 to x = 2. Our final answer for the volume is that

V = 1.2204033 x 1039 units3

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 by Endercreeper01, 10 January 2014 - 12:05 AM.

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#24 Endercreeper01

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Posted 10 January 2014 - 03:45 PM

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 dA among the length of the solid, written as A=\int_a^b dA. The infinitesimal dA is the circumference times ds, i.e dA=Cds where C is the circumference and s is length. The circumference is 2\pi y, making A_s=2 \pi \int_a^b y ds when we plug in the value of C. Since ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx, we plug in the value of ds, giving us the equation for surface area of a solid of revolution:

A_s=2\pi \int_a^b y\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx

If it is rotated about the y axis, it will be:

A_s=\int_a^b x\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy

 

Our next tutorial will be on double integrals.

Remember, if you need any help with this, you can just ask our calculus forum.


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#25 Endercreeper01

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Posted 10 January 2014 - 07:56 PM

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:

V=\int_a^b \int_c^d f(x, y) dx dy

If you are finding the volume between 2 areas, you would use the equation:

V=\int_a^b \int_c^d f(x, y) - g(x, y) dx dy

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

\int_1^3 \int_5^9 e^{xy} + sin(7x+2y) - 4xy dx dy

The antiderivative of this function (not including C) is \frac{e^{xy}}{xy} - x^2 y^2 - \frac{cos(7x+2y)}{14xy}.

Substituting the values of the limits and evaluating gives us our answer:

V = 19705489659     
This can be extended to n dimensions. Generally, the n volume over an n dimensional region is:

V_n = \int_a^b \int_c^d \int_e^f ... f(x, y, z... x_n) dx dy dz... dx_n

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. 

 


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#26 Endercreeper01

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Posted 11 January 2014 - 07:57 PM

This is the last tutorial on integral calculus. If I forgot to do a tutorial on anything in integral calculus, remind me, but if not, this will be the final integral calculus tutorial. I will also provide practice problems upon request.

If anyone wants more tutorials, I will make them. I have lots of time for tutorials, so time will not be a problem.

 

Part 15: Surface area/ Surface volume of functions in n dimensions

 

We will start off in 3 dimensions just for simplicity.

To find the surface area, we have to integrate each infinitesimal area over the entire region, i.e A=\int_a^b dA. This is equal to the double integral over the infinitesimal length \int_a^b \int_c^d ds because \int_c^d ds = dA.

In 3 dimensions, ds is given by

ds = \sqrt{dx^2+dy^2+dz^2}

which is equivalent to

\sqrt {1+\left(\frac{dz}{dx}\right)^2+\left(\frac{dz}{dy}\right)^2} dxdy 

Inserting the value of ds gives us the equation for the surface area of a function z=f(x, y):

A=\int_a^b \int_c^d \sqrt{1+\left(\frac{dz}{dx}\right)^2+\left(\frac{dz}{dy}\right)^2}dxdy

This can be generalized to n dimensions by saying

S=\int_a^b \int_c^d \int_e^f... \sqrt{1+\left(\frac{df}{dx}\right)^2+\left(\frac{df}{dy}\right)^2+\left(\frac{df}{dz}\right)^2...\left(\frac{df}{dx_n}\right)^2} dxdydz...dx_n

Where S is the n surface area/volume of the function in n-1 dimensions.

For example, we will work this integral out when a = 6, b = 12,  c = 4, and d = 7 of the function f(x, y)=\sqrt x + \sqrt y

The square of the partial derivative of f with respect to x or y will be x when taken with respect to x and y when taken with respect to y. This now becomes

\int_6^{12} \int_4^7 \sqrt{1+x+y} dx dy

The second antiderivative is

\frac{4(x+y+1)^{\frac{5}{2}}}{15}

and evaluating this expression for the surface area gives us

A = 370.011408631

That concludes the tutorial on integral calculus. 

Remember, if you need any help with this, you can just ask our calculus forum.

 


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#27 Endercreeper01

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Posted 13 January 2014 - 01:50 AM

The actual final tutorial will be on line integrals. 

Part 16. Line Integrals

With line integrals, x and y are coordinates on a curve C. First, we must parametrize the curve, i.e x=g(t) and y=h(t). The parametrized curve will be written as a vector function r(t)=g(t)i + h(t)j. It is also assumed to be smooth. When a curve is smooth, r(t) is continuous and never 0 for any value of t.

When you have a function f(x, y) along a curve C, the line integral is

\int_C f(x, y) \ ds = \oint f(x, y) \ ds

The ds means that the integral is with respect to the arc length, not x or y.

In parametric form,

ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}dt

Rewriting the equation in parametric form gives us:

\oint f(x, y) \ ds = \int_a^b f(g(t), h(t)) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}dt

Because

\sqrt{\left(\frac{dx}{dt}\right) ^2 + \left(\frac{dy}{dt}\right) ^2}=\left|\left|\frac{dr}{dt}\right|\right|

we obtain the equation for the line integral:

\oint f(x, y) \ ds = \int_a^b f(g(t), h(t))\left|\left|\frac{dr}{dt}\right|\right|dt

For example, we will work out the line integral from 2 to 6 of f(x, y) = x2 + y2, where x = cos(t) and y = sin(t). For this integral,

\left|\left|\frac{dr}{dt}\right|\right|=\sqrt{1-2sin^2(t)}

and

f(g(t), h(t))=sin^2(t)+cos^2(t)=1

The integral now becomes

\int_2^6 \sqrt{1-sin^2(t)} \ dt

The antiderivative of this function is 

\frac{tan(t)}{\sqrt{tan^2(t)+1}}
Evaluating the antiderivative gives us

\int_2^6 \sqrt{1-sin^2(t)} \ dt=0.63

We can also find line integrals of piecewise smooth curves. A piecewise smooth curve is a smooth curve made up of parts.

The line integral for a piecewise smooth curve is:

\oint f(x, y) \ ds = \sum \int_C f(x, y) \left|\left|\frac{dr}{dt}\right|\right| dt

i.e it is the sum of it's parts.

 Part 17. Line Integrals with Respect to x and y

Line integrals can also be taken with respect to x or y.

Because 

dx=\frac{dx}{dt} dt

and 

dy=\frac{dy}{dt}dt

We can definite the line integral with respect to x or y as

\oint f(x, y) dx = \int_a^b f(x(t), y(t))\frac{dx][dt}dt

and

\oint f(x, y) dy=\int_c^d f(x(t), y(t)) \frac{dy}{dt}dt

For example, we will work out the integral from 6 to 9 of f(x, y) = x^2 + y^2 where x=sin(t) and y=cos(t).

f(x, y) becomes 1, leaving us with 

\int_6^9 cos(t) dt

The antiderivative of the function is -sin(t).

Evaluating this gives us

\int_6^9 cos(t) dt=-0.69

That concludes the tutorial on integral calculus.

What will happen to these tutorials?


Edited by Endercreeper01, 13 January 2014 - 01:53 AM.

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#28 hypervalent_iodine

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Posted 15 January 2014 - 02:57 AM

Just FYI: I've pinned some old tutorials related to this subject in this sub-forum. They may be found here:

 

http://www.sciencefo...differentation/

 

http://www.sciencefo...rst-principles/

 

Edit: actually, I decided to move them to the tutorials forum: http://www.sciencefo...tics-tutorials/


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#29 rahul_26

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Posted 24 September 2016 - 06:08 AM

Try Actucation and Khan academy


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