Finding the Area of the Graph

[lookagain]

Of course I know the answer is 12 regardless of the method used.

When you stated your message in post # 18, one part of me thought you might be
facetious and trying to humor nasi112.

 

[lookagain]

#2: \( \displaystyle 2\int_0^{\frac{\pi }{4}} {\left( {\cos (x) - \sin (x)} \right)dx} \) why?

nasi112, the 2 in front of the integral is a mistake. If I had to guess, pka has
already known the correct way to set up the integral every day continuously
for the past several decades, but a distracting thought lasting a fraction of
a second caused him to type that "2" that he never had an intention of writing.

 

[nasi112]

That's a strange question! Why would you think the area of a region depends upon what method you use to find it?
Or, for that matter whether anyone finds that area at all!

This is a good question. If I can find the area without Calculus, let us find all areas without Calculus. Why the headache using Calculus? Another point is that, Jomo has never shown a solution for an integral, so when he talks about areas, someone has to doubt his thoughts.

You are referring to nasi112, correct?

The area bounded by y = 5, y = 2, x = 4, x = 8

"Top curve" minus the "bottom curve": y = 5 - 2 = 3

The integrand is 3. The antiderivative is 3x.

You're integrating from x = 4 to x = 8.

3(8) - 3(4) = 24 - 12 = 12, which is also the method done graphically.

Very nice. I wished if Jomo has done it.

Of course I know the answer is 12 regardless of the method used.

Is this always the case? You cannot prove what you say. You need someone else to prove it. I knew this would happen.

 

[HallsofIvy]

Of course I know the answer is 12 regardless of the method used.

So mathematics is not flawed?

 

[Deleted member 4993]

This is a good question. If I can find the area without Calculus, let us find all areas without Calculus. Why the headache using Calculus? Another point is that, Jomo has never shown a solution for an integral, so when he talks about areas, someone has to doubt his thoughts.




Very nice. I wished if Jomo has done it.




Is this always the case? You cannot prove what you say. You need someone else to prove it. I knew this would happen.

You said

It is a question for you Khan if you wanna answer it. Because I have a doubt that those are equal to each other​

You need to review first year introductory Calculus.

I strongly suggest you do that before you try to be sarcastic with senior tutors with multiple advanced degrees in theoretical and/or applied mathematics and/or physical sciences. Be polite at least to hide your ignorance!!

 

[Steven G]

So mathematics is not flawed?

Well there is Banach Tarski theorem. I still have a problem with this but then again I do not see a problem with the proof.

 

[Steven G]

This is a good question. If I can find the area without Calculus, let us find all areas without Calculus. Why the headache using Calculus?

Areas for rectangles, triangles, circles, trapezoids, etc can be found without integrals. Come on you knew this! You can, if you choose to, use integrals to find the area of these shapes. You are correct, why bother. However, there are other shapes that can not be found by simple formulas and must be done by using integration.