Area under y=x^2: Calculate the Antiderivative

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Discussion Overview

The discussion revolves around finding the antiderivative of the function y=x² and calculating the area under the curve from 0 to 1. It includes explanations of the fundamental theorem of calculus, the power rule for derivatives, and methods for finding antiderivatives.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant asks how the antiderivative of f(x) = x² was obtained, referencing a textbook example.
  • Another participant explains the second part of the fundamental theorem of calculus, stating that if F is an antiderivative of f, then f(x) = F'(x).
  • A different participant presents a formula for finding antiderivatives, suggesting that for a variable x, the antiderivative can be calculated as (1/n+1)x^(n+1) where n is the original power.
  • Another contribution emphasizes the process of thinking of a function whose derivative returns the original function, applying the power rule in reverse to find F(x) = (1/3)x³.
  • One participant expresses understanding of the specific example but seeks a more general method for finding antiderivatives, noting that complexity increases with more complicated functions.
  • A participant lists four methods for finding antiderivatives: substitution, using a table of integration, integration by parts, and a method referred to as "friction," although the spelling is noted as incorrect.

Areas of Agreement / Disagreement

Participants generally agree on the methods for finding antiderivatives and the application of the fundamental theorem of calculus, but there is no consensus on a singular method or approach, especially regarding more complex functions.

Contextual Notes

Some methods mentioned may depend on the specific form of the function being integrated, and the discussion does not resolve which method is universally applicable or preferable in all cases.

winston2020
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This is out of my textbook:

EXAMPLE 6:
Find the area under the parabola y=x2 from 0 to 1.

SOLUTION:
An antiderivative of f(x) = x2 is F(x) = 1/3x3. The required area is found using Part 2 of the Fundamental Theorem...

My question is: how was the antiderivative obtained?
 
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The second part of the fundamental theorem of calculus says, essentially, that if we have a continuous function f over some interval, [a,b]. Then let F be an antiderivative of f.

So it follows that:
f(x) = F'(x)

-----

Take the derivative of F(x) and you will indeed obtain f(x). All that happened was the derivative in reverse.
 
When you have a variable (let's use x) you can use this formula:

(1/n+1)x^n+1.
Read as: "One over n plus one, times the variable, that variable raised to n plus one. "n" is the original power from the problem.
Example: x^4 would be 1/5 x^5. To prove this, you could find the derivative of my solution. (which would be 5*(1/5) x^5-1 = x^4.
Hope this helped a little.
 
Basically, to find an antiderivative, you have to think up a function that, if you take its derivative, you would get your original function back again.

Now, the Power Rule of derivative says that if f(x) = ax^n, then f'(x) = anx^{n-1}. So use this backwards:

You have f(x) = x^2. Your task is to find F(x).

In this case, you the exponent in f(x) is n-1 = 2, so the exponent in F(x) is n = 3.

You also know that in f(x), an = 1, so a = 1/3.

Now, you can write F(x) = ax^n = 1/3x^3.
 
Thanks everyone. I actually do understand how to get this particular example's anitderivative. I wanted to know if there was a more general way of achieving this. I can only imagine once f(x) gets a little more complicated, finding the antiderivative could get quite painful (although finding the derivative can also be quite painful :wink:).
 
there is four way to antiderivative (as i have known ) ...

first one is subtution
second using the table of intgration
third is by part
fourth is friction



excuse my splling
 

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