Area under y=x^2: Calculate the Antiderivative

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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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