# Area under y=x^2

1. Sep 15, 2008

### winston2020

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?

2. Sep 15, 2008

### Feldoh

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)

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Take the derivative of F(x) and you will indeed obtain f(x). All that happened was the derivative in reverse.

3. Sep 15, 2008

### StingerManB

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.

4. Sep 15, 2008

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 $$F(x) = ax^n = 1/3x^3$$.

5. Sep 15, 2008

### winston2020

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 ).

6. Sep 15, 2008