Area under y=x^2: Calculate the Antiderivative

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In summary, the antiderivative of f(x) = x2 is F(x) = 1/3x3. The required area is found using Part 2 of the Fundamental Theorem...
  • #1
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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  • #2
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.
 
  • #3
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
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 [tex]f(x) = ax^n[/tex], then [tex]f'(x) = anx^{n-1}[/tex]. So use this backwards:

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

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

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

Now, you can write [tex]F(x) = ax^n = 1/3x^3[/tex].
 
  • #5
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:).
 
  • #6
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
 

FAQ: Area under y=x^2: Calculate the Antiderivative

What is the meaning of "area under y=x^2"?

The "area under y=x^2" refers to the region bounded by the curve of the function y=x^2 and the x-axis.

What is the purpose of calculating the antiderivative of y=x^2?

The antiderivative of y=x^2 represents the function whose derivative is y=x^2. It is useful in finding the original function when only its derivative is known.

How do you calculate the antiderivative of y=x^2?

The antiderivative of y=x^2 can be calculated by adding 1 to the exponent, dividing by the new exponent, and adding a constant term. In this case, the antiderivative is 1/3 x^3 + C.

Can the area under y=x^2 be negative?

No, the area under y=x^2 cannot be negative since it represents the region bounded by the curve and the x-axis, which cannot have a negative area. However, the value of the antiderivative may be negative for certain values of x.

Why is the antiderivative of y=x^2 a parabola?

The antiderivative of y=x^2 is a parabola because the derivative of a parabola is y=x^2. This means that the original function had a rate of change that was represented by a parabola, and the antiderivative represents the original function.

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