Antiderivatives & Indefinite Integrals

In summary, the homework statement asks for an integral of the form (x^(1/2)-2*x^(2/3)+1)*x^(-1/4)dx. Differential Calculus (calculus 1) is required for this problem.
  • #1
MitsuShai
159
0

Homework Statement



Compute the following integral:



Homework Equations


none


The Attempt at a Solution


x^(1/2)-2(x^2)^(1/3) +1*x^(-1/4)dx =
(2/3)x^(3/2)-...I don't know this part...+x * (4/3)x^(3/4)
and this is where I stopped...

btw sorry I didn't put the symbols and stuff, I'm slow when it comes to technology...
 
Last edited:
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  • #2
what level is this at?
 
  • #3
Gregg said:
what level is this at?

Differential Calculus (calculus 1)
 
  • #4
is this the right integral?

[tex]\int \frac{x^{\frac{1}{2}}-2\left(x^2+1\right)^{\frac{1}{3}}}{x^{1/4}} \, dx[/tex]
 
  • #5
I don't think that integral is going to be possible without special functions.
 
  • #6
Cyosis said:
I don't think that integral is going to be possible without special functions.

O_O special funcitons?
 
  • #7
Where did you get get this integral from and what is the question? Finding a 'normal' anti-derivative is not going to happen. So if this is a calculus one exercise there is more to it than you're telling us.
 
  • #8
Cyosis said:
Where did you get get this integral from and what is the question? Finding a 'normal' anti-derivative is not going to happen. So if this is a calculus one exercise there is more to it than you're telling us.

The question was "Compute the following integral: "
 
  • #9
I don't see how you will be able to compute that integral at your level.
 
  • #10
I see, while you may think this integral is nearly the same it is in fact much much easier. Just do the division and you get three trivial expressions.
 
  • #11
Cyosis said:
I see, while you may think this integral is nearly the same it is in fact much much easier. Just do the division and you get three trivial expressions.

ok then, what do you think?

∫ [x^(1/2) – 2*x^(2/3) +1] * x^(-1/4) dx

Distribute the x^(-1/4)
∫ x^(-1/8) – 2*x^(-1/6) + x^(-1/4) dx

Solve
(8/7)x^(7/8) – (12/5)x^(5/6) + (4/3)x^(3/4) + C , I entered that in but it's wrong...
 
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  • #12
The method is correct, but the arithmetic is not. For example 1/2-1/4 is not -1/8 and 2/3-1/4 is not -1/6.
 
  • #13
Cyosis said:
The method is correct, but the arithmetic is not. For example 1/2-1/4 is not -1/8 and 2/3-1/4 is not -1/6.


(x^(1/2))(x(-1/4) - (2x^(2/3))(x(-1/4)) + (x(-1/4)) = (x^(1/4)) - (2x^(5/12)+ x^(-1/4)
x^(1/4)-2x(5/12)+x^(-1/4)
(4/5)x^(5/4) - (24/17)x^(17/12) +(4/3)x^(3/4) + c

it's right, thank you :)
 

1. What is an antiderivative?

An antiderivative, also known as the primitive or indefinite integral, is the reverse operation of differentiation. It is a function whose derivative is the original function.

2. How is an antiderivative different from a definite integral?

An antiderivative does not have any specific limits of integration, while a definite integral has both an upper and lower limit. This means that an antiderivative represents a family of functions, while a definite integral represents a single value.

3. Can any function have an antiderivative?

No, not all functions have an antiderivative that can be expressed using elementary functions. Some functions, like e^x or sin(x^2), do not have an antiderivative that can be written in terms of common functions. These functions are called non-elementary functions.

4. How do you find the antiderivative of a function?

To find the antiderivative of a function, you can use the reverse rules of differentiation. For example, if the original function is 2x, the antiderivative would be x^2 + C, where C is a constant. You can also use integration techniques such as substitution or integration by parts.

5. What is the significance of the constant of integration in an antiderivative?

The constant of integration, represented by C, is added to the antiderivative because when a function is differentiated, the constant disappears. This means that there are multiple antiderivatives for a given function, but they all differ by a constant. The constant of integration is necessary to account for all possible antiderivatives.

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