Integrating Inverse Powers: Solving ∫1/∛(x^2) dx

[priscilla98]

Homework Statement

Evaluate the given integral::

∫1/∛(x^2 ) dx

The Attempt at a Solution

∫(x^2)^-1/3 (1) dx

1/2 ∫(x^2)^-1/3 (2) dx


But somehow I'm stuck here. I know this is an easy problem but I just can't figure it out. Any help is appreciated. Thanks a lot

 

[priscilla98]

Please any help is appreciated. Am i right so far?

 

[gb7nash]

Homework Statement

Evaluate the given integral::

∫1/∛(x^2 ) dx

The Attempt at a Solution

∫(x^2)^-1/3 (1) dx

1/2 ∫(x^2)^-1/3 (2) dx


But somehow I'm stuck here. I know this is an easy problem but I just can't figure it out. Any help is appreciated. Thanks a lot

I can't tell what's going on. You might want to use latex. Is the problem:

$$\int \frac{1}{x^{2/3}}dx$$ ?

 

[cepheid]

You're right that the function you have is x^-2/3.

Remember that when differentiating a function of the form xⁿ, you get the result:

nx^n-1

In other words, the procedure for differentiation is

1. multiply the expression by a factor equal to the exponent
2. subtract 1 from the exponent

Therefore, the procedure for integrating a power function is to reverse these steps. In other words do the exact opposite steps in the exact opposite order:

1. Add 1 to the exponent
2. Divide the expression by this new exponent

If you started with xⁿ, you end up with:

[ 1/(n+1) ]xⁿ⁺¹

You know that this procedure works, because if you apply it to your differentiated expression nx^n-1, you get back what you started with:

(n/n)xⁿ = xⁿ

 

[priscilla98]

I can't tell what's going on. You might want to use latex. Is the problem:

$$\int \frac{1}{x^{2/3}}dx$$ ?

I'm sorry. The problem states to evaluate the given integral which is

1 / 3√x^2 dx

 

[priscilla98]

I can't tell what's going on. You might want to use latex. Is the problem:

$$\int \frac{1}{x^{2/3}}dx$$ ?

You're right that the function you have is x^-2/3.

Remember that when differentiating a function of the form xⁿ, you get the result:

nx^n-1

In other words, the procedure for differentiation is

1. multiply the expression by a factor equal to the exponent
2. subtract 1 from the exponent

Therefore, the procedure for integrating a power function is to reverse these steps. In other words do the exact opposite steps in the exact opposite order:

1. Add 1 to the exponent
2. Divide the expression by this new exponent

If you started with xⁿ, you end up with:

[ 1/(n+1) ]xⁿ⁺¹

You know that this procedure works, because if you apply it to your differentiated expression nx^n-1, you get back what you started with:

(n/n)xⁿ = xⁿ

$$\int$$ $$\frac{1}{3\sqrt[]{x}^{2}}$$

This is the integral, thanks a lot

 

[gb7nash]

Is that the third root or 3 times the square root?

 

[priscilla98]

It's the third root.

 

[cepheid]

Is that the third root or 3 times the square root?

You can see from the original post that it is clearly supposed to be the third root. Let's not complicate the issue. It is not necessary for the OP to learn LaTeX just for this thread.

priscilla98:

Try typing out a solution in plain text using the steps I provided. It's all just exponents, multiplying and dividing, so you can just use ^ * and / like you were before. Or you can try using the buttons marked X² and X₂ above the reply text box in order to get superscripts and subscripts.

 

[priscilla98]

It's the third root.

$$\int$$ $$\frac{1}{\3sqrt[]{x}^{2}$$

 

[HallsofIvy]

$$\frac{1}{\sqrt[3]{x^2}}= x^{-2/3}$$

So just integrate using
$$\int x^n dx= \frac{1}{n+1}x^{n+1}+ C$$

 

[I like Serena]

Here's a few definitions and rules that may be useful since I'm not sure you already know these:

$$\sqrt[3]{x} = x^{\frac 1 3}$$

$$\frac 1 x = x^{-1}$$

$$(x^2)^{\frac 1 3}=x^{2 \cdot \frac 1 3}=x^{\frac 2 3}$$

$$\frac 1 {\sqrt[3](x^2)} = \frac 1 {(x^2)^{\frac 1 3}} = \frac 1 {x^{2 \cdot \frac 1 3}} = \frac 1 {x^{\frac 2 3}} = x^{-\frac 2 3}$$