Integrate ln(cube root (2+y^3)) dydx

In summary, the given problem is to integrate ln(cube root (2+y^3)) dydx with limits of y =sqrt (x) to 1 and x from 1 to 0. By changing the order of integration, the problem can be simplified to -\int_{y= 0}^1\int_{x= 0}^{y^2} ln(\sqrt[3]{2+ y^3})dxdy. Integrating with respect to x first and using a substitution will make the problem more manageable.
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Homework Statement



integrate ln(cube root (2+y^3)) dydx with limits
of y =sqrt (x) to 1
of x from 1 to 0


Homework Equations



integral of ln x= x ln x.

This ends up being very convoluted, haven't even begun to insert limits






The Attempt at a Solution

 
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  • #2
So this is
[tex]\int_{x= 1}^0\int_{y= \sqrt{x}}^1 ln(\sqrt[3]{2+ y^3})dydx= -\int_{x= 1}^0\int_{y= \sqrt{x}}^1 ln(\sqrt[3]{2+ y^3})dydx[/tex]

If you change the order of integration, that becomes
[tex]-\int_{y= 0}^1\int_{x= 0}^{y^2} ln(\sqrt[3]{2+ y^3})dxdy[/tex]

Integrate with respect to x first and you should see a simple substitution.

(Writing [itex]ln(\sqrt[3]{2+ y^3})= (1/3)ln(2+ y^3)[/itex] will also simplify.)
 

Related to Integrate ln(cube root (2+y^3)) dydx

1. What is the purpose of integrating ln(cube root (2+y^3)) dydx?

The purpose of integrating ln(cube root (2+y^3)) dydx is to find the antiderivative or the indefinite integral of the given function. This helps in solving various mathematical problems related to finding the area under the curve or calculating the total change in a quantity over a given interval.

2. How do you approach solving an integral involving ln(cube root (2+y^3)) dydx?

To solve an integral involving ln(cube root (2+y^3)) dydx, we first use the appropriate substitution to simplify the integrand. In this case, we can use the substitution u = 2+y^3 to rewrite the integral as ln(cube root u) dy. Then, we can use integration by parts or other integration techniques to solve for the antiderivative.

3. Can the integral of ln(cube root (2+y^3)) dydx be evaluated exactly?

No, the integral of ln(cube root (2+y^3)) dydx cannot be evaluated exactly. It is a non-elementary integral, which means it cannot be expressed in terms of elementary functions such as polynomials, exponential and trigonometric functions. Hence, we can only approximate its value using numerical methods.

4. What is the domain of ln(cube root (2+y^3)) dydx?

The domain of ln(cube root (2+y^3)) dydx is all real numbers excluding the values of y that make the argument of the natural logarithm negative or zero. In this case, the argument 2+y^3 must be greater than zero, which gives the domain as y > -2^(1/3).

5. How is the integral of ln(cube root (2+y^3)) dydx used in real-life applications?

The integral of ln(cube root (2+y^3)) dydx has various applications in physics, engineering, and economics. For example, it can be used to calculate the work done in a thermodynamic process or the total cost of production in economics. It is also used in solving differential equations and other mathematical models in various fields.

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