How to Integrate a Polynomial Under a Square Root?

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In summary, the conversation is discussing how to integrate the given function and the use of trigonometric substitution to solve the problem. It is recommended to break the integral into two parts, 3x^2 and (4-x^2)^(1/2), and use trig substitution for the second part. It is also mentioned that the product rule cannot be used for integration and that a certain trigonometric identity needs to be used to eliminate the radical.
  • #1
LinearAlgebra
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How do you integrate this??

Integral of 3x^2 + (4-x^2)^(1/2) dx ??

I tried a u substitution for the 4-x^2 but what do you do with 3x^2? If someone could walk we through this, i'd greatly appreciate it...i hate being stuck on something so trivial in the middle of a problem.
 
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  • #2
Break it into two integrals, 3x^2 and (4-x^2)^(1/2).

Then use maybe x=2sin(u)
 
  • #3
sin? where did sin come into the picture? Is that the only way to do this?
 
  • #4
[tex]\int 3x^2 + \sqrt{4-x^2} \,dx[/tex]

Yes, break into two integrals:

[tex]\int 3x^2 \, dx + \int (4-x^2)^\frac{1}{2} \, dx[/tex]

I think theperthvan is saying the second integral needs trig substitution.
 
Last edited:
  • #5
so is the second part equal to (2/3)(4-x^2)^3/2 ?? you don't have to do anything with the internal part, 4-x^2 like you do with the product rule in differentiation?
 
  • #6
  • #7
LinearAlgebra said:
so is the second part equal to (2/3)(4-x^2)^3/2 ?? you don't have to do anything with the internal part, 4-x^2 like you do with the product rule in differentiation?

No, that will most definitely not work. Here you have a polynomial under a square root. You need to get rid of the square root. Here's what you do.

For [tex]\int \sqrt{a^{2}-x^{2}} \,dx[/tex], use [tex]x = a \, sin(\theta)[/tex]

Note that you also need to substitute the differential, [tex]dx = a \, cos(\theta)d\theta[/tex]

I hope I'm not saying too much, but also remember to use a certain trigonometric identity to eliminate the radical.

Having said this, if you don't know what a trigonometric substitution is, then my guess is that your calculus class hasn't yet covered it. Do you need to compute an antiderivative, or are you trying to compute a definite integral? Because if you're doing the definite integral from 0 to 2, you can do this simply by using the formula for the area of a circle.
 

1. How do you integrate this function?

Integrating a function involves finding the antiderivative of the function. This can be done using various integration techniques such as substitution, integration by parts, or trigonometric substitution.

2. What is the purpose of integration?

The purpose of integration is to find the total accumulation or area under a curve. It is also used to solve various problems in physics, engineering, and other fields.

3. Can integration be done for all functions?

No, integration cannot be done for all functions. Some functions are non-integrable or do not have an antiderivative that can be expressed in terms of elementary functions. In such cases, numerical methods are used to approximate the value of the integral.

4. How do you solve definite integrals?

To solve definite integrals, you need to first evaluate the indefinite integral and then substitute the upper and lower limits of integration into the result. The difference between the two values will give the final result of the definite integral.

5. What are the applications of integration in real life?

Integration has various applications in real life, such as calculating distance traveled by an object with varying velocity, finding the area under a demand and supply curve to determine the equilibrium point in economics, and determining the displacement of a moving object by finding the area under its velocity-time graph.

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