How do you integrate this?

  • #1

Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
184
0
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
469
2
[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
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  • #7
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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.
 

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