Integration Question (2 Viewers)

[frazzle777]

How do I integrate (4-x^2)^0.5 or is there a formula that I can use?

 

[_ShiFTy_]

You would need to use a trig substition

In this case, let x = 2sinu

 

[Mountain.Dew]

in the exam, you will be given a Table of Standard Integrals, which covers integrating that specific polynomial function ==> eventually, your answer will involve logs

 

[Riviet]

In the last case, if the limits don't involve the edge of the semi-circle or centre of it, you will definitely need to use the trig substitution, because the area is irregular.

 

[Shady01]

isnt it just -1(4-x^2)^-.5 ????
becuase u bring down that index and multiply it by the derivative in tha brackets? Im not sure if ive read the ques rite.. but cant u do that?

 

[Mikearun]

IS the answer -4/3.(4-^2)^(3/2) ?

 

[Riviet]

You may be confusing yourself with this integral:

integral of (ax+b)ⁿ=(ax+b)ⁿ⁺¹/[a(n+1)]

This only works when the inside function in brackets is linear or of degree 1.

 

[frazzle777]

I see. I've never done substitution using trig, only substitution involving u=(whatever you want).

 

[frazzle777]

Ok, normally with substitution when x is replaced with u, dx has to be changed to du but I'm not quite sure how to do this with this particular equation. After using _ShiFTy_'s idea the equation now looks like this:

(4-4(sin^2)u)^0.5

 

[insert-username]

This integration uses a second form of integration by substitution. It goes:

∫f(x) dx = ∫f(x) dx/d@.d@, where @ is usually an angle. Like in normal integration by substitution, the two d@'s cancel out and leave you with the original thing, except in a form which you can integrate.

So:

f(x) = (4 - x²)^1/2

Let x = 2sin@

Hence, @ = sin^-1x/2 and dx/d@ = 2cos@

So f(x) = (4 - 4sin²@)^1/2

= [4(1 - sin²@]^1/2

= 2(1 - sin²@)^1/2 -take 4 out from under the square root

= 2(cos²@)^1/2

= 2cos@ -take the square root of cos²@

Using the rule:

∫(4 - x²)^1/2 dx = ∫2cos@.2cos@.d@

= ∫4cos²@ d@

= 2∫2cos²@ d@

= 2∫(1 + cos2@) d@

= 2@ + sin2@ + c

Then, for definite integrals, you can transform the limits and integrate. Note that in 3 unit, you will be TOLD what substitution to use. However, you will need to execute the steps yourself. Just remember:

∫f(x) dx = ∫f(x) dx/d@.d@


I_F

 

[_ShiFTy_]

= 2∫(1 + cos2@) d@

= 2@ + 1/4sin2@ + c

Just a minor error
The end result should be
= 2@ + sin2@ + c

 

[frazzle777]

ahhhh... now i see what i was doing wrong. After putting in the values for the definite integral it worked out nicely. Thanks to everyone who helped me understand the question and also taught me how to substitute with trig functions. hehe Something i better learn sometime.

 

[Mikearun]

Oh i see now !

 

[insert-username]

Just a minor error
The end result should be
= 2@ + sin2@ + c

Yeah, sorry. Fitzpatrick has that example and they did it wrong - I just checked with my teacher. Silly me.


I_F