How to determine this integral? Thank you

  • Thread starter Thread starter williamwong0402
  • Start date Start date
  • Tags Tags
    Integral Thank you
williamwong0402
Messages
9
Reaction score
0

Homework Statement


upload_2016-12-12_17-11-48.png


Homework Equations


k∫[ƒ(x)]n ƒ'(x) dx

The Attempt at a Solution


i tried to using algebraic substitution to determine that i had let u = 1-x or X2-2x+1 or x or root(x) but it still cannot solve it.
Please give me hint how to solve it.
Thank you
[/B]
 
Physics news on Phys.org
I'd just expand it as a sum of the three terms, each of which is a half-integer power of x, integrate each separately and then simplify the result sum.
 
Expand ##(1-x)^2=1-2x+x^2## and then treat each term separately.
 
  • Like
Likes williamwong0402
The substitution ##u = \sqrt{x}## should have simplified it. Perhaps you ought to post what you did to see where you went wrong.
 
$$ ∫ \frac{ (1-x)^2 }{ 2 \sqrt{x} } dx$$
$$ =∫ \frac{ (x^2+2x+1) }{ 2 \sqrt{x} } dx$$
$$ Let u = \sqrt{x} $$
$$ dx= \frac{ 2du }{ x^{\frac{ -1 }{ 2 }} } $$
$$ ∫\frac{ (x^2+2x+1) }{ 2u } \frac{ 2du }{ x^{\frac{ -1 }{ 2 }} }$$

and then i have no idea what is next step
thank you very much
 
williamwong0402 said:
$$ ∫ \frac{ (1-x)^2 }{ 2 \sqrt{x} } dx$$
$$ =∫ \frac{ (x^2+2x+1) }{ 2 \sqrt{x} } dx$$
$$ Let u = \sqrt{x} $$
$$ dx= \frac{ 2du }{ x^{\frac{ -1 }{ 2 }} } $$
$$ ∫\frac{ (x^2+2x+1) }{ 2u } \frac{ 2du }{ x^{\frac{ -1 }{ 2 }} }$$

and then i have no idea what is next step
thank you very much

The point of a substitution is to replace one variable with another, not to mix the two. You need to replace all the terms in ##x## with the relevant term in ##u##.
 
williamwong0402 said:
$$ ∫ \frac{ (1-x)^2 }{ 2 \sqrt{x} } dx$$
$$ =∫ \frac{ (x^2+2x+1) }{ 2 \sqrt{x} } dx$$
$$ Let u = \sqrt{x} $$

and then i have no idea what is next step
thank you very much

Using any kind of substitution in this problem is unhelpful; it just makes the problem harder, not easier! Just integrate the three terms separately.
 
  • Like
Likes williamwong0402
thank you ~:smile:
The answer:
$$ \frac{ x \frac{ 5 }{ 2 } }{ 5 } + \frac{ 2x \frac{ 3 }{ 2 } }{ 3 } + \sqrt{x} $$
Am i right ?
 
Ray Vickson said:
Using any kind of substitution in this problem is unhelpful; it just makes the problem harder, not easier! Just integrate the three terms separately.

If the OP doesn't understand how to do substitution, it's as well to get that fixed. He'll need to get to grips with it sooner or later!
 
  • #10
williamwong0402 said:
thank you ~:smile:
The answer:
$$ \frac{ x \frac{ 5 }{ 2 } }{ 5 } + \frac{ 2x \frac{ 3 }{ 2 } }{ 3 } + \sqrt{x} $$
Am i right ?
You lost a minus sign when you expanded the square, and your formatting is a bit scrambled.
 
  • Like
Likes williamwong0402
  • #11
williamwong0402 said:
thank you ~:smile:
The answer:
$$ \frac{ x \frac{ 5 }{ 2 } }{ 5 } + \frac{ 2x \frac{ 3 }{ 2 } }{ 3 } + \sqrt{x} $$
Am i right ?

You should never (or rarely) have to ask whether an indefinite integral answer is correct. You should differentiate your answer and see whether you get the original function back. You have, however, forgotten the constant of integration.
 
  • Like
Likes williamwong0402
  • #12
ok ~ thank you so much
 

Similar threads

Back
Top