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Integration by parts

  • Thread starter aleao
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  • #1
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Formula for integration by parts:

[tex]\int f(x)dx = \int u dv = uv - \int v du[/tex]



Use integration by parts to find the following integrals:

a) [tex]\int x e^{1-x} dx [/tex]

b) [tex]\int_1^4 \frac {ln \sqrt x} {\sqrt x} dx[/tex]

c) [tex]\int_{-2}^1 (2x+1)(x+3)^{3/2} dx[/tex]

d) [tex]\int x^3 \sqrt{3x^2+2} dx[/tex]

Answers in back of the book:

a) [tex]-e^{1-x}(x+1)+C[/tex]

b) 4 ln 2 - 2

c) [tex]\frac{74}{4}[/tex]

d) [tex]\frac {1}{9}x^2(3x^2+2)^{\frac {3}{2}}-\frac{2}{135}(3x^2+2)^{\frac {5}{2}} +C[/tex]

My attempts:

a) [tex]\int xe^{(1-x)} dx[/tex]

[tex]u = x[/tex]

[tex]dv = e^{1-x} dx[/tex]

[tex]du = 1 [/tex]

[tex]v = -e^{1-x}[/tex]

which gives:

[tex]-xe^{1-x} - \int -e^{1-x}(1)[/tex]

[tex]-xe^{1-x} - e^{1-x}[/tex]

[tex]-e^{1-x}(x+1) +C[/tex]

GOT IT! Thanks Melawrghk and dx!


b) [tex]\int_1^4 \frac {ln \sqrt x} {\sqrt x} dx[/tex]

[tex]u = ln x^{{1}{2}}[/tex]

[tex]dv = x^{\frac {-1}{2}}[/tex]

[tex]du = \frac { \frac {1}{2}\sqrt x}{\sqrt x} = \frac {1}{2x}[/tex]

[tex]v = 2x^{\frac{1}{2}}[/tex]

which gives:

[tex] ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - \int 2x^{\frac{1}{2}} \frac {1}{2x}[/tex]

[tex] ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - \int \frac{x^{\frac {1}{2}}}{x}[/tex]

[tex] ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - \int x^{\frac {-1}{2}}[/tex]

[tex] ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - 2x^{\frac{1}{2}}[/tex]

[tex] 2x^{\frac{1}{2}}(ln x^{\frac {1}{2}}-1)[/tex]

Plug in 4 and 1:

[tex]4(ln2-1)-2(ln1-1)[/tex]

[tex] = 4ln2-4+2[/tex]

[tex] = 4ln2-2[/tex]

GOT IT! Thanks again dx!!

I'm looking at the other two and I'm blanking.

I've been out of school for three years, and jumped right into business calculus since it's a requirement. So I apologize if this is elementary, but I've been banging my head on this for hours now... and of course the teacher is nowhere to be found. :cry:

Thanks in advance!

EDIT: Got a correct now, thanks to Melawrghk and dx.
 
Last edited:

Answers and Replies

  • #2
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For a), switch your u and dv and try again :)

EDIT: and for b), check your derivative of ln(sqrt(x)). sqrt(x) can be written as x^(1/2), so when you take the derivative you'll get (1/2)x^(3/2). Coupled with the square root of x on the bottom it will cancel nicely.
 
  • #3
dx
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For part (a), v = e1-x would be a better choice.
 
  • #4
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Is [tex] \int e^{1-x} = e^{1-x}[/tex] ?

Edit: I'm only assuming this because

[tex] \int e^{x} = e^{x}[/tex]

... but now that I think about it...

[tex] \int e^{1-x} = \frac {1}{1-x} e^{1-x+1} = \frac {1}{1-x} e^{x}[/tex]

right?
 
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  • #5
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EDIT: and for b), check your derivative of ln(sqrt(x)). sqrt(x) can be written as x^(1/2), so when you take the derivative you'll get (1/2)x^(3/2). Coupled with the square root of x on the bottom it will cancel nicely.
Wouldn't the derivative of x^(1/2) = (1/2)x^(-1/2)?
 
  • #6
dx
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Is [tex] \int e^{1-x} = e^{1-x}[/tex] ?

Edit: I'm only assuming this because

[tex] \int e^{x} = e^{x}[/tex]

... but now that I think about it...

[tex] \int e^{1-x} = \frac {1}{1-x} e^{1-x+1} = \frac {1}{1-x} e^{x}[/tex]

right?
Nope. ∫ e1-x dx = -e1-x. Use the substitution t = 1-x to see why.
 
  • #7
dx
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Wouldn't the derivative of x^(1/2) = (1/2)x^(-1/2)?
Yes, that's correct.
 
  • #8
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Nope. ∫ e1-x dx = -e1-x. Use the substitution t = 1-x to see why.
Aha! Great, now I got part a and edited up there. Thanks :)
 
  • #9
dx
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For part (b), use v = √x.
 
  • #10
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For part (b), use v = √x.
Could you walk me through this method, please?

I'm taught to look at the equation and assign u and dv. Then derive u to get du and integrate dv to get v.

The book says:

Step 1: Choose functions u and v so that f(x)dx = u dv. Try to pick u so that du is simpler than u and a dv that is easy to integrate.

Step 2: Organize the computation... (they put it in a neat box :tongue:)

Step 3: Complete the integration by finding ∫f(x) dx = ∫ u dv = uv - ∫ v du

So I'm not sure how I assign v = √x directly.

Here's the problem again:

[tex]
\int_1^4 \frac {ln \sqrt x} {\sqrt x} dx
[/tex]
 
  • #11
dx
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Hint: If v = √x, then

[tex] dv = \frac{dx}{2\sqrt{x}}[/tex]
 
  • #12
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Got part b! Dx, you rock!
 
  • #13
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OK so for part d I'm running into a seemingly basic problem. I'm not sure how to integrate √(3x2+2)

I'm just fried....
 
  • #14
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OK so for part d I'm running into a seemingly basic problem. I'm not sure how to integrate √(3x2+2)

I'm just fried....
You have a direct formula for that.
 
  • #15
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What is the formula? I'm trying to study for tomorrow morning's final, doing these practice questions and papers are everywhere.
 
  • #16
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http://planetmath.org/encyclopedia/IntegrationFormulas.html [Broken]
 
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  • #17
Cyosis
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Don't use a formula, splitting it up in x^3 and the root is a poor way to split things. Do it like this instead.

[tex]x^3 \sqrt{3x^2+2}=x^2*x\sqrt{3x^2+2}[/tex]. Now use partial integration and note that [itex]x\sqrt{3x^2+2}[/itex] is really easy to integrate by substituting u=3x^2+2, or even "guessing" the primitive.
 
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  • #18
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Oops.. My apologies... I just read the post no 13 and not the original post. So i just thought its [itex]sqrt{3x^2+2}[/itex] and didnt notice the x cube
 

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