Definite Integral Confusion

  • Thread starter verd
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  • #1
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Hi,

So this might be overwhelmingly stupid... But the fundamental theorem of calculus states:
[tex]\int_{a}^{b}f(x)dx=F(b)-F(a)[/tex]
Where F is any antiderivative of f.

So I have this very simple integral that I'm trying to solve...:

[tex]2\pi\int_{0}^{2}x^3\sqrt{1+9x^4}dx\rightarrow \ u=1+9x^4[/tex]
[tex]=\frac{\pi}{18}*\frac{2}{3}u^{3/2}\mid_{1}^{145}=\frac{\pi}{27}(1+9x^4)^{3/2}\mid_{1}^{145}[/tex]

For the answer, I'm given the following:
[tex]\frac{\pi}{27}(145\sqrt{145}-1)[/tex]

However, when I use the fundamental theorem of calculus, I get something nasty:
[tex]\frac{\pi}{27}[(1+9(145)^4)^{3/2}-(1+9(1)^4)^{3/2}][/tex]
[tex]\mbox{calculator gives}=\frac{2}{27}[(1989227813\sqrt{1989227813}-5\sqrt{5})*\pi*\sqrt{2}]}[/tex]

I don't know why I keep getting the wrong thing. I'm obviously making some sort of stupid error. Any suggestions?
 
Last edited:

Answers and Replies

  • #2
0rthodontist
Science Advisor
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Plug the two expressions (yours and the calculator's) into your calculator to evaluate them as numbers.
 
  • #3
146
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The long one is 2.91x10^13
The shorter one with the sqrt(145) comes to 203.044...
 
  • #4
dav2008
Gold Member
609
1
Why are you evaluating from 1 to 145? Those are the limits for u, not x.

Either keep it in terms of u and evaluate from 1 to 145 or change it back to x (extra work) and evaluate from 0 to 2:

[tex]=\frac{\pi}{27}u^{3/2}\mid_{1}^{145}=\frac{\pi}{27}(1+9x^4)^{3/2}\mid_{0}^{2}[/tex]

In that line it's pointless to change back to x since it's a definite integral. Just evaluate u from 1 to 145 and you're done.
 
Last edited:
  • #5
146
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Ah.

There's the confusion. ...For some reason I guess I thought I had to replace the u by x as with indefinite integrals.

Thanks much for helping me realize my mistake.


Thanks,
Heather
 
  • #6
dav2008
Gold Member
609
1
There's nothing stopping you from rewriting it in terms of x, but like I said you have to be careful to write the limits in terms of x. Plus, it's just easier to keep things in u.
 
  • #7
146
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Here's another problem where I'm screwing up the substitution... I've got the limits straight, (thanks to you), but I don't know why I'm getting this wrong.

[tex]2\pi\int_{4}^{9}\sqrt{x+\frac{1}{4}dx}\rightarrow \ u=x+\frac{1}{4} \\ = 2\pi\int_{17/4}^{37/4}\sqrt{u}du \\ =2\pi*\frac{2}{3}u^{3/2}\mid_{17/4}^{37/4}[/tex]

Somewhere along the line, my integral is supposed to come out to this:
[tex]2\pi*\frac{1}{6}(4x+1)^{3/2}[/tex]

...Incorrect substitution...??
u=x+1/4
du=dx

Either way, my answer is off by 1/2 when I get down to it with what I've done.

My answer:
[tex]\frac{\pi}{3}(37\sqrt{37}-17\sqrt{17})[/tex]
The book's answer:
[tex]\frac{\pi}{6}(37\sqrt{37}-17\sqrt{17})[/tex]
 
  • #8
146
0
Nevermind. I figured this out. Quick multiplication happened as a result of me not moving the denominators out of the expression with the limits plugged in.

With the correction, I get the correct pi/6 multiplier.

I would however like to know how the book and how maple is getting the answer with the 1/6 multiplier... Can someone explain that?
...It's different from what I got.
 
  • #9
t!m
147
6
Is
[tex]2\pi*\frac{1}{6}(4x+1)^{3/2}[/tex]
what you got, or what your book/maple got?
 
  • #10
146
0
what the book/maple got for the integral w/out limits
 

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