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kneaiak

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Jan 11, 2013
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y=2x^(3/2), Integral limits are [0,5]

y'=3x^(1/2)

(y')^2=9x^(3/2)

int from 0 to 5 of sqr-root(1+9x^(3/2)) dx

I raised the sqr-root to the (1/2) power:

int from 0 to 5 of (1+9x^(3/2))^(1/2)

when I integrate I get ((1+9x^(3/2))^(3/2))/2 evaluated from 0 to 5

when I plug in my limits I get (28^(3/2))/2 - 1/2

Did I mess up on the integration?
 
Good! You should be aware that your integration is wrong as well. You can't just apply the power rule to anything you wish. The integral you attempted to do is highly nontrivial
 
Fortunately, after you have corrected the square of y' to \(\displaystyle (3x^{1/2})^2= 9x\), your integral becomes
\(\displaystyle \int_0^5\sqrt{1+ 9x}dx\)
and the substitution u=1+9x works nicely.
 
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