# definite integral problem

• May 6th 2010, 03:59 AM
cummings15
definite integral problem
can someone help me solve

integral from 0 to 4 of

(x)/ (square root of x^2+9) dx
• May 6th 2010, 04:06 AM
piglet
Quote:

Originally Posted by cummings15
can someone help me solve

integral from 0 to 4 of

(x)/ (square root of x^2+9) dx

Have you made any attempt at it yourself? This is a pretty straight-forward problem and im sure you've covered problems similar to this in school or uni

If ur really stuck, Hint: substitute u = x^2+9 and hence get a substitution for dx
• May 6th 2010, 04:06 AM
skeeter
Quote:

Originally Posted by cummings15
can someone help me solve

integral from 0 to 4 of

(x)/ (square root of x^2+9) dx

use the method of substitution ...

$\displaystyle u = x^2+9$
• May 6th 2010, 04:10 AM
Failure
Quote:

Originally Posted by cummings15
can someone help me solve

integral from 0 to 4 of

(x)/ (square root of x^2+9) dx

Just note that $\displaystyle x$ is (almost) the derivatived of $\displaystyle x^2+9$. Therefore you'd better substitute $\displaystyle z := x^2+9$ right away, which gives you that $\displaystyle dz = 2xdx$.

Plugging this into your integral allows you to convert it to the form $\displaystyle \tfrac{1}{2}\int \frac{1}{\sqrt{z}}\, dz$.
• May 6th 2010, 04:20 AM
cummings15
so far i got

u=x^2+9
du=2xdx
(1/2)du=xdx

so i pluged it in and got

(1/2) integral of (du/u)

what do i do with the 0 to 4 part?
• May 6th 2010, 04:40 AM
Failure
Quote:

Originally Posted by cummings15
so far i got

u=x^2+9
du=2xdx
(1/2)du=xdx

so i pluged it in and got

(1/2) integral of (du/u)

No, that's not quite correct, you should have $\displaystyle \tfrac{1}{2}\int\frac{1}{\sqrt{u}}\, du$

Quote:

what do i do with the 0 to 4 part?
Finding the anti-derivative is the main problem, you have two choices as regards the limits of the integral: you can transform the limits along with the integrand:

$\displaystyle \int_0^4\frac{x}{\sqrt{x^2+9}}\, dx=\ldots =\tfrac{1}{2}\int_{u(0)}^{u(9)}\frac{du}{\sqrt{u}} \, du = \Big[\sqrt{u}\Big]_{u=9}^{25} = \sqrt{25}-\sqrt{9}=2$

or you can do a back-substitution before you apply the old limits:
$\displaystyle \int_0^4\frac{x}{\sqrt{x^2+9}}\, dx=\ldots =\tfrac{1}{2}\int_{u(0)}^{u(4)}\frac{du}{\sqrt{u}} \, du = \Big[\sqrt{u}\Big]_{u=u(0)}^{u(4)} = \Big[\sqrt{x^2+9}\Big]_{x=0}^4=\ldots = 2$