can someone help me solve

integral from 0 to 4 of

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

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- May 6th 2010, 03:59 AMcummings15definite 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 AMpiglet
- May 6th 2010, 04:06 AMskeeter
- May 6th 2010, 04:10 AMFailure
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 AMcummings15
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 AMFailure
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?

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

or you can do a back-substitution before you apply the old limits:

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