Thread: Integration with substitution I think, but how?

1. Integration with substitution I think, but how?

Hi guys, just found this site today and was hoping somebody here would have the time to help me out on what's probably a relatively simple problem...

I'm asked to evaluate the integral x/sqrt(1+4x) from 0 to 3. This seems innocent enough, and seeing as we've worked with substitutions recently I suspect I'm intended to head down that path.

However.. I can't get anything out of substituting u=1+4x; du ends up just being a constant (4) which isn't helping me any. I wonder if there's some trig substitution I'm just not seeing here.

Also, where can I get a guide on using the math tags?

thanks!

2. Originally Posted by storpotaten
Hi guys, just found this site today and was hoping somebody here would have the time to help me out on what's probably a relatively simple problem...

I'm asked to evaluate the integral x/sqrt(1+4x) from 0 to 3. This seems innocent enough, and seeing as we've worked with substitutions recently I suspect I'm intended to head down that path.

However.. I can't get anything out of substituting u=1+4x; du ends up just being a constant (4) which isn't helping me any. I wonder if there's some trig substitution I'm just not seeing here.

Also, where can I get a guide on using the math tags?

thanks!

$\int_{0}^{3}\frac{x}{\sqrt{1+4x}}dx$

$u=1+4x \iff \frac{u-1}{4}=x \to du=4dx$

using this we get

$\int_{0}^{3}\frac{x}{\sqrt{1+4x}}dx = \int_{1}^{13}\frac{\frac{u-1}{4}}{\sqrt{u}}\frac{du}{4}=\frac{1}{16}\int_{1}^ {13}u^{1/2}-u^{-1/2}du$

3. Math tags

You can hoover you cursor over anyone's code to see it.

I use this wiki site

http://en.wikipedia.org/wiki/Help Formula

4. Hello, storpotaten!

Welcome aboard!

$\int^3_0\frac{x}{\sqrt{1+4x}}\,dx$
With the square root of a linear expression, we can let $u$ equal the entire radical . . .

Let $u \,=\,\sqrt{1+4x}\quad\Rightarrow\quad x\:=\:\frac{1}{4}(u^2-1) \quad\Rightarrow\quad dx \:=\:\frac{1}{2}u\,du$

Substitute: . $\int\frac{\frac{1}{4}(u^2-1)}{u}\left(\frac{1}{2}u\,du\right) \;=\;\frac{1}{8}\int(u^2-1)\,du$

. . . . . $= \;\frac{1}{8}\left(\frac{u^3}{3} - u\right) + C \;=\;\frac{1}{24}u(u^2-3) + C$

Back-substitute: . $\frac{1}{24}\sqrt{1+4x}\left(\left[\sqrt{1+4x}\right]^2-3\right) + C \;=\;\frac{1}{24}\sqrt{1+4x}\,(1 + 4x - 3) +C$

. . . . $= \;\frac{1}{24}\sqrt{1+4x}\,(4x-2) + C \;=\;\frac{1}{12}\sqrt{1+4x}\,(2x-1) + C$

Now evaluate the definite integral . . .

5. You guys are the best -- thanks!