Integration prob.: int [6 to 1] [x / sqrt x + 3] dx

[mooshupork34]

here's a problem that i'm having trouble with so if anyone could help, i'd appreciate.

the definite integral of 6 to 1 for the function x divided by the square root of x+3 represents an area and a net change.

i have to find out the following
a) the area of what?
b) the net change of what function?

 

[galactus]

Is this what you have:

\(\displaystyle \L\\\int_{1}^{6}\frac{x}{\sqrt{x+3}}dx\)

If so, let \(\displaystyle \L\\u=\sqrt{x+3}, \;\ du=\frac{1}{2\sqrt{x+3}}dx, \;\ 2du=\frac{1}{\sqrt{x+3}}dx\)

This is the area you're finding.

 

[mooshupork34]

Thanks! But how do you use the

2du = (1)/(square root of x + 3) to solve the problem?

 

[mooshupork34]

So here's what I got so far....and I don't know if this is correct.

I substituted sqrt[x+3] for you. Then, I got that du = 2 * sqrt[x+3] dx.

I realized that du is just 2 * u.

So then this left me with the integral of:

x/u multiplied by 2u

That's where I got stuck.

 

[galactus]

You got off on the wrong footing.

\(\displaystyle \L\\\int_{1}^{6}\frac{x}{\sqrt{x+3}}dx\)

We are going to use \(\displaystyle \L\\u=\sqrt{x+3}\) ...[1]

Differentiating, we get \(\displaystyle \L\\u=\frac{1}{2\sqrt{x+3}}dx\)..OK so far?.

Therefore, \(\displaystyle \L\\2du=\frac{1}{\sqrt{x+3}}dx\)...See?.


Solve [1] for x: \(\displaystyle \L\\x=u^{2}-3\)

As you can see(hopefully), you now have your substitutions:

Now, because of the this, we must change the limits of integration.

We have that x=1 and x=6:

\(\displaystyle \L\\u=\sqrt{1+3}=2\)
\(\displaystyle \L\\u=\sqrt{6+3}=3\)

Now, we have:

\(\displaystyle \L\\2\int_{2}^{3}(u^{2}-3)du\)